The Problem Is the Problem: Towards Scalable Mathematical Discovery

# The Problem Is the Problem: Towards Scalable Mathematical Discovery
Source: [https://arxiv.org/html/2608.16977](https://arxiv.org/html/2608.16977)
Shengtong Zhang11footnotemark:1Affiliation:Anysphere Co\.\{zeyuzhen, avigad, ptetali, swelleck\}@andrew\.cmu\.edu,shengtong@anysphere\.co[https://github\.com/zeyu\-zheng/FAR](https://github.com/zeyu-zheng/FAR)Jeremy AvigadAffiliation:Carnegie Mellon UniversityPrasad TetaliAffiliation:Carnegie Mellon UniversitySean WelleckAffiliation:Carnegie Mellon University

###### Abstract

AI systems are increasingly capable of contributing to mathematical research\. In research practice, frontier\-model reasoning is a limited resource, and expert mathematical review is even more sharply constrained\. Allocating these scarce resources well is therefore central to making AI\-assisted mathematical discovery efficient\. In most current AI\-for\-math workflows, human effort is concentrated at the beginning and end, in selecting suitable research problems and later reviewing the resulting artifacts\. These two stages are becoming bottlenecks for research\-level mathematics\. We address them by proposing a new human\-AI discovery paradigm\. The human input is no longer a single problem selected in advance, but a research direction in which the experts have interest and expertise\. The system then searches a broad literature corpus for candidate problems in that direction\. Inspired by search and recommender systems, we buildFind, Attempt, and Recommend \(FAR\), a literature\-to\-review cascade that automates the search for suitable problems and focuses human attention on artifacts that have passed several stages of filtering\. In a combinatorics pilot, the pipeline starts from 5,245 combinatorics papers, recovers 6,453 candidate conjectures or open problems, and filters them to 4,717 apparently well\-posed and still\-open conjectures\. Subsequent reasoning and automated triage stages surface 598 potential resolutions111Available at[https://probxiv\.com](https://probxiv.com/)\.and select 77 items for author\-team review\. Among them, we identify many interesting discoveries, including results on conjectures and questions of Davies–Jenssen–Perkins–Roberts, Erdős–Straus, Ikenmeyer–Pak–Panova, and Lund–Saraf–Wolf\. These results demonstrate the effectiveness of this new mode of human\-AI collaboration for mathematical discovery\.

## 1Introduction

AI systems are increasingly capable of contributing to mathematical research, with recent progress on mathematical reasoning benchmarks\([Hendrycks et al\. 2021](https://arxiv.org/html/2608.16977#bib.bib33);[Zheng et al\. 2021](https://arxiv.org/html/2608.16977#bib.bib74);[Guo et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib30);[Shao et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib60)\), formal theorem proving\([Trinh et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib66);[Chervonyi et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib16);[Hubert et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib34);[Ren et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib55);[Xin et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib70);[Chen et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib13);[Seed 2026](https://arxiv.org/html/2608.16977#bib.bib58)\), and selected research\-level problems\([OpenAI 2026](https://arxiv.org/html/2608.16977#bib.bib51);[Alon et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib2);[Tsoukalas et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib68);[Team 2026](https://arxiv.org/html/2608.16977#bib.bib65)\)\. Most of these systems operate with a problem\-level interface, in which a researcher supplies a theorem, conjecture, or formal goal, the system attempts it, and the output is checked\. This interface is useful, but it leaves out an essential part of research: deciding which problems are worth attempting in the first place\.

We study the decision of which problems to attempt in terms of effort allocation for AI\-assisted mathematical discovery\. Frontier\-model reasoning and expert mathematical review are scarce, and their value depends on which conjectures receive them\. Rather than concentrating reasoning and review effort on a few problems chosen in advance, we propose a new workflow in which experts specify a research direction, and the AI system automatically finds problems to focus effort on \(Figure[1](https://arxiv.org/html/2608.16977#S1.F1)\)\. We buildFind, Attempt, and Recommend \(FAR\), a literature\-to\-review cascade\. Given a broad mathematical topic,FARfinds relevant open conjectures from a large literature corpus, attempts to prove or disprove them, and recommends promising conjecture\-resolution pairs for expert review\.

![Refer to caption](https://arxiv.org/html/2608.16977v1/paradigm.png)Figure 1:From choosing a problem to choosing a direction\.The upper part shows the problem\-level interface; the lower part shows our approach\. Figure[3](https://arxiv.org/html/2608.16977#S3.F3)gives the details ofFAR\.We instantiate the workflow in combinatorics\. Starting from 51,110 mathematics papers, the pipeline identifies 5,245 combinatorics papers, extracts 6,453 candidate conjectures or open problems from 2,742 papers, and filters these to 4,717 apparently well\-posed and still\-open conjectures\. A first broad attempt run covers all 4,717 of them\. Automated triage surfaces 598 potential resolutions, and a final selection step chooses 77 items for internal author\-team review\. We manually checked 15 of the resulting artifacts, chosen by our own interest\. They include counterexamples to conjectures of Davies–Jenssen–Perkins–Roberts and of Lund–Saraf–Wolf, a proof of Ikenmeyer–Pak–Panova’s conjecture on symmetric\-group characters, and an answer to a question of Erdős–Straus on divisibility among binomial coefficients\.

We study various strategies for allocating a budget of solving attempts in our pipeline\. We frame allocation as a constrained optimization problem, and derive strategies for maximizing notions of quality and importance\. We show that these strategies can lead to more successful artifacts than with a uniform baseline\. Furthermore, the optimal strategy depends on the objective: for instance, the strategy differs if we wish to maximize the number of successful artifacts versus maximizing the importance across all artifacts\.

In summary, our contributions are as follows:

- •We formulate scalable AI\-assisted mathematical discovery as an effort\-allocation problem over a pool of interesting mathematical problems, leading to a human–AI collaboration paradigm for using frontier\-model reasoning and expert mathematical review efficiently\.
- •Inspired by search and recommender systems, we introduceFind, Attempt, and Recommend \(FAR\), which builds this problem pool from the mathematical literature and turns model attempts into reviewable artifacts\.
- •We instantiate the workflow in a combinatorics pilot and analyze the resulting literature\-to\-review funnel, including strategies for allocating model attempts\.
- •We obtain author\-reviewed solutions to problems from the combinatorics literature, spanning proofs, counterexamples, and answers to open\-ended questions\. The write\-ups are collected in Appendix[C](https://arxiv.org/html/2608.16977#A3)\.

## 2Motivation and Related Work

### 2\.1AI for Mathematical Research

Recent AI systems have made rapid progress on the part of mathematical research that begins after a problem or objective has been specified\. For example, FunSearch and AlphaEvolve search for new mathematical constructions, programs, and algorithms\([Romera\-Paredes et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib57);[Novikov et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib48)\)\. AlphaProof Nexus studies formal proof search on open research\-level problems\([Tsoukalas et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib68)\)\. Aletheia, Rethlas, QED, and other recent pipelines explore autonomous or semi\-autonomous workflows for attempting open mathematical problems\([Feng et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib27);[Ju et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib37);[An et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib6);[Ju et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib37);[Peng et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib52)\)\. Frontier models have also contributed to individually selected problems of long\-standing research interest, including OpenAI’s disproof of the Unit Distance Problem and the Fable\-assisted counterexample to the Jacobian conjecture\([OpenAI 2026](https://arxiv.org/html/2608.16977#bib.bib51);[Alon et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib2);[Alpöge 2026](https://arxiv.org/html/2608.16977#bib.bib4);[Bukh et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib12)\)\. Related work on an AI co\-mathematician\([Zheng et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib73)\)develops a collaborative framework in which AI agents pursue parallel workstreams while mathematicians steer the research process\. Taken together, these works point toward what Tao describes as a transition from proof scarcity to proof abundance\([Tao 2026](https://arxiv.org/html/2608.16977#bib.bib64)\)\.

Mathematical research, however, rarely starts from an isolated problem statement\. Before trying to solve a problem, researchers often need to study a direction of interest, find problems that matter within it, and decide whether those problems are worth sustained thought\. These steps are ordinary parts of mathematical work, but they mostly sit outside the scope of today’s AI\-for\-math systems\. In current uses of AI for mathematics, the human work before and after model reasoning—problem selection and expert review—is becoming the narrow part of the pipeline\.

Our approach moves the starting point of AI assistance to this earlier stage of mathematical research\. Rather than selecting a problem for the system, mathematicians specify a research direction\. The system searches an available literature corpus, recovers and attempts candidate problems in that direction at scale, and returns a small set of artifacts for mathematical review\. This shift is analogous to the move from task\-driven agents, which act on a specified task, toward more proactive systems that help surface what tasks are worth pursuing\([Zhou et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib75);[Lu et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib43);[Lab 2026](https://arxiv.org/html/2608.16977#bib.bib39)\)\. The human\-specified direction bounds the system’s initiative and aligns the search with the mathematicians’ interests and domain expertise\. Mathematicians remain responsible for validating and reporting any resulting mathematical claims\([Alper et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib3);[Shan et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib59)\)\.

### 2\.2Constructing an Attemptable Pool

A research direction does not by itself provide the system with a set of attemptable problems\. Mathematical benchmarks such as PutnamBench, FrontierMath, and FirstProof provide clean, self\-contained problem pools\([Tsoukalas et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib67);[Glazer et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib29);[Abouzaid et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib1)\)\. In software engineering, another major domain for LLM agents\([Yang et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib71)\), GitHub provides a centralized and structured collection of repositories, issues, documentation, and executable software from which benchmarks such as SWE\-bench and ProgramBench can be constructed\([Jimenez et al\. 2024](https://arxiv.org/html/2608.16977#bib.bib36);[Yang et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib72)\)\.

Mathematics does have valuable collections, including the Open Problem Garden, AIM Problem Lists, and Formal Conjectures\([Open Problem Garden 2026](https://arxiv.org/html/2608.16977#bib.bib50);[American Institute of Mathematics 2026](https://arxiv.org/html/2608.16977#bib.bib5);[Firsching et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib28)\)\. Their coverage is selective, however, and they are not the primary infrastructure in which mathematical problems and their surrounding research context are recorded\. Much of this information remains dispersed across the literature\. A conjecture may appear as a numbered statement, a question, a remark, an unresolved case, or a sentence embedded in local notation, and its status may change after publication\. Constructing an attemptable pool of problems therefore requires recovering candidate statements from their source context and checking whether they remain well posed and unresolved\.

We draw on search and recommender systems to organize this process\. Building a pool of attemptable problems is akin to candidate retrieval, i\.e\., recovering statements from a large corpus using imperfect signals and filtering them for provenance, well\-posedness, and current status\([Belkin & Croft 1992](https://arxiv.org/html/2608.16977#bib.bib8);[Liu 2009](https://arxiv.org/html/2608.16977#bib.bib41);[Liu et al\. 2022](https://arxiv.org/html/2608.16977#bib.bib42)\)\. We also draw on the idea of recommendation cascades, in which progressively more selective stages reduce a large collection of candidates to a small set of artifacts for expert attention\([Ricci et al\. 2010](https://arxiv.org/html/2608.16977#bib.bib56);[Covington et al\. 2016](https://arxiv.org/html/2608.16977#bib.bib19);[Wang et al\. 2011](https://arxiv.org/html/2608.16977#bib.bib69);[Chen et al\. 2017](https://arxiv.org/html/2608.16977#bib.bib14);[Zhu et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib76)\)\. Our techniques also relate to literature\-based discovery, which searches published knowledge for research opportunities not visible from a single paper\([Swanson 1986b](https://arxiv.org/html/2608.16977#bib.bib62);[Swanson 1986a](https://arxiv.org/html/2608.16977#bib.bib61)\)\.

Our objective differs from automated conjecturing, a complementary line of work that creates new mathematical conjectures rather than surfacing existing ones\. This line includes automated theory formation, the Ramanujan Machine, and the data\-driven TxGraffiti system\([Colton 2012](https://arxiv.org/html/2608.16977#bib.bib17);[Raayoni et al\. 2021](https://arxiv.org/html/2608.16977#bib.bib54);[Davila 2026](https://arxiv.org/html/2608.16977#bib.bib22)\)\. Recent LLM work has used generated conjectures to expand formal training data and couple conjecturing with proving, as in LeanConjecturer and STP, while Moonshine makes conjecture generation the organizing objective of an autonomous mathematical research agent\([Onda et al\. 2025](https://arxiv.org/html/2608.16977#bib.bib49);[Dong & Ma 2025](https://arxiv.org/html/2608.16977#bib.bib23);[Chen & Jiang 2026](https://arxiv.org/html/2608.16977#bib.bib15)\)\. We instead recover unresolved statements that authors have already placed in the literature\. Each candidate retains its source paper, statement text, local context, and status evidence, so that later attempts and reviews can be checked against what the source actually claimed\. After status checking, the result is an attemptable pool𝒫\\mathcal\{P\}of source\-grounded conjectures that appear well posed and still open\.

### 2\.3Effort Allocation under Uncertainty

Let𝒰\\mathcal\{U\}be the universe of mathematical questions that can be expressed in natural language\. The human practice of mathematical research can be viewed as a large effort\-allocation process over𝒰\\mathcal\{U\}\. Mathematicians search this space for questions worth exploring, and then try to answer them or make progress on them\. As AI agents become primary sources of mathematical attempts, allocating agent compute raises a similar problem\. At the same time, the Leiden Declaration emphasizes that mathematicians remain responsible for validating and reporting mathematical claims\([Alper et al\. 2026](https://arxiv.org/html/2608.16977#bib.bib3)\), so the allocation of expert review effort also matters\. The attemptable pool𝒫\\mathcal\{P\}described in Section[2\.2](https://arxiv.org/html/2608.16977#S2.SS2)forms a small part of𝒰\\mathcal\{U\}, but it is a natural starting point because its conjectures and open problems have already been selected, stated, and discussed by mathematicians\.

Even for human mathematicians, deciding where to spend effort is difficult\. Prior impressions of difficulty often differ from difficulty in hindsight\. Some simply stated problems resist solution for a long time, while some long\-standing problems eventually admit unexpectedly simple arguments\. In an AI\-assisted setting there is an additional source of uncertainty: what is difficult for human mathematicians need not be difficult in the same way for the current model\. In turn, an important question is how to use limited model attempts to discover which conjectures from the attemptable pool𝒫\\mathcal\{P\}are likely to produce artifacts that are worth expert review\.

Figure 2:Schematic view of the current reachable region\.The figure is illustrative, and both axes should be read qualitatively\. A single pass over the pool probes which conjectures can produce artifacts worth review under the current model\. As model capability improves, more conjectures may become reachable\.The question of allocating limited attempts to a set of problems invites a simple bandit interpretation\([Bubeck & Cesa\-Bianchi 2012](https://arxiv.org/html/2608.16977#bib.bib11);[Lattimore & Szepesvári 2019](https://arxiv.org/html/2608.16977#bib.bib40)\)\. From this perspective, a conjecturec∈𝒫c\\in\\mathcal\{P\}is an arm, and spending reasoning and review effort on it is a pull\. A pull yields an assessed outcomeY⁡\(c\)Y\(c\), which may be no reliable result, a known resolution, or a candidate proof or counterexample\. In our experiments we pull each arm only once, as in the initialization stage of the UCB algorithm\([Auer et al\. 2002](https://arxiv.org/html/2608.16977#bib.bib7)\), and then use subsequent steps to concentrate human review on a much smaller set of outputs\. Further investigating bandit algorithms for our setting is left for future work\. Our effort allocation problem differs from the systems discussed in Section[2\.1](https://arxiv.org/html/2608.16977#S2.SS1), which amount to concentrating reasoning effort on one or a few problems that are selected in advance\. Finally, we note that effort allocation is dynamic: as models improve, conjectures that previously produced no useful progress may enter the current system’s reachable region, thereby demanding different allocations of effort than the older models\. Figure[2](https://arxiv.org/html/2608.16977#S2.F2)illustrates this reachable region\.

## 3Method

Our methodFind, Attempt, and Recommend \(FAR\)instantiates the allocation view above as a literature\-to\-review pipeline\. Analogous to search and recommender systems\([Wang et al\. 2011](https://arxiv.org/html/2608.16977#bib.bib69);[Chen et al\. 2017](https://arxiv.org/html/2608.16977#bib.bib14);[Liu et al\. 2022](https://arxiv.org/html/2608.16977#bib.bib42), e\.g\.,\),FARprogressively narrows a large collection through increasingly costly stages, with later stages receiving more compute per item\. As illustrated in Figure[3](https://arxiv.org/html/2608.16977#S3.F3), given a literature corpus and a research direction,FAR*finds*relevant open problems and filters them into an attemptable pool𝒫\\mathcal\{P\}\(the Label, Extract, and Check stages\),*attempts*candidate resolutions \(Solve\), and*recommends*selected conjecture\-resolution pairs for expert review \(Judge\)\.

Figure 3:From papers to recommendations for expert review\.The upper row shows a search or recommender pipeline that recalls and filters candidates from a large corpus\. Its numbers indicate typical orders of magnitude\. The lower row shows the analogousFARpipeline\. Numbers in the lower row are counts from our pilot run detailed in Section[4](https://arxiv.org/html/2608.16977#S4)\.### 3\.1Finding Relevant Open Problems

The finding stage takes a literature corpus and a research direction as input and returns an attemptable pool𝒫\\mathcal\{P\}of relevant open problems\. It identifies papers in the research direction, extracts unresolved statements from them, and finally checks whether those statements are valid and still open\. The corpus may be arXiv, a topic\-specific collection, or another large\-scale source of mathematical literature\. The prompts for the three stages are given in Appendix[A](https://arxiv.org/html/2608.16977#A1)\.

##### Finding relevant papers via labeling\.

Before the pipeline runs, a mathematician fixes a research direction at whatever granularity they need\. An agent then reads each paper, labels whether it lies in that direction, and keeps the papers that do\. Labeling plays the part of recall in a retrieval cascade\. It sees the entire corpus, so it runs the cheapest model in the pipeline\.

##### Extracting and recovering conjectures\.

An agent extracts unresolved statements from the papers that survive labeling\. A paper may yield several such statements or none\. These may be labeled conjectures, questions, or open problems, or they may appear in prose\. The extractor excludes future work that poses no specific mathematical question, and statements resolved within the same paper\. Beyond those exclusions it is permissive, since a statement it passes over cannot be recovered later while a spurious one is dropped by the next stage\. Each extracted statement becomes a candidate, recorded with its source paper\. Table[1](https://arxiv.org/html/2608.16977#S3.T1)gives a representative source\-to\-candidate example\.

##### Checking validity and status\.

An agent searches for later work on each candidate and records the supporting sources as status evidence\. It labels the candidate*open*when its source paper states a concrete unresolved problem and no credible resolution is found,*solved*when credible evidence resolves it, and*invalid*when the extracted text does not state a concrete open problem\. Open candidates form𝒫\\mathcal\{P\}, meaning that the pool contains conjectures that appear relevant, well posed, and still open under the available evidence\. Like the eligibility filters that the drop items that a recommender can no longer serve, this stage removes the candidates that turn out to be solved or invalid\.

Table 1:A recovered candidate, from source text to the pool\.

### 3\.2Attempting for Candidate Resolutions

Every conjecture in the pool𝒫\\mathcal\{P\}receives one attempt, which is the unit of effort that this stage allocates\. An attempt may be a single agent run, a longer harness, a multi\-agent workflow, or repeated sampling under a selection rule\. Performing one attempt per conjecture simply allocates the attempt budget uniformly across problems in the pool\. As discussed in Section[2\.3](https://arxiv.org/html/2608.16977#S2.SS3), this can be seen as the initialization step of a bandit algorithm, and we study other strategies in Section[4\.3](https://arxiv.org/html/2608.16977#S4.SS3)\.

To perform an attempt, an agent is given each conjecture as extracted, together with its source paper, so that its notation is read against the text that introduced it\. Although the checking stage already searched for whether the problem is open, the checking stage ran a weaker model\. Determining that a conjecture is equivalent to something that is already settled can take non\-trivial reasoning, and hence may benefit from the stronger model used in the attempting stage\. The agent therefore searches before it attempts, and labels the outcomeKNOWNwhen a credible source already resolves the conjecture,NEWwhen it produces a complete proof or counterexample of its own,FIXwhen the statement as written is defective and the minimal repair it proposes cannot be settled, andNONEwhen it reaches none of these\. The prompt is given in Appendix[A](https://arxiv.org/html/2608.16977#A1)\. OnlyNEWoutcomes go on to the next stage\.

Thefindstages narrowed down the pool of problems, meaning that theattemptstage can use more resources per problem\. For the attempt stage we run the most capable model in the pipeline, and it returns a set𝒴\\mathcal\{Y\}pairing every conjecture in𝒫\\mathcal\{P\}with an outcome\.

### 3\.3Recommendation for Expert Review

Expert review is the scarcest resource in the pipeline, and only a few of the conjecture\-resolution pairs in𝒴\\mathcal\{Y\}can receive it\. Therecommendstage decides which pairs receive expert review\. It judges whether the result is correct and whether it is significant enough to publish, similar to what a referee determines in human peer review\. The judging and grading in this stage are akin to the final filters in a search or recommender cascade\. Here, a mathematician is the user that the results are served to\.

##### Judging\.

One or more agents check eachNEWoutcome in𝒴\\mathcal\{Y\}for correctness, asking whether it addresses the statement that it targets, whether the argument or construction is complete, and whether every step holds\. Each agent marks the outcomePASSorFAIL, and an outcome passes only if every agent passes it\.

##### Recommending for review\.

A second agent takes the outcomes that passed and sorts each one as already known, as new but too minor to stand alone, or as substantial enough to publish on its own\. Deciding the first requires a fresh literature search, since an existing resolution may have escaped both earlier stages\. The last group forms the set𝒜\\mathcal\{A\}of artifacts\. The judging and grading prompts are given in Appendix[A](https://arxiv.org/html/2608.16977#A1)\.

##### Expert review\.

By this point𝒜\\mathcal\{A\}is small, and everything in it lies in the direction that the mathematician set at the start\. The mathematician\(s\) read the artifacts that interest them, check that each is correct and not already known, and write up those that hold\.

## 4A Pilot Run in Combinatorics

We instantiateFind, Attempt, and Recommendin combinatorics, a domain whose results the authors have the expertise to verify\. The remainder of this section reports the setup of the run, the outcomes it produced, and an analysis of those outcomes\.

### 4\.1Setup

For this pilot we assemble a corpus of 51,110 mathematics papers fromOpenAlexmetadata and source links\([Priem et al\. 2022](https://arxiv.org/html/2608.16977#bib.bib53)\)\. The pipeline is not tied to that source, and arXiv or another large\-scale collection of mathematical literature would serve as well\. Labeling keeps 5,245 papers\. Extraction recovers 6,453 candidates from 2,742 of them, and checking leaves 4,717 conjectures drawn from 2,206 papers\. These form the attemptable pool𝒫\\mathcal\{P\}\.

The stages run different models\. Labeling usesgpt\-oss\-120b, extractiongemini\-3\.5\-flash, and checkinggemini\-3\.1\-prowith web search\. Attempting, judging, and grading all usegpt\-5\.5atxhighreasoning effort, and each conjecture in𝒫\\mathcal\{P\}received one attempt, instantiated here as a single run of theopencodeagent in a working directory holding the paper and the statement\. Each claimed resolution was put to three independent judges\. This is the ordering Section[3](https://arxiv.org/html/2608.16977#S3)describes, with progressively more capable models as the set narrows and the task becomes harder\.

### 4\.2Outcomes

Every conjecture in𝒫\\mathcal\{P\}receives one attempt and one of four outcomes\. OnlyNEWoutcomes are judged, and only those that pass are graded\. The tree below gives the count at each step\.

4,717conjectures attempted=𝒫=\\mathcal\{P\}2,905NONEno result443KNOWNalready resolved in the literature319FIXstatement defective as written1,050NEWclaimed resolution452FAILfails judging598PASSpasses judging75already known446too minor to stand alone77publishable=𝒥=\\mathcal\{J\}=𝒜=\\mathcal\{A\}

The run produced claimed resolutions for 1,050 of the 4,717 conjectures\. Judging accepted 598 of them, which we write𝒥\\mathcal\{J\}, and grading left 77 of those to form𝒜\\mathcal\{A\}\.

The authors reviewed 15 of the artifacts in𝒜\\mathcal\{A\}that they found particularly interesting, and every one of them is mathematically correct\. One, an asymptotic bound on a divisor\-difference problem of Erdős recorded in Guy’s miscellany, had been settled a few months before our run by a route that none of the searches in the cascade turned up\. Section[5](https://arxiv.org/html/2608.16977#S5)discusses several of these results, and Appendix[C](https://arxiv.org/html/2608.16977#A3)collects the write\-ups\.

### 4\.3Analysis: Allocating Effort for Discovery

Beyond the mathematics that the run produced, the outcomes of the run let us study alternative effort allocation strategies to the uniform strategy that we used in the pilot\.

To do so, we use the outcomes of each stage \(namely, the conjectures𝒥\\mathcal\{J\}that ended up passing judging and the final artifacts𝒜\\mathcal\{A\}\) along withdifficultyandimportancescores that were collected during the run\. Concretely, the agent’s prompt in the checking stage asked it to include two estimates:

- •a*difficulty*dd, anchored at00for a conjecture whose resolution would be an unpublishable exercise and at11for one publishable in a top journal;
- •an*importance*ii, anchored at00for a statement with no substantive mathematical content and at11for a Fields\-Medal\-level problem\.

In both cases the prompt asks for the scores to follow a roughly normal distribution centered on0\.50\.5\. The scores are fixed before any reasoning budget is spent, and the run then attempted every conjecture in𝒫\\mathcal\{P\}, so the two scores can be read against outcomes that they preceded\.

#### 4\.3\.1Validity of the Agent’s Difficulty and Importance Scores

We first study the validity of the difficulty and importance scores, before using them within allocation strategies\. For a group of conjectures𝒮⊆𝒫\\mathcal\{S\}\\subseteq\\mathcal\{P\}, write

- •δ⁡\(𝒮\)=\|𝒮∖𝒥\|/\|𝒮\|\\delta\(\\mathcal\{S\}\)=\|\\mathcal\{S\}\\setminus\\mathcal\{J\}\|\\,/\\,\|\\mathcal\{S\}\|, the fraction of𝒮\\mathcal\{S\}with no accepted resolution;
- •ι⁡\(𝒮\)=\|𝒜∩𝒮\|/\|𝒥∩𝒮\|\\iota\(\\mathcal\{S\}\)=\|\\mathcal\{A\}\\cap\\mathcal\{S\}\|\\,/\\,\|\\mathcal\{J\}\\cap\\mathcal\{S\}\|, the fraction of its accepted resolutions graded publishable\.

Theδ⁡\(S\)\\delta\(S\)metric is viewed as an empirical estimate of the difficulty, andι⁡\(S\)\\iota\(S\)as an empirical estimate of the importance based on the outcomes of the run\. Therefore, we use these metrics to validate the model\-estimated difficulty scoreddand importance scoreii\. The tree in Section[4\.2](https://arxiv.org/html/2608.16977#S4.SS2)gives both over the whole pool,δ⁡\(𝒫\)=4,119/4,717\\delta\(\\mathcal\{P\}\)=4\{,\}119/4\{,\}717andι⁡\(𝒫\)=77/598\\iota\(\\mathcal\{P\}\)=77/598\. Readingddandiias maps𝒫→\[0,1\]\\mathcal\{P\}\\to\[0,1\], Figure[4](https://arxiv.org/html/2608.16977#S4.F4)plotsδ\(d−1\[a,b\)\)\\delta\(d^\{\-1\}\[a,b\)\)in \(a\) andι\(i−1\[a,b\)\)\\iota\(i^\{\-1\}\[a,b\)\)in \(b\), over the intervals\[a,b\)\[a,b\)marked on each axis\.

\(a\) difficultyddagainstδ\\delta

\(b\) importanceiiagainstι\\iota

Figure 4:Each score against the quantity it judges\.Panel \(a\) plotsδ\(d−1\[a,b\)\)\\delta\(d^\{\-1\}\[a,b\)\)on an axis starting at50%50\\%, panel \(b\) plotsι\(i−1\[a,b\)\)\\iota\(i^\{\-1\}\[a,b\)\)\.nncounts attempts in \(a\) and accepted resolutions in \(b\)\. Candidates that a later status recheck reclassified as solved or invalid are excluded\.Figure[4](https://arxiv.org/html/2608.16977#S4.F4)shows thatδ\\delta, the fraction of attempts with no accepted resolution, andι\\iota, the publishable fraction of the accepted ones, both correlate positively with the difficulty and importance scores the agent assigned during the checking stage\. We measure each association by the area under the ROC curve \(AUC\)\([Hanley & McNeil 1982](https://arxiv.org/html/2608.16977#bib.bib32)\), the probability that the score ranks a randomly chosen conjecture that has the outcome above a randomly chosen conjecture that does not, ties counting half\. Difficulty achieves an AUC of0\.690\.69against having no accepted resolution, and importance0\.600\.60against being graded publishable among the accepted\. Appendix[B\.1](https://arxiv.org/html/2608.16977#A2.SS1)repeats the figure with intervals attached and gives these statistics in full\. A Mann\-Whitney test\([Mann & Whitney 1947](https://arxiv.org/html/2608.16977#bib.bib45)\)puts the first atp<10−40p<10^\{\-40\}and the second atp=0\.008p=0\.008, both showing strong positive correlation\. We therefore use them below in analyzing allocation strategies\.

We also note that the two scores are not independent of each other\. Their Spearman rank correlation is0\.830\.83, and the figure below illustrates that dependence\. Over 60% of the conjectures lie strictly above the diagonal, while almost nothing falls strictly below\. In other words, a problem that the agent calls important is almost always one that it also calls hard\. This reflects a selection effect, since the pool holds only problems that are still open, and an important question stays open only while it remains hard\.

![[Uncaptioned image]](https://arxiv.org/html/2608.16977v1/score_joint.png)The two scores do not carry the same information, however\. We demonstrate this by comparing each conjecture only against others that received the same importance score\. For each valuevvofii, writeAUCv\\mathrm\{AUC\}\_\{v\}for the AUC ofddwithin the conjectures scoredvv, andnvn\_\{v\}for the number of pairs it compares\. The stratified AUC is then,

AUCstrat=∑vnv​AUCv∑vnv=0\.56,p<10−5\.\\mathrm\{AUC\}\_\{\\mathrm\{strat\}\}=\\frac\{\\sum\_\{v\}n\_\{v\}\\,\\mathrm\{AUC\}\_\{v\}\}\{\\sum\_\{v\}n\_\{v\}\}=0\.56,\\qquad p<10^\{\-5\}\.Therefore, among conjectures the LLM scored equally important, the difficulty score still effectively separates the attempts that returned an accepted resolution from those that did not\.

#### 4\.3\.2Where to Spend a Limited Budget

Section[2\.3](https://arxiv.org/html/2608.16977#S2.SS3)posed the question: once agents are a primary source of mathematical attempts, how should their compute be allocated? We take it up here in a simple setting\. Suppose the attempting stage of Section[3\.2](https://arxiv.org/html/2608.16977#S3.SS2)is given a budget ofBBattempts, and each conjecture receives at most one attempt\. Writing𝒮\\mathcal\{S\}for the set of conjectures that the stage attempts andfffor the value of the artifacts that it produces, the stage should maximize the following:

maximize𝒮⊆𝒫\\displaystyle\\text\{maximize\}\_\{\\;\\mathcal\{S\}\\subseteq\\mathcal\{P\}\}𝔼⁡\[f⁡\(𝒜∩𝒮\)\]\\displaystyle\\mathbb\{E\}\\big\[\\,f\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\\,\\big\]\(1\)subject to\\displaystyle\\text\{subject to\}\|𝒮\|=B,\\displaystyle\|\\mathcal\{S\}\|=B,where the expectation is over the outcomes of the attempts\. Three choices offfare natural:

- •f1=\|𝒜∩𝒮\|f\_\{1\}=\|\\mathcal\{A\}\\cap\\mathcal\{S\}\|: the total number of publishable artifacts\.
- •f2=∑c∈𝒜∩𝒮i⁡\(c\)f\_\{2\}=\\sum\_\{c\\in\\mathcal\{A\}\\cap\\mathcal\{S\}\}i\(c\): the total importance of those artifacts\.
- •f3=maxc∈𝒜∩𝒮⁡i⁡\(c\)f\_\{3\}=\\max\_\{c\\in\\mathcal\{A\}\\cap\\mathcal\{S\}\}i\(c\): the importance of the single best artifact\.

##### Optimal strategies\.

Consider the outcome of a single attempt on conjectureccas random\. Then𝔼​δ​\(c\)\\mathbb\{E\}\\,\\delta\(c\)is the probability that it returns no accepted resolution,𝔼​ι​\(c\)\\mathbb\{E\}\\,\\iota\(c\)the probability that a resolution it does return is graded publishable, andp⁡\(c\)=\(1−𝔼​δ​\(c\)\)​𝔼​ι​\(c\)p\(c\)=\\big\(1\-\\mathbb\{E\}\\,\\delta\(c\)\\big\)\\,\\mathbb\{E\}\\,\\iota\(c\)the probability thatccends in𝒜\\mathcal\{A\}\. If𝔼​δ​\(c\)\\mathbb\{E\}\\,\\delta\(c\)and𝔼​ι​\(c\)\\mathbb\{E\}\\,\\iota\(c\)are known in advance, we can solve the optimization problem \([1](https://arxiv.org/html/2608.16977#S4.E1)\) exactly forf1f\_\{1\}andf2f\_\{2\}, and within a constant factor forf3f\_\{3\}\.

Forf1f\_\{1\}andf2f\_\{2\}, by linearity of expectation:

𝔼⁡\[f1​\(𝒜∩𝒮\)\]=∑c∈𝒮p⁡\(c\),𝔼⁡\[f2​\(𝒜∩𝒮\)\]=∑c∈𝒮i⁡\(c\)​p​\(c\)\.\\mathbb\{E\}\\,\[f\_\{1\}\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\]=\\sum\_\{c\\in\\mathcal\{S\}\}p\(c\),\\qquad\\mathbb\{E\}\\,\[f\_\{2\}\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\]=\\sum\_\{c\\in\\mathcal\{S\}\}i\(c\)\\,p\(c\)\.Sorting𝒫\\mathcal\{P\}byp⁡\(c\)p\(c\), respectively byi⁡\(c\)​p​\(c\)i\(c\)\\,p\(c\), and keeping the firstBBtherefore solves \([1](https://arxiv.org/html/2608.16977#S4.E1)\)\.

The argument forf3f\_\{3\}runs through the classical problem of monotone submodular maximization\([Krause & Golovin 2014](https://arxiv.org/html/2608.16977#bib.bib38)\)\. Formally, a set functionFFon subsets of𝒫\\mathcal\{P\}is*submodular*ifF⁡\(𝒮\)\+F⁡\(𝒯\)≥F⁡\(𝒮∪𝒯\)\+F⁡\(𝒮∩𝒯\)F\(\\mathcal\{S\}\)\+F\(\\mathcal\{T\}\)\\geq F\(\\mathcal\{S\}\\cup\\mathcal\{T\}\)\+F\(\\mathcal\{S\}\\cap\\mathcal\{T\}\)for all𝒮,𝒯⊆𝒫\\mathcal\{S\},\\mathcal\{T\}\\subseteq\\mathcal\{P\}, and*monotone*ifF⁡\(𝒮\)≤F⁡\(𝒯\)F\(\\mathcal\{S\}\)\\leq F\(\\mathcal\{T\}\)whenever𝒮⊆𝒯\\mathcal\{S\}\\subseteq\\mathcal\{T\}\.

Maximizing a monotone submodularFFunder a cardinality constraint contains maximum coverage as a special case and is therefore NP\-hard\. It is known, however, that a simple greedy algorithm achieves at least a\(1−1/e\)\(1\-1/e\)fraction of the optimum\([Nemhauser et al\. 1978](https://arxiv.org/html/2608.16977#bib.bib47)\), and that no polynomial algorithm beats it unless P=\\,=\\,NP\([Feige 1998](https://arxiv.org/html/2608.16977#bib.bib26)\)\. The algorithm starts with𝒮0=∅\\mathcal\{S\}\_\{0\}=\\emptyset, and at each iterationjjit adds the element of largest gainF⁡\(c∣𝒮j−1\)=F⁡\(𝒮j−1∪\{c\}\)−F⁡\(𝒮j−1\)F\(c\\mid\\mathcal\{S\}\_\{j\-1\}\)=F\(\\mathcal\{S\}\_\{j\-1\}\\cup\\\{c\\\}\)\-F\(\\mathcal\{S\}\_\{j\-1\}\), so that

𝒮j=𝒮j−1∪\{arg​maxc∈𝒫∖𝒮j−1F\(c∣𝒮j−1\)\},j=1,…,B\.\\mathcal\{S\}\_\{j\}=\\mathcal\{S\}\_\{j\-1\}\\cup\\Big\\\{\\operatorname\*\{arg\\,max\}\_\{c\\in\\mathcal\{P\}\\setminus\\mathcal\{S\}\_\{j\-1\}\}F\(c\\mid\\mathcal\{S\}\_\{j\-1\}\)\\Big\\\},\\qquad j=1,\\dots,B\.
We show that𝒮↦𝔼⁡\[f3​\(𝒜∩𝒮\)\]\\mathcal\{S\}\\mapsto\\mathbb\{E\}\\,\[f\_\{3\}\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\]is monotone submodular, which is what the guarantee above requires\. The mapf3f\_\{3\}is monotone by definition\. For submodularity, suppose without loss of generality thatf3​\(𝒰\)≥f3​\(𝒱\)f\_\{3\}\(\\mathcal\{U\}\)\\geq f\_\{3\}\(\\mathcal\{V\}\); thenf3​\(𝒰∪𝒱\)=f3​\(𝒰\)f\_\{3\}\(\\mathcal\{U\}\\cup\\mathcal\{V\}\)=f\_\{3\}\(\\mathcal\{U\}\)andf3​\(𝒰∩𝒱\)≤f3​\(𝒱\)f\_\{3\}\(\\mathcal\{U\}\\cap\\mathcal\{V\}\)\\leq f\_\{3\}\(\\mathcal\{V\}\), sof3​\(𝒰\)\+f3​\(𝒱\)≥f3​\(𝒰∪𝒱\)\+f3​\(𝒰∩𝒱\)f\_\{3\}\(\\mathcal\{U\}\)\+f\_\{3\}\(\\mathcal\{V\}\)\\geq f\_\{3\}\(\\mathcal\{U\}\\cup\\mathcal\{V\}\)\+f\_\{3\}\(\\mathcal\{U\}\\cap\\mathcal\{V\}\)\. The map𝒮↦𝒜∩𝒮\\mathcal\{S\}\\mapsto\\mathcal\{A\}\\cap\\mathcal\{S\}preserves unions and intersections, so it carries both properties over to𝒮↦f3​\(𝒜∩𝒮\)\\mathcal\{S\}\\mapsto f\_\{3\}\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\. Taking expectations then gives the same two properties for𝒮↦𝔼⁡\[f3​\(𝒜∩𝒮\)\]\\mathcal\{S\}\\mapsto\\mathbb\{E\}\\,\[f\_\{3\}\(\\mathcal\{A\}\\cap\\mathcal\{S\}\)\]\.

##### Building a strategy from the scores\.

In practice, neither𝔼​δ​\(c\)\\mathbb\{E\}\\,\\delta\(c\)nor𝔼​ι​\(c\)\\mathbb\{E\}\\,\\iota\(c\)are known\. Before any reasoning is spent, the only signals available are the model\-based difficulty and importance scores that the pool building stage recorded\. By the arguments above, if we can approximateppfrom these scores then we obtain near\-optimal strategies forf1f\_\{1\}andf2f\_\{2\}\. Forf3f\_\{3\}an approximation ofppis not enough, since each greedy step needs the value ofF⁡\(c∣𝒮\)F\(c\\mid\\mathcal\{S\}\), which depends on the joint distribution of the attempt outcomes\. Hence we propose an alternative algorithm forf3f\_\{3\}that, like those forf1f\_\{1\}andf2f\_\{2\}, needs only an approximation ofppand does well in our run\.

We fit the two factors ofppby least squares,1−δ1\-\\deltaonddover𝒫\\mathcal\{P\}andι\\iotaoniiover𝒥\\mathcal\{J\}, and writeδ^\\hat\{\\delta\}andι^\\hat\{\\iota\}for the resulting estimates of𝔼​δ\\mathbb\{E\}\\,\\deltaand𝔼​ι\\mathbb\{E\}\\,\\iota\. We writeδ^\\hat\{\\delta\}andι^\\hat\{\\iota\}for the two estimates of𝔼​δ\\mathbb\{E\}\\,\\deltaand𝔼​ι\\mathbb\{E\}\\,\\iota\. Their productp^=\(1−δ^\)​ι^\\hat\{p\}=\(1\-\\hat\{\\delta\}\)\\,\\hat\{\\iota\}estimatespp\. In theory, ranking𝒫\\mathcal\{P\}onp^\\hat\{p\}and taking the topBBis near\-optimal forf1f\_\{1\}, and ranking oni​p^i\\,\\hat\{p\}is near\-optimal forf2f\_\{2\}\. Forf3f\_\{3\}, we choose a top fraction of𝒫\\mathcal\{P\}based on importance, and then rank onp^\\hat\{p\}\. The least squares fits underlying these strategies do not require the whole pool, so a round may attempt a small part of the pool first, fit and compare strategies on the returned outcomes, and allocate the remaining budget with the strategy that performed best\.

##### Empirical results\.

We compare how these strategies perform on the outcomes that our pilot run produced\. As the baseline strategy, we take𝒮\\mathcal\{S\}to be a uniform random subset of𝒫\\mathcal\{P\}of sizeBB\. Against it we compare ranking the whole of𝒫\\mathcal\{P\}onp^\\hat\{p\}, ranking it oni​p^i\\,\\hat\{p\}, and ranking onp^\\hat\{p\}inside a top fraction of𝒫\\mathcal\{P\}by importance \(inside such a restrictioniivaries too little to reorder anything, so ranking there oni​p^i\\,\\hat\{p\}gives an almost same strategy\)\. Figure[5](https://arxiv.org/html/2608.16977#S4.F5)shows the baseline, the two rankings, and the two restrictions that returned the most artifacts\.

Each strategy is evaluated by55\-fold cross\-validation\. We split𝒫\\mathcal\{P\}into five parts at random, fit both factors on four of them, rank the fifth by the resultingp^\\hat\{p\}and keep its topB/5B/5, and combine the five selections, so that no conjecture is ever ranked by a fit that saw its own outcome\. Each point of the figure averages10001000such partitions, atB=10B=10,2525,5050,7575,100100,200200and300300\. Appendix[B\.2](https://arxiv.org/html/2608.16977#A2.SS2)tabulates every plotted point\.

Figure 5:Allocation strategies against the budget\.Each fit is made on four fifths of𝒫\\mathcal\{P\}and applied to the remaining fifth, from whichB/5B/5conjectures are drawn\. The five selections together make one set ofBBconjectures, and each point averages what that set returns over10001000random partitions\. Ties are broken at random, and the uniform baseline is computed exactly from its closed form\. Here,BBonly counts the allocated attempts\.Onf1f\_\{1\}, the run follows the theoretical analysis closely\. Ranking𝒫\\mathcal\{P\}onp^\\hat\{p\}outperforms every other strategy at every budget\. Onf2f\_\{2\}, the analysis prescribes ranking oni​p^i\\,\\hat\{p\}, but ranking onp^\\hat\{p\}instead beats it narrowly at every budget\. Onf3f\_\{3\}, neither ranking of the whole pool does well, and at the larger budgets both even fall behind the uniform random baseline\. What works best in practice is to discard all but the most important part of𝒫\\mathcal\{P\}and rank what remains onp^\\hat\{p\}\. Of the fractions we tried, keeping the most important1/101/10did best on this run\. Note that the results above depend on the scoring model, on the sources that the pool was built from, and on the models in the cascade\. Nevertheless, the results suggest that using strategies derived from model\-based importance and difficulty scores, it is possible to allocate the budget more effectively than with a uniform strategy in our problem setting\. Furthermore, the optimal allocation strategy differs betweenf1f\_\{1\}andf3f\_\{3\}, providing evidence that the best strategy can depend on the objective that the mathematician is optimizing for\.

## 5Selected Reviewed Results

We manually reviewed fifteen of the artifacts from the pilot run, chosen by our own interest, and found no mathematical error in any of them\. We discuss some of them below, which show three different degrees of novelty\. Full write\-ups of these results and of the others we reviewed are collected in Appendix[C](https://arxiv.org/html/2608.16977#A3)\.

### 5\.1A Known Result Graded as New

##### Sets with no large divisor difference \(Appendix[C\.4](https://arxiv.org/html/2608.16977#A3.SS4)\)\.

Erdős\([Guy 1983](https://arxiv.org/html/2608.16977#bibd.bib3)\)asks how large a set of integers can be if no two of its elements have a large difference dividing the larger one\. Formally, fort≥1t\\geq 1, letF⁡\(n,t\)F\(n;t\)be the largest size of a setA⊆\{1,…,n\}A\\subseteq\\\{1,\\dots,n\\\}in which no two elementsx<yx<ysatisfy both\(y−x\)\|y\(y\-x\)\\mid yandy−x≥ty\-x\\geq t\. The question is whetherF⁡\(n,t\)≤\(12\+o⁡\(1\)\)​nF\(n;t\)\\leq\(\\tfrac\{1\}\{2\}\+o\(1\)\)nfor every fixedtt\.

The artifact returned for this problem proves thatF⁡\(n,t\)/n→12F\(n;t\)/n\\to\\tfrac\{1\}\{2\}\. The odd numbers give the lower bound, since the difference of two odd numbers is even and cannot divide the larger\. For the upper bound it fixes a finite setPPof odd primes exceedingttwith∑p∈P1/p\\sum\_\{p\\in P\}1/plarge\. For eachp∈Pp\\in Pthe set cannot hold both2​r​p2rpand\(2​r−1\)​p\(2r\-1\)p, whose difference isp≥tp\\geq tand divides2​r​p2rp, so the even elements ofAAare divisible by no more primes ofPP, in total, than the odd integers outsideAAare\. A second\-moment estimate on each parity class turns this into the matching upper bound\.

The recommend stage judged the proof correct and graded it as a publishable result\. Reviewing it ourselves we found the mathematics correct but the result already known\. Four months before our run a proof of the same statement had been recorded on the Erdős problems site, obtained by Liam Price with ChatGPT\-5\.2\([Bloom 2026](https://arxiv.org/html/2608.16977#bibd.bib1)\), and Tao also observed there that the bound follows quickly from an inequality of[Elliott 2012](https://arxiv.org/html/2608.16977#bibd.bib2)\. The cascade never found this record\.

### 5\.2A Connection Not Previously Made

##### Small unions of lines closing a route to the Nikodym bound \(Appendix[C\.10](https://arxiv.org/html/2608.16977#A3.SS10)\)\.

For a setLLof affine lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}letP⁡\(L\)=⋃ℓ∈LℓP\(L\)=\\bigcup\_\{\\ell\\in L\}\\ellbe the union of its lines\.[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)conjecture that for every constantC\>0C\>0and everyα\\alphawithα⁡\(q\)/q→∞\\alpha\(q\)/q\\to\\infty, a setLLof at leastC​q3Cq^\{3\}lines in which no plane containsα⁡\(q\)\\alpha\(q\)lines must satisfy\|P⁡\(L\)\|≥\(1−o⁡\(1\)\)​q3\|P\(L\)\|\\geq\(1\-o\(1\)\)q^\{3\}\. They show that this would give an optimal bound on the size of a Nikodym set in three dimensions, and[Tao 2025](https://arxiv.org/html/2608.16977#bibj.bib7)still cites it as open in that role\.

The artifact returned for this problem disproves the conjecture for every oddqq\. It takes the affine paraboloidz=x2−ν​y2z=x^\{2\}\-\\nu y^\{2\}withν\\nua nonsquare, and keeps at each of its points the\(q\+1\)/2\(q\+1\)/2tangent lines on whichx2−ν​y2−zx^\{2\}\-\\nu y^\{2\}\-zis always a square\. The family hasq2​\(q\+1\)/2q^\{2\}\(q\+1\)/2lines, no affine plane holds more thanq\+1q\+1of them, and its union has exactlyq2​\(q\+1\)/2q^\{2\}\(q\+1\)/2points, a density of1/2\+o⁡\(1\)1/2\+o\(1\)\. The same family also refutes the stronger Conjecture 1\.5 of that paper\.

We found the construction mathematically correct, and verified the counts exhaustively forq≤13q\\leq 13\. The recommend stage, however, informed us that the construction already exists in other contexts\. In projective language the family is a classical object of finite geometry, the half\-tangent partition of an elliptic quadric\([Bruen & Drudge 1999](https://arxiv.org/html/2608.16977#bibj.bib1);[Cossidente & Pavese 2017](https://arxiv.org/html/2608.16977#bibj.bib2)\)\. The contribution here is to link the existing construction to this conjecture\.

### 5\.3Results with No Precedent Found

The three results below have each been checked by an author or by a domain expert, and are new so far as we could determine\. They represent three kinds of discovery\. The first proves a conjecture, the second refutes one by counterexample, and the third answers an open\-ended question\.

##### Many\-one𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-completeness for binary symmetric group characters \(Appendix[C\.6](https://arxiv.org/html/2608.16977#A3.SS6)\)\.

For partitionsλ\\lambdaandμ\\muof an integernn, writeχλ​\(μ\)\\chi^\{\\lambda\}\(\\mu\)for the irreducible character value of the symmetric group\.[Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)consider the problem of computingχλ​\(μ\)\\chi^\{\\lambda\}\(\\mu\)fromλ\\lambdaandμ\\mugiven as lists of parts in binary\. They prove that it is𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under Turing reductions and conjecture that it is𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under many\-one reductions\.

The artifact returned for it proves that conjecture\. It reduces the difference of two counts of exact covers of a finite set to a single character value at a two\-row partitionλ=\(n−s,s\)\\lambda=\(n\-s,s\), and checks membership in𝖦𝖺𝗉𝖯\\mathsf\{GapP\}separately\. A two\-row character value is itself a difference,Nμ​\(s\)−Nμ​\(s−1\)N\_\{\\mu\}\(s\)\-N\_\{\\mu\}\(s\-1\), whereNμ​\(t\)N\_\{\\mu\}\(t\)is the number of ways to choose parts ofμ\\musumming tott\.

##### Maximum versus average independent set size in triangle\-free graphs \(Appendix[C\.3](https://arxiv.org/html/2608.16977#A3.SS3)\)\.

For a graphGGletα⁡\(G\)\\alpha\(G\)be its independence number andα¯​\(G\)\\bar\{\\alpha\}\(G\)the expected size of a uniformly random independent set\.[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)conjecture thatα⁡\(G\)/α¯​\(G\)≥2−od​\(1\)\\alpha\(G\)/\\bar\{\\alpha\}\(G\)\\geq 2\-o\_\{d\}\(1\)for every triangle\-freeGGof minimum degreedd, and show that this would giveR⁡\(3,k\)≤\(12\+o⁡\(1\)\)​k2/log⁡kR\(3,k\)\\leq\(\\tfrac\{1\}\{2\}\+o\(1\)\)k^\{2\}/\\log k\. It is restated as open in recent surveys of the hard\-core model and Ramsey theory\([Davies & Kang 2025](https://arxiv.org/html/2608.16977#bibc.bib2);[Morris 2026](https://arxiv.org/html/2608.16977#bibc.bib7)\)\.

The artifact returned as a counterexample the Cartesian productC5​□​Km,mC\_\{5\}\\,\\square\\,K\_\{m,m\}, which is triangle\-free and\(m\+2\)\(m\+2\)\-regular\. An independent set of the product picks an independent subset ofC5C\_\{5\}at each vertex ofKm,mK\_\{m,m\}, and the two sides must pick disjoint subsets\. This makesα=4​m\\alpha=4mand the expected size computable exactly, and the ratio tends to24/1324/13\. ReplacingC5C\_\{5\}by the circulantC13​\(1,5\)C\_\{13\}\(1,5\)further lowers the limit to32/1932/19\.

##### Divisibility among binomial coefficients \(Appendix[C\.5](https://arxiv.org/html/2608.16977#A3.SS5)\)\.

For a fixed integern≥2n\\geq 2,[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2)ask for the natural densityd∗​\(n\)d^\{\*\}\(n\)of thosemmthat admit somekkwith1≤k≤m−n1\\leq k\\leq m\-nand\(n\+kn\)\|\(m\+kk\)\\binom\{n\+k\}\{n\}\\mid\\binom\{m\+k\}\{k\}, that is, the proportion of suchmmamong the integers up toxxasx→∞x\\to\\infty\. They settlen=1n=1themselves and record thatn=2n=2“seems much more difficult to decide”\.

The returned artifact shows thatd∗​\(n\)=1d^\{\*\}\(n\)=1for every fixedn≥2n\\geq 2\. Given a boundB\>nB\>nit takeskB=∏p≤Bpepk\_\{B\}=\\prod\_\{p\\leq B\}p^\{e\_\{p\}\}withepe\_\{p\}least such thatpep\>np^\{e\_\{p\}\}\>n, so that by Kummer’s theorem every prime factor ofNB=\(n\+kBn\)N\_\{B\}=\\binom\{n\+k\_\{B\}\}\{n\}exceedsBB\. Each such primeqqtherefore exceedsnn, and so divides exactly one ofkB\+1,…,kB\+nk\_\{B\}\+1,\\dots,k\_\{B\}\+n, saykB\+i⁡\(q\)k\_\{B\}\+i\(q\)\. By Legendre’s formula the conditionmmodq≥i⁡\(q\)m\\bmod q\\geq i\(q\)makes\(m\+kBkB\)\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}divisible by the full power ofqqthat dividesNBN\_\{B\}\. The Chinese remainder theorem turns these residue conditions into a set of density∏q\(1−i⁡\(q\)/q\)\\prod\_\{q\}\(1\-i\(q\)/q\)that tends to11asB→∞B\\to\\infty\.

## 6Conclusion

We have argued that an important–yet understudied–aspect of AI for mathematics concerns selecting which problems to solve\. We introducedFAR, a literature\-to\-review cascade that recovers open problems from a corpus, attempts each of them, and recommends promising research outcomes for expert review\. In a combinatorics pilot,FARrecovered 6,453 candidate statements, checked 4,717 of them into an attemptable pool, and returned 77 artifacts graded as substantial enough to publish, 15 of which the authors reviewed and found to be correct\. Using the outcomes of our run, we studied strategies for allocating effort, i\.e\., selecting a subset of conjectures to attempt given a budget\. We framed this problem as constrained optimization and derived strategies that are favorable both theoretically and in practice\. We showed that simply asking an agent in our pipeline to output difficulty and importance scores before any reasoning is spent yields scores that can be used effectively within the allocation strategies\. We hope thatFARis a starting point for new techniques that help explore mathematics, which consists of not just solving problems, but of deciding which problems are worth spending effort on in the first place\.

## Acknowledgments

We thank Sylvester W\. Zhang for his help with expert review\. Shengtong Zhang thanks Anysphere co\. for providing compute for this project\. This work was supported in part by NSF Grant DMS\-2434614 and DARPA ExpMath Grant HR0011262E028\.

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## Appendix APrompt Templates and Operational Details

This appendix includes the prompts used in our implementation\. The released code also carries the schema validators and the retry logic\.

### A\.1Finding Relevant Open Problems

In the pilot the direction was “combinatorics”\.

ReturnoneJSONobjectwiththisschema:

\{

”comment”:”…”,

”in\_direction”:false

\}

Rules:

\-‘comment‘mustnamethepaper’sprimarysubjectinafewwords\.

\-‘in\_direction‘mustbeaJSONboolean\.

\-Use‘true‘whenthepaper’sprimarycontentliesintheresearchdirection\.

\-Use‘false‘whenitdoesnot,whenthecontentisnotmathematical,orwhenthepaperappearsmislabeled\.

Researchdirection:\{direction\}

Papercontent:

\{text\}

ReturnoneJSONobjectwiththisschema:

\{

”title”:”…”,

”authors”:\[”…”,”…”\],

”decision\_basis”:”…”,

”has\_open\_conjecture”:false,

”conjectures”:\[

\{

”conjecture\_label”:”…”,

”conjecture\_text”:”…”,

”conjecture\_section”:”…”

\}

\]

\}

Rules:

\-‘title‘mustbeanon\-emptystring\.

\-‘authors‘mustbeaJSONarrayofnon\-emptyauthor\-namestrings\.

\-‘decision\_basis‘mustbeoneshortEnglishsentence\.

\-‘has\_open\_conjecture‘mustbeaJSONboolean\.

\-‘conjectures‘mustbeaJSONarray\.If‘has\_open\_conjecture‘isfalse,itmustbe‘\[\]‘\.

\-Set‘has\_open\_conjecture‘totrueiffthepapercontainsatleastoneexplicitunresolvedmathematicalstatement\.

\-Counttheseashits:

1\.labeled‘Conjecture‘/‘Question‘/‘OpenProblem‘

2\.sentenceswithmarkerslike‘openquestion‘,‘openproblem‘,‘openissue‘,‘remainsunknownwhether‘,or‘wesuspect…althoughwehavebeenunabletoestablish…‘

3\.adirectstatementthataspecificmathematicalproperty,existenceclaim,orclassificationproblem‘stillremainsanopenissue‘

\-DoNOTcount:

1\.genericfutureworkthatdoesnotposeaspecificmathematicalquestion

2\.resultsthathavealreadybeenprovedorresolvedwithinthepaperitself

\-Ifasentencesaysaspecificclaimorpropertyis‘stillanopenissue‘,countitevenifitisnotwrittenasaformalquestion\.

\-If‘has\_open\_conjecture‘istrue,extractonlytheexplicitunresolvedstatementsthemselves,notnearbyspeculation\.

\-‘conjecture\_label‘shouldusethepaper’slabelwhenpresent,otherwiseuseashortfallbacklike‘Unlabeledopenproblem1‘\.

\-‘conjecture\_text‘shouldcopythepaper’sunresolvedstatementasfaithfullyaspossibleandpreservenotation\.

\-‘conjecture\_section‘shouldbethevisiblesection/subsectiontitle,or‘””‘ifunavailable\.

Papercontent:

\{text\}

ReturnoneJSONobjectwiththisschema:

\{

”sources”:\[

\{”title”:”…”,”url”:”…”,”claim”:”…”\}

\],

”reason”:”…”,

”status”:”solved”,

”importance”:0\.5,

”difficulty”:0\.5

\}

Rules:

\-Verifythecandidate’scurrentstatususingcurrentwebinformation\.

\-‘status‘mustbeoneof:‘open‘,‘solved‘,‘invalid‘\.

\-Use‘open‘whenthecandidateisaconcreteopenprobleminthesourceandnocrediblesolvedevidenceisfound\.

\-Use‘solved‘whenacrediblesourceappearstosolveit\.

\-Use‘invalid‘whenitisnotaconcreteopenprobleminthesource\.

\-‘sources‘shouldlistonlysourcesdirectlysupportingthestatus\.

\-each‘claim‘mustbewhatthatsourcesaysaboutthiscandidate\.

\-for‘solved‘,‘sources‘mustnameatleastonesourcethatresolvesthecandidate\.

\-‘reason‘mustbeoneconciseEnglishsentence\.

\-‘importance‘mustbeanumberin\[0,1\]forthecandidateitself:candidateswithnosubstantivemathematicalcontentshouldbescored0;Fields\-Medal\-levelproblemsshouldbescored1;mostordinaryresearchproblemsshouldfollowaroughlynormaldistributioncenteredaround0\.5\.

\-‘difficulty‘mustbeanumberin\[0,1\]:solvingitwouldbeanunpublishableexerciseshouldbescored0;solvingitwouldbepublishableinatopjournal\(Annals,Inventiones,JAMS,Acta\)shouldbescored1;mostproblemsshouldfollowaroughlynormaldistributioncenteredaround0\.5\.

\-For‘solved‘or‘invalid‘,set‘importance‘and‘difficulty‘to0\.

Papertitle:\{title\}

Paperauthors:\{authors\}

Candidatelabel:\{conjecture\_label\}

Candidatesection:\{conjecture\_section\}

Candidatetext:

\{conjecture\_text\}

### A\.2Attempting for Candidate Resolutions

Youarearesearch\-levelmathematicalreasoner\.Thisisatesttoseehowwellyoucancraftnon\-trivial,novelandcreativeproofsgivenamathproblem\.

Givenanatural\-languageproblem,conjecture,orpapermetadata,reconstructthemostlikelyformalmathematicalstatementandresolveit\.

First,statethereconstructedconjectureprecisely,includingallhypotheses,definitions,notation,quantifiers,ambientcategory,andaxiomsystemwhenrelevant\.Explainbrieflywhatinformationsupportsthisreconstruction\.Ifthereconstructionisambiguous,listtheplausibleformalizationsandchooseonetoanalyze,explicitlynotingtheambiguity\.

Donottreatthefactthatthesourcelabelsthestatementopen,conjectural,unresolved,oraproblemasareasontostop\.Thetaskistoattackthestatementmathematically\.However,donotlowerthestandardofproof\.Neverpresentanincomplete,heuristic,orspeculativeargumentasacompleteproof\.

Beforecommittingtoaproof,testthestatementagainstdegenerate,extremal,low\-dimensional,finite,infinite,andstandardmodelexamplesappropriatetothefield\.Lookactivelyforcounterexamplesaswellasproofs\.

Iftheliteralstatementisfalsebecauseofadegenerate,boundary,vacuous,ortypo\-likecase,donotstopaftergivingthecounterexample\.Instead:

\-Statetheliteralcounterexampleclearlyandexplainwhyitfalsifiestheliteralstatement\.

\-Diagnosewhetherthefailureappearstocomefromasmallformulationdefect,suchasamissingnonzero/nonempty/nontrivialassumption,awronginequalitydirection,anomittedendpointcondition,amissingconnectednessorfinitenesshypothesis,aconfusionbetweenstrictandnon\-strictinequalities,amissingregularitycondition,oraconventionmismatch\.

\-Proposetheminimalnaturalrepairorrepairstothestatement,usingthefewestandmoststandardchangesconsistentwiththepaper’sterminology,surroundingcontext,andapparentmathematicalintent\.

\-Checkthattheproposedrepairisnotmerelyadhoc,vacuous,orsoweakenedthatitnolongercapturestheintendedconjecture\.

\-Retesttherepairedstatementagainsttheoriginalcounterexampleandnearbydegeneratecases\.

\-Thenproveorrefutethemostplausiblerepairedstatement\.

Acompleteanswermustbearigorousprooforarigorouscounterexample\.

Presentthereasoninginalocallycheckableform:definitions,lemmas,propositions,andproofs\.Foreveryinvokedtheorem,verifyitshypothesesinthepresentsetting\.Trackdependenciesofconstants,choices,witnesses,bases,subsequences,exceptionalsets,embeddings,isomorphisms,andparameters\.

Iftheprooforcounterexampleisknownintheliterature,statethathonestlyandprovideareliablereference\.Distinguishexactresolutionsfromstrongertheorems,weakerpartialresults,equivalentreformulations,andmerelyrelatedwork\.Donotinventreferences\.

Aftertheprooforcounterexample,includeaverificationauditconfirmingthattheformalizedstatementmatchesthereconstructedconjecture,thatnoextraassumptionswereintroduced,thatalltheoremhypotheseswerechecked,andthattheconclusionexactlymatchesthetargetstatement\.

Responseformat:

Thefirstlinemustbeexactlyoneof:KNOWN,NEW,FIX,NONE\.

\-KNOWN:areliableexistingsourceintheliteraturealreadyprovestheconjectureorgivesacounterexample/disproof\.Citethesource\.

\-NEW:youranswergivesacompleteresolutionthatisnotpresentedasknownliterature\.UseNEWforeitheracompleteproofthattheconjectureistrueoracompletecounterexample/disproofthattheconjectureisfalse\.

\-FIX:youhaveidentifiedasmallformulationdefectandproposedaminimalnaturalrepair,butyouareunabletoproveorrefutetherepairedstatement\.UseFIXtoindicatethatyouhavedonethis\.

\-NONE:youfoundneitheraknownresolutionnorareliablecompleteproof/counterexampledespiteallefforts\.

Thenusethesesectionsexactly:

Problem:

Result:

Citation:

Readinput\.jsoninthecurrentdirectory\.Itcontainsthepapertitle,authors,papertext,thesourcesastatuscheckturnedup,andtargetconjecture\.Thetargetconjectureisinconjecture\.text\.

Resolvethattargetconjectureandreturnonlytherequiredlabeledanswer\.

### A\.3Judging and Grading

KNOWNandNEWoutcomes are sent to the judge, and its label routes the outcome rather than certifying it\. Only the results it accepts as new reach the grader\.

Youareastrictrefereefornatural\-languagemathematicsproofs\.Thisisatesttoseehowwellyoucanrefereeaproposednatural\-languagemathematicsproofgivenamathproblem\.

Checktheclaimedresolutionordisproofagainstthetargetconjecturesuppliedintheusertask\.

Acceptonlyiftheclaimedresolutionordisproofattacksthecorrectstatementandismathematicallyrigorousandcomplete\.

Avalidcounterexampleordisproofmaypassifitrigorouslydisprovestheconjecture\.

Rejectifithasfatalproofgaps,hallucinateddependencies,hiddenassumptions,oramismatchbetweenthestatedtheoremandtheoriginalconjecture\.

Donotrejectmerelybecausetheoriginalpapercalledtheconjectureopen\.

InthecasewhentheclaimedresolutionordisproofisNEW,youshouldalsoconductaverythoroughliteraturesearchusingthewebsearchtooltoseeifasimilarorstrongerresultalreadyexistsintheliterature\.

Onthefirstline,writeexactlyoneword:PASSorFAILorKNOWN\.

\-PASS:theclaimedresolutionismathematicallycompleteandattacksthecorrectstatement,andinthecaseofNEW,asimilarorstrongerresultdoesnotexistintheliteraturedespiteyourbestsearchefforts\.

\-KNOWN:theclaimedresolutionisNEW,butasimilarorstrongerresultalreadyexistsintheliterature\.

\-FAIL:ifneitheroftheaboveconditionsaremet\.

Thenbrieflyexplainyourverdict,includingthemostimportantgapifyoufailit\.

Readinput\.jsonandsolution\.mdinthecurrentdirectory\.input\.jsoncontainspapermetadata,thepapertext,thesourcesastatuscheckturnedup,andthetargetconjecture\.solution\.mdcontainstheclaimedresolutiontocheck\.

ReturnonlyPASSorFAILorKNOWNfollowedbyyourexplanation\.

Youareaseniorcombinatoricsrefereeperformingafinalquality\-controlpassonaresultthataproverproducedandajudgealreadyacceptedasacorrectresolution\.

YourjobisNOTtore\-verifycorrectnessfromscratch\(assumetheproofiscorrectunlessaliteraturesearchclearlycontradictsit\)\.Yourjobistoclassifytheresultbyitsnoveltyandpublishablesignificance,soahumancantriageitafterwards\.

Dotwothings:

1\.Literaturecheck\.

Conductaverythoroughwebsearchtodeterminewhethertheresolution,orasimilarorstrongerstatement,isalreadyknownintheliterature\.Gobeyondjustsearchingforpapersthatcitetheoriginalpaper;youshouldsearchforallopen\-accessnotes,surveys,forums,andothersourcesthatmightcontaintheresult\.Theproverandearlierjudgesmayhavemissedanexistingreference;catchingsuchcasesisaprimarygoalofthispass\.

2\.Significancegrading\.Iftheresultisgenuinelynotintheliterature,assesshowsignificantitisasacontributiontocombinatorics:howhard,hownovel,howinterestingtothecommunity,andwhatvenueitwouldplausiblymerit\.

Onthefirstline,writeexactlyonetoken:KNOWN,TYPE1,TYPE2,orTYPE3\.

\-KNOWN:theresult\(orasimilarorstrongerresult\)isinfactalreadyknownintheliterature,despitetheproverandearlierjudgestreatingitasnew\.Citethereference\.

\-TYPE1:genuinelynewbutminorandunpublishableonitsown\(e\.g\.aroutineexercise,atrivialspecialcase,animmediatecorollaryofstandardresults\)\.

\-TYPE2:genuinelynewandsubstantialenoughtosupportastandalonepaperinastandardcombinatoricsormathematicsjournal\.

\-TYPE3:genuinelynewandstrongenoughtomeritpublicationinatopcombinatoricsjournal\(amajoradvance,aresolvedwell\-knownconjecture,oraresultofbroadinterest\)\.

Theseboundariesaredeliberatelyrough;whenuncertainbetweentwogrades,picktheloweroneandexplaintheuncertainty\.

Afterthefirstline,usethesesectionsexactly:

Classificationrationale:

Literaturecheck:

Citation:

Readinput\.json,solution\.md,andjudge\.mdinthecurrentdirectory\.input\.jsoncontainsthepapermetadata,thepapertext,thesourcesastatuscheckturnedup,andthetargetconjecture,solution\.mdcontainstheresolutionthatwasacceptedasnew,andjudge\.mdcontainstheverdictsoftheearlierjudges\.

ClassifytheresultandreturnonlyKNOWN,TYPE1,TYPE2,orTYPE3onthefirstline,followedbytherequiredsections\.

The artifacts put forward for expert review are theTYPE2andTYPE3items\.

## Appendix BAnalysis Details

This appendix gives the data behind the two figures of Section[4\.3](https://arxiv.org/html/2608.16977#S4.SS3)\.

### B\.1Score validity

Figure[6](https://arxiv.org/html/2608.16977#A2.F6)gives the same two associations as Figure[4](https://arxiv.org/html/2608.16977#S4.F4)at a uniform bin width of0\.10\.1\. Each point is the measured rate in its bin and each bar its95%95\\%Wilson interval\.

Figure 6:Each score against the quantity it judges, with Wilson intervals\.Panel \(a\) givesδ\\deltaagainst the difficulty score, panel \(b\) givesι\\iotaagainst the importance score\.
### B\.2Allocation curves

The tables below give the value of each point plotted in Figure[5](https://arxiv.org/html/2608.16977#S4.F5)\. Every entry is an average over10001000random partitions of the pool into five parts, with both factors ofp^\\hat\{p\}fitted on four of them and the remaining part ranked by the result, from whichB/5B/5conjectures are taken\. The five selections together form the set ofBBconjectures that the entry scores\. Ties are broken at random, and the uniform baseline is evaluated from its closed form rather than sampled\.

Table 2:Expected number of artifacts, the objectivef1f\_\{1\}\.Table 3:Expected total importance of the artifacts returned, the objectivef2f\_\{2\}\.Table 4:Expected maximum importance among the artifacts returned, the objectivef3f\_\{3\}\.

## Appendix CReviewed Solutions

This appendix collects the write\-ups of the results the authors reviewed, ordered alphabetically by the mathematicians who posed the problems\.

### C\.1Paley graphs with a prescribed adjacency property

This problem concerns the number of vertices of a Paley graph that are adjacent to one prescribed vertex and to neither of two others\. Ananchuen and Caccetta proved that the Paley graph onqqvertices has at leastkksuch vertices, for every choice of the three, as soon asq\>\(1\+2​2​k\)2q\>\(1\+2\\sqrt\{2k\}\)^\{2\}, and conjectured that this threshold is exact\. We show that it is not: the Paley graph on55=31255^\{5\}=3125vertices has at least377377such vertices for every choice, although3125<\(1\+2​754\)23125<\(1\+2\\sqrt\{754\}\)^\{2\}\. The proof is a quadratic character count in which the only obstruction is an elliptic curve over𝔽55\\mathbb\{F\}\_\{5^\{5\}\}of trace110110, and Waterhouse’s classification of elliptic\-curve traces forbids a nonzero trace divisible by the characteristic over a field of odd degree in characteristic greater than33\.

#### C\.1\.1Introduction

Letm,nm,nbe nonnegative integers and letkkbe a positive integer\. Following[Ananchuen et al\. 1992](https://arxiv.org/html/2608.16977#biba.bib5), a graphGGhas*propertyP⁡\(m,n,k\)P\(m,n,k\)*if for every pair of disjoint setsA,B⊆V⁡\(G\)A,B\\subseteq V\(G\)with\|A\|=m\|A\|=mand\|B\|=n\|B\|=nthere are at leastkkvertices outsideA∪BA\\cup Bthat are adjacent to every vertex ofAAand to no vertex ofBB\. We write𝒢⁡\(m,n,k\)\\mathcal\{G\}\(m,n,k\)for the class of graphs with propertyP⁡\(m,n,k\)P\(m,n,k\)\. These classes are quantitative refinements of the adjacency axioms of[Blass & Harary 1979](https://arxiv.org/html/2608.16977#biba.bib7), who showed by a probabilistic argument that almost all graphs have propertyP⁡\(n,n,1\)P\(n,n,1\), from which the same follows for everyP⁡\(m,n,k\)P\(m,n,k\); a graph isnn\-existentially closed exactly when it lies in𝒢⁡\(i,n−i,1\)\\mathcal\{G\}\(i,n\-i,1\)for all0≤i≤n0\\leq i\\leq n\. Explicit examples are much harder to come by, and the extremal question of how few vertices a graph in𝒢⁡\(m,n,k\)\\mathcal\{G\}\(m,n,k\)can have was raised by[Exoo 1981](https://arxiv.org/html/2608.16977#biba.bib10)and studied systematically by[Ananchuen et al\. 1992](https://arxiv.org/html/2608.16977#biba.bib5)\.

For a prime powerq≡1\(mod4\)q\\equiv 1\\pmod\{4\}, the*Paley graph*GqG\_\{q\}has vertex set𝔽q\\mathbb\{F\}\_\{q\}, two distinct verticesu,vu,vbeing adjacent exactly whenu−vu\-vis a nonzero square in𝔽q\\mathbb\{F\}\_\{q\}; this is well defined because−1\-1is a square in𝔽q\\mathbb\{F\}\_\{q\}\. Paley graphs are the standard supply of explicit graphs with prescribed adjacency properties\.[Blass et al\. 1981](https://arxiv.org/html/2608.16977#biba.bib8)showed thatGp∈𝒢⁡\(n,n,1\)G\_\{p\}\\in\\mathcal\{G\}\(n,n,1\)for every primep≡1\(mod4\)p\\equiv 1\\pmod\{4\}withp\>n2​24​np\>n^\{2\}2^\{4n\}, and[Ananchuen & Caccetta 1993](https://arxiv.org/html/2608.16977#biba.bib4)obtained thresholds for the full range of parameters by estimating the relevant character sums, among them

q\>\(1\+2​2​k\)2⟹Gq∈𝒢⁡\(1,2,k\)\.q\>\\bigl\(1\+2\\sqrt\{2k\}\\bigr\)^\{2\}\\quad\\Longrightarrow\\quad G\_\{q\}\\in\\mathcal\{G\}\(1,2,k\)\.This same threshold reappears in[Ananchuen & Caccetta 1995](https://arxiv.org/html/2608.16977#biba.bib1), where it is deduced, together with the companion conclusionGq∈𝒢⁡\(2,1,k\)G\_\{q\}\\in\\mathcal\{G\}\(2,1,k\), from thenn\-parameter implication

q\>\{\(n−2\)​2n\+2\}​q\+\(n\+2​k−1\)​2n−2​n−1⟹Gq∈𝒢⁡\(1,n,k\)∩𝒢⁡\(n,1,k\),q\>\\bigl\\\{\(n\-2\)2^\{n\}\+2\\bigr\\\}\\sqrt\{q\}\+\(n\+2k\-1\)2^\{n\}\-2n\-1\\quad\\Longrightarrow\\quad G\_\{q\}\\in\\mathcal\{G\}\(1,n,k\)\\cap\\mathcal\{G\}\(n,1,k\),which atn=2n=2reduces to the threshold above\.

Forn=1n=1thenn\-parameter threshold is exactly right\. Indeed it then readsq\>4​k−3q\>4k\-3, which forq≡1\(mod4\)q\\equiv 1\\pmod\{4\}meansq≥4​k\+1q\\geq 4k\+1, while[Exoo 1981](https://arxiv.org/html/2608.16977#biba.bib10)proved that every graph in𝒢⁡\(1,1,k\)\\mathcal\{G\}\(1,1,k\)has at least4​k\+14k\+1vertices; henceGq∈𝒢⁡\(1,1,k\)G\_\{q\}\\in\\mathcal\{G\}\(1,1,k\)if and only ifq≥4​k\+1q\\geq 4k\+1\([Ananchuen & Caccetta 1995](https://arxiv.org/html/2608.16977#biba.bib1)\)\.[Ananchuen & Caccetta 1995](https://arxiv.org/html/2608.16977#biba.bib1)leave the analogous exactness forn=2n=2as a conjecture, writing:

*We have verified, by computer, that ifq≡1\(mod4\)q\\equiv 1\\pmod\{4\}is a prime power less than or equal to10091009andkkis a positive integer withq<\(1\+2​2​k\)2q<\\bigl\(1\+2\\sqrt\{2k\}\\bigr\)^\{2\}, thenGq∉𝒢⁡\(1,2,k\)G\_\{q\}\\notin\\mathcal\{G\}\(1,2,k\)\. We conjecture that this is true for allqq\.*

It is convenient to restate the conjecture as a single inequality\. The Paley graph is vertex\-transitive under translation, so in testing propertyP⁡\(1,2,k\)P\(1,2,k\)we may always takeA=\{0\}A=\\\{0\\\}\. Writingη\\etafor the quadratic character of𝔽q\\mathbb\{F\}\_\{q\}, extended byη⁡\(0\)=0\\eta\(0\)=0, the number of vertices to be counted forB=\{b,c\}B=\\\{b,c\\\}is

N\(b,c\)=\#\{x∈𝔽q∖\{0,b,c\}:η\(x\)=1,η\(x−b\)=η\(x−c\)=−1\}\.N\(b,c\)=\\\#\\bigl\\\{x\\in\\mathbb\{F\}\_\{q\}\\setminus\\\{0,b,c\\\}:\\eta\(x\)=1,\\ \\eta\(x\-b\)=\\eta\(x\-c\)=\-1\\bigr\\\}\.\(2\)ThusGq∈𝒢⁡\(1,2,k\)G\_\{q\}\\in\\mathcal\{G\}\(1,2,k\)if and only ifk≤Nmin​\(q\)k\\leq N\_\{\\min\}\(q\), where

Nmin\(q\)=min\{N\(b,c\):b,c∈𝔽q∗,b≠c\}\.N\_\{\\min\}\(q\)=\\min\\bigl\\\{N\(b,c\):b,c\\in\\mathbb\{F\}\_\{q\}^\{\*\},\\ b\\neq c\\bigr\\\}\.Sinceq<\(1\+2​2​k\)2q<\(1\+2\\sqrt\{2k\}\)^\{2\}is equivalent to8​k\>\(q−1\)28k\>\(\\sqrt\{q\}\-1\)^\{2\}, the conjecture asserts precisely that

8​Nmin​\(q\)≤\(q−1\)2for every prime power​q≡1\(mod4\)\.8\\,N\_\{\\min\}\(q\)\\leq\(\\sqrt\{q\}\-1\)^\{2\}\\qquad\\text\{for every prime power \}q\\equiv 1\\pmod\{4\}\.\(3\)Ananchuen and Caccetta go on, in the same remark, to choose the three vertices in their character\-sum estimate so that it yields an upper bound as well, concluding thatGq∉𝒢⁡\(1,2,k\)G\_\{q\}\\notin\\mathcal\{G\}\(1,2,k\)wheneverq<\(−1\+2​2​\(k\+1\)\)2q<\\bigl\(\-1\+2\\sqrt\{2\(k\+1\)\}\\bigr\)^\{2\}\. The conjecture, they note, therefore has content only in the window

\(−1\+2​2​\(k\+1\)\)2≤q≤\(1\+2​2​k\)2,\\bigl\(\-1\+2\\sqrt\{2\(k\+1\)\}\\bigr\)^\{2\}\\leq q\\leq\\bigl\(1\+2\\sqrt\{2k\}\\bigr\)^\{2\},an interval of length\(8\+o⁡\(1\)\)​2​k\(8\+o\(1\)\)\\sqrt\{2k\}\. Our counterexample lies inside it: fork=377k=377the window is2915\.01​…≤q≤3126\.83​…2915\.01\\ldots\\leq q\\leq 3126\.83\\ldots, andq=3125q=3125\.

[Ananchuen 2001](https://arxiv.org/html/2608.16977#biba.bib2)and[Ananchuen & Caccetta 2006](https://arxiv.org/html/2608.16977#biba.bib3)carry the character\-sum method of the note\([Ananchuen & Caccetta 1995](https://arxiv.org/html/2608.16977#biba.bib1)\)over to the graphs built from cubic and quartic residues, and[Bonato 2009](https://arxiv.org/html/2608.16977#biba.bib9)surveys the explicit constructions ofnn\-existentially closed graphs\. The note’s companion paper\([Australia 1994](https://arxiv.org/html/2608.16977#biba.bib6)\)takes up𝒢⁡\(1,2,1\)\\mathcal\{G\}\(1,2,1\)directly, exhibiting a graph in that class of every order at least1010except1111and noting that𝒢⁡\(1,2,1\)\\mathcal\{G\}\(1,2,1\)contains no graph of any other order\. We are aware of no earlier counterexample to the conjecture and, beyond the two bounds above, of no partial result on the exactness of the threshold for𝒢⁡\(1,2,k\)\\mathcal\{G\}\(1,2,k\)\.

###### Theorem C\.1\.1\.

The Paley graphG3125G\_\{3125\}on555^\{5\}vertices belongs to𝒢⁡\(1,2,377\)\\mathcal\{G\}\(1,2,377\), while

3125<\(1\+2​2⋅377\)2\.3125<\\bigl\(1\+2\\sqrt\{2\\cdot 377\}\\bigr\)^\{2\}\.In particular equation[3](https://arxiv.org/html/2608.16977#A3.E3)fails atq=55q=5^\{5\}, so the conjecture of[Ananchuen & Caccetta 1995](https://arxiv.org/html/2608.16977#biba.bib1)is false\.

Expanding the three character conditions in equation[2](https://arxiv.org/html/2608.16977#A3.E2)turns8​N​\(b,c\)8N\(b,c\)intoq\+1\+S⁡\(b,c\)−R⁡\(b,c\)q\+1\+S\(b,c\)\-R\(b,c\), where

S⁡\(b,c\)=∑x∈𝔽qη⁡\(x⁡\(x−b\)​\(x−c\)\)S\(b,c\)=\\sum\_\{x\\in\\mathbb\{F\}\_\{q\}\}\\eta\\bigl\(x\(x\-b\)\(x\-c\)\\bigr\)is minus the trace of the elliptic curvey2=x⁡\(x−b\)​\(x−c\)y^\{2\}=x\(x\-b\)\(x\-c\)andR⁡\(b,c\)∈\{0,4,8\}R\(b,c\)\\in\\\{0,4,8\\\}is a correction coming from the three points0,b,c0,b,c\. Forq=3125q=3125Hasse’s bound givesS⁡\(b,c\)≥−111S\(b,c\)\\geq\-111and henceN⁡\(b,c\)≥376N\(b,c\)\\geq 376, one short of what is needed\. The crucial observation is that the equality caseN⁡\(b,c\)=376N\(b,c\)=376forces the curve to have trace exactly110110, a nonzero multiple of the characteristic, and over𝔽55\\mathbb\{F\}\_\{5^\{5\}\}no elliptic curve has such a trace\.

#### C\.1\.2The character count

Throughout the rest of this sectionqqis a prime power withq≡1\(mod4\)q\\equiv 1\\pmod\{4\}, andη\\etais the quadratic character of𝔽q\\mathbb\{F\}\_\{q\}, extended byη⁡\(0\)=0\\eta\(0\)=0\. Thusη⁡\(x\)=1\\eta\(x\)=1ifxxis a nonzero square,η⁡\(x\)=−1\\eta\(x\)=\-1ifxxis a nonsquare, andη⁡\(−1\)=1\\eta\(\-1\)=1\. In particular distinctu,v∈𝔽qu,v\\in\\mathbb\{F\}\_\{q\}are adjacent inGqG\_\{q\}if and only ifη⁡\(u−v\)=1\\eta\(u\-v\)=1\.

Only one character sum evaluation is needed:

∑x∈𝔽qη⁡\(\(x−u\)​\(x−v\)\)=−1\(u≠v\)\.\\sum\_\{x\\in\\mathbb\{F\}\_\{q\}\}\\eta\\bigl\(\(x\-u\)\(x\-v\)\\bigr\)=\-1\\qquad\(u\\neq v\)\.\(4\)Indeed, substitutingx=u\+\(v−u\)​yx=u\+\(v\-u\)yturns the left side into∑yη⁡\(\(v−u\)2​y​\(y−1\)\)=∑yη⁡\(y⁡\(y−1\)\)\\sum\_\{y\}\\eta\\bigl\(\(v\-u\)^\{2\}y\(y\-1\)\\bigr\)=\\sum\_\{y\}\\eta\\bigl\(y\(y\-1\)\\bigr\)\. The termy=0y=0vanishes, and fory≠0y\\neq 0we haveη⁡\(y⁡\(y−1\)\)=η⁡\(y2​\(1−y−1\)\)=η⁡\(1−y−1\)\\eta\\bigl\(y\(y\-1\)\\bigr\)=\\eta\\bigl\(y^\{2\}\(1\-y^\{\-1\}\)\\bigr\)=\\eta\(1\-y^\{\-1\}\)\. Asyyruns over𝔽q∗\\mathbb\{F\}\_\{q\}^\{\*\}the element1−y−11\-y^\{\-1\}runs over𝔽q∖\{1\}\\mathbb\{F\}\_\{q\}\\setminus\\\{1\\\}, so the sum equals∑w≠1η⁡\(w\)=−η⁡\(1\)=−1\\sum\_\{w\\neq 1\}\\eta\(w\)=\-\\eta\(1\)=\-1\.

###### Lemma C\.1\.3\.

Letb,c∈𝔽q∗b,c\\in\\mathbb\{F\}\_\{q\}^\{\*\}be distinct and letN⁡\(b,c\)N\(b,c\)be as in equation[2](https://arxiv.org/html/2608.16977#A3.E2)\. Then

8​N​\(b,c\)=q\+1\+S⁡\(b,c\)−R⁡\(b,c\),8N\(b,c\)=q\+1\+S\(b,c\)\-R\(b,c\),\(5\)where

S⁡\(b,c\)=∑x∈𝔽qη⁡\(x⁡\(x−b\)​\(x−c\)\)S\(b,c\)=\\sum\_\{x\\in\\mathbb\{F\}\_\{q\}\}\\eta\\bigl\(x\(x\-b\)\(x\-c\)\\bigr\)and

R⁡\(b,c\)=\(1−η⁡\(b\)\)​\(1−η⁡\(c\)\)\+\(1\+η⁡\(b\)\)​\(1−η⁡\(b−c\)\)\+\(1\+η⁡\(c\)\)​\(1−η⁡\(b−c\)\)\.R\(b,c\)=\\bigl\(1\-\\eta\(b\)\\bigr\)\\bigl\(1\-\\eta\(c\)\\bigr\)\+\\bigl\(1\+\\eta\(b\)\\bigr\)\\bigl\(1\-\\eta\(b\-c\)\\bigr\)\+\\bigl\(1\+\\eta\(c\)\\bigr\)\\bigl\(1\-\\eta\(b\-c\)\\bigr\)\.MoreoverR⁡\(b,c\)∈\{0,4,8\}R\(b,c\)\\in\\\{0,4,8\\\}\.

###### Proof\.

Put

T=∑x∈𝔽q\(1\+η⁡\(x\)\)​\(1−η⁡\(x−b\)\)​\(1−η⁡\(x−c\)\)\.T=\\sum\_\{x\\in\\mathbb\{F\}\_\{q\}\}\\bigl\(1\+\\eta\(x\)\\bigr\)\\bigl\(1\-\\eta\(x\-b\)\\bigr\)\\bigl\(1\-\\eta\(x\-c\)\\bigr\)\.Forx∉\{0,b,c\}x\\notin\\\{0,b,c\\\}each ofη⁡\(x\)\\eta\(x\),η⁡\(x−b\)\\eta\(x\-b\),η⁡\(x−c\)\\eta\(x\-c\)is±1\\pm 1, so the summand equals88whenxxis one of the points counted byN⁡\(b,c\)N\(b,c\)and equals00otherwise\. The three excluded points contribute

\(1−η⁡\(−b\)\)​\(1−η⁡\(−c\)\),\(1\+η⁡\(b\)\)​\(1−η⁡\(b−c\)\),\(1\+η⁡\(c\)\)​\(1−η⁡\(c−b\)\)\\bigl\(1\-\\eta\(\-b\)\\bigr\)\\bigl\(1\-\\eta\(\-c\)\\bigr\),\\qquad\\bigl\(1\+\\eta\(b\)\\bigr\)\\bigl\(1\-\\eta\(b\-c\)\\bigr\),\\qquad\\bigl\(1\+\\eta\(c\)\\bigr\)\\bigl\(1\-\\eta\(c\-b\)\\bigr\)respectively, and these sum toR⁡\(b,c\)R\(b,c\)becauseη⁡\(−u\)=η⁡\(u\)\\eta\(\-u\)=\\eta\(u\)\. Hence

T=8​N​\(b,c\)\+R⁡\(b,c\)\.T=8N\(b,c\)\+R\(b,c\)\.
On the other hand, expanding the product gives

T=∑x1\\displaystyle T=\\sum\_\{x\}1\+∑xη\(x\)−∑xη\(x−b\)−∑xη\(x−c\)\\displaystyle\+\\sum\_\{x\}\\eta\(x\)\-\\sum\_\{x\}\\eta\(x\-b\)\-\\sum\_\{x\}\\eta\(x\-c\)\+∑xη\(\(x−b\)\(x−c\)\)−∑xη\(x\(x−b\)\)−∑xη\(x\(x−c\)\)\+S\(b,c\),\\displaystyle\+\\sum\_\{x\}\\eta\\bigl\(\(x\-b\)\(x\-c\)\\bigr\)\-\\sum\_\{x\}\\eta\\bigl\(x\(x\-b\)\\bigr\)\-\\sum\_\{x\}\\eta\\bigl\(x\(x\-c\)\\bigr\)\+S\(b,c\),all sums being overx∈𝔽qx\\in\\mathbb\{F\}\_\{q\}\. The first sum isqq, and the three single\-character sums vanish becauseη\\etais nonprincipal\. By equation[4](https://arxiv.org/html/2608.16977#A3.E4)each of the three quadratic sums equals−1\-1, so together they contribute−1\+1\+1=1\-1\+1\+1=1\. ThereforeT=q\+1\+S⁡\(b,c\)T=q\+1\+S\(b,c\), and comparing the two expressions forTTgives equation[5](https://arxiv.org/html/2608.16977#A3.E5)\.

It remains to determine the possible values ofR⁡\(b,c\)R\(b,c\)\. Writeα=η⁡\(b\)\\alpha=\\eta\(b\),β=η⁡\(c\)\\beta=\\eta\(c\)andγ=η⁡\(b−c\)\\gamma=\\eta\(b\-c\), all of which lie in\{±1\}\\\{\\pm 1\\\}, so that

R⁡\(b,c\)=\(1−α\)​\(1−β\)\+\(1\+α\)​\(1−γ\)\+\(1\+β\)​\(1−γ\)\.R\(b,c\)=\(1\-\\alpha\)\(1\-\\beta\)\+\(1\+\\alpha\)\(1\-\\gamma\)\+\(1\+\\beta\)\(1\-\\gamma\)\.Ifγ=1\\gamma=1thenR⁡\(b,c\)=\(1−α\)​\(1−β\)R\(b,c\)=\(1\-\\alpha\)\(1\-\\beta\), which is44whenα=β=−1\\alpha=\\beta=\-1and00otherwise\. Ifγ=−1\\gamma=\-1thenR⁡\(b,c\)=\(1−α\)​\(1−β\)\+2​\(1\+α\)\+2​\(1\+β\)R\(b,c\)=\(1\-\\alpha\)\(1\-\\beta\)\+2\(1\+\\alpha\)\+2\(1\+\\beta\), which is88whenα=β=1\\alpha=\\beta=1and44in the three remaining cases\. This exhausts all possibilities\. ∎

#### C\.1\.3Traces of elliptic curves

Two facts about elliptic curves over finite fields are needed\. The first is Hasse’s bound\([Silverman 2009](https://arxiv.org/html/2608.16977#biba.bib12), Chapter V, Theorem 1\.1\): ifEEis an elliptic curve over𝔽q\\mathbb\{F\}\_\{q\}, then its tracet=q\+1−\#​E​\(𝔽q\)t=q\+1\-\\\#E\(\\mathbb\{F\}\_\{q\}\)satisfies\|t\|≤2​q\|t\|\\leq 2\\sqrt\{q\}\. The second is the classification of the integers that occur as traces, due to[Waterhouse 1969](https://arxiv.org/html/2608.16977#biba.bib13)\.

###### Theorem C\.1\.4\.

Letppbe a prime, letr≥1r\\geq 1, and letttbe an integer with\|t\|≤2​pr/2\|t\|\\leq 2p^\{r/2\}\. There is an elliptic curve over𝔽pr\\mathbb\{F\}\_\{p^\{r\}\}of tracettif and only if at least one of the following holds:

1. 1\.gcd⁡\(t,p\)=1\\gcd\(t,p\)=1;
2. 2\.rris even andt=±2​pr/2t=\\pm 2p^\{r/2\};
3. 3\.rris even,p≢1\(mod3\)p\\not\\equiv 1\\pmod\{3\}andt=±pr/2t=\\pm p^\{r/2\};
4. 4\.rris odd,p∈\{2,3\}p\\in\\\{2,3\\\}andt=±p\(r\+1\)/2t=\\pm p^\{\(r\+1\)/2\};
5. 5\.t=0t=0, and eitherrris odd orrris even withp≢1\(mod4\)p\\not\\equiv 1\\pmod\{4\}\.

The same classification is restated by[Schoof 1987](https://arxiv.org/html/2608.16977#biba.bib11), where it is the starting point for counting the isomorphism classes of elliptic curves in a fixed isogeny class\. We use it only through the following consequence\.

###### Corollary C\.1\.5\.

Letp\>3p\>3be a prime and letrrbe odd\. IfEEis an elliptic curve over𝔽pr\\mathbb\{F\}\_\{p^\{r\}\}whose tracettis divisible bypp, thent=0t=0\.

###### Proof\.

Of the five cases of Theorem[C\.1\.4](https://arxiv.org/html/2608.16977#A3.SS1.Thmtheorem4), the first is excluded byp\|tp\\mid t, the second and third byrrbeing odd, and the fourth byp\>3p\>3\. Only the fifth remains, and it givest=0t=0\. ∎

#### C\.1\.4Proof of the main theorem

###### Proof of Theorem[C\.1\.1](https://arxiv.org/html/2608.16977#A3.SS1.Thmtheorem1)\.

Setq=55=3125q=5^\{5\}=3125\. By vertex\-transitivity it suffices to prove thatN⁡\(b,c\)≥377N\(b,c\)\\geq 377for every pair of distinctb,c∈𝔽q∗b,c\\in\\mathbb\{F\}\_\{q\}^\{\*\}, sinceN⁡\(b,c\)N\(b,c\)counts exactly the vertices adjacent to00and to neitherbbnorcc\.

Fix such a pair and consider

Eb,c:y2=x⁡\(x−b\)​\(x−c\)\.E\_\{b,c\}:\\quad y^\{2\}=x\(x\-b\)\(x\-c\)\.The cubicx​\(x−b\)​\(x−c\)x\(x\-b\)\(x\-c\)has three distinct roots andchar⁡𝔽q=5≠2\\operatorname\{char\}\\mathbb\{F\}\_\{q\}=5\\neq 2, soEb,cE\_\{b,c\}is nonsingular, that is, an elliptic curve\. For eachx∈𝔽qx\\in\\mathbb\{F\}\_\{q\}the number ofy∈𝔽qy\\in\\mathbb\{F\}\_\{q\}withy2=x⁡\(x−b\)​\(x−c\)y^\{2\}=x\(x\-b\)\(x\-c\)is1\+η⁡\(x⁡\(x−b\)​\(x−c\)\)1\+\\eta\\bigl\(x\(x\-b\)\(x\-c\)\\bigr\), so summing overxxand adding the point at infinity gives

\#​Eb,c​\(𝔽q\)=q\+1\+S⁡\(b,c\)\.\\\#E\_\{b,c\}\(\\mathbb\{F\}\_\{q\}\)=q\+1\+S\(b,c\)\.Hence the trace ofEb,cE\_\{b,c\}ist=q\+1−\#​Eb,c​\(𝔽q\)=−S⁡\(b,c\)t=q\+1\-\\\#E\_\{b,c\}\(\\mathbb\{F\}\_\{q\}\)=\-S\(b,c\), and Hasse’s bound gives

\|S⁡\(b,c\)\|≤2​q=50​5<112,\|S\(b,c\)\|\\leq 2\\sqrt\{q\}=50\\sqrt\{5\}<112,so thatS⁡\(b,c\)≥−111S\(b,c\)\\geq\-111\. Combining this with Lemma[C\.1\.3](https://arxiv.org/html/2608.16977#A3.SS1.Thmtheorem3)andR⁡\(b,c\)≤8R\(b,c\)\\leq 8yields

8​N​\(b,c\)=q\+1\+S⁡\(b,c\)−R⁡\(b,c\)≥3126−111−8=3007,8N\(b,c\)=q\+1\+S\(b,c\)\-R\(b,c\)\\geq 3126\-111\-8=3007,and thereforeN⁡\(b,c\)≥376N\(b,c\)\\geq 376\.

It remains to exclude the equality case\. SupposeN⁡\(b,c\)=376N\(b,c\)=376\. Then equation[5](https://arxiv.org/html/2608.16977#A3.E5)reads

3008=3126\+S⁡\(b,c\)−R⁡\(b,c\),3008=3126\+S\(b,c\)\-R\(b,c\),soS⁡\(b,c\)=R⁡\(b,c\)−118S\(b,c\)=R\(b,c\)\-118\. AsR⁡\(b,c\)∈\{0,4,8\}R\(b,c\)\\in\\\{0,4,8\\\}this forcesS⁡\(b,c\)∈\{−118,−114,−110\}S\(b,c\)\\in\\\{\-118,\-114,\-110\\\}, and the boundS⁡\(b,c\)≥−111S\(b,c\)\\geq\-111leaves only

R⁡\(b,c\)=8,S⁡\(b,c\)=−110\.R\(b,c\)=8,\\qquad S\(b,c\)=\-110\.The curveEb,cE\_\{b,c\}would then have tracet=−S⁡\(b,c\)=110t=\-S\(b,c\)=110\. But110110is a nonzero multiple of55, andq=55q=5^\{5\}hasp=5\>3p=5\>3withr=5r=5odd, so Corollary[C\.1\.5](https://arxiv.org/html/2608.16977#A3.SS1.Thmtheorem5)rules this out\. ThereforeN⁡\(b,c\)≠376N\(b,c\)\\neq 376, and the preceding bound givesN⁡\(b,c\)≥377N\(b,c\)\\geq 377for every pair of distinctb,c∈𝔽q∗b,c\\in\\mathbb\{F\}\_\{q\}^\{\*\}\. HenceG3125∈𝒢⁡\(1,2,377\)G\_\{3125\}\\in\\mathcal\{G\}\(1,2,377\)\.

Finally,

\(1\+2​2⋅377\)2=\(1\+2​754\)2=3017\+4​754\>3017\+108=3125,\\bigl\(1\+2\\sqrt\{2\\cdot 377\}\\bigr\)^\{2\}=\\bigl\(1\+2\\sqrt\{754\}\\bigr\)^\{2\}=3017\+4\\sqrt\{754\}\>3017\+108=3125,because272=729<75427^\{2\}=729<754\. This completes the proof\. ∎

## References\.

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### C\.2A 4\-uniform tree that is not 5\-good

This problem asks whetherrr\-uniform trees are optimal in the Ramsey problem against a completerr\-uniform hypergraph, as ordinary trees are against a clique\. They are not\. The unique44\-uniform tree on seven vertices fails to be55\-good, and the obstruction is a red/blue coloring ofK8\(4\)K\_\{8\}^\{\(4\)\}whose red edges are exactly the fourteen affine planes of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}\.

#### C\.2\.1Introduction

All hypergraphs here are finite and simple\. For anrr\-uniform hypergraphHHwe writeV⁡\(H\)V\(H\)andE⁡\(H\)E\(H\)for its vertex and edge sets, andKn\(r\)K\_\{n\}^\{\(r\)\}for the completerr\-uniform hypergraph onnnvertices\. The Ramsey numberR⁡\(G,H,r\)R\(G,H;r\)is the leastppsuch that every red/blue coloring ofE⁡\(Kp\(r\)\)E\(K\_\{p\}^\{\(r\)\}\)contains a red copy ofGGor a blue copy ofHH\. A*Berge cycle*of lengthk≥2k\\geq 2inHHis an alternating sequencev1,e1,v2,e2,…,vk,ekv\_\{1\},e\_\{1\},v\_\{2\},e\_\{2\},\\dots,v\_\{k\},e\_\{k\}of distinct verticesviv\_\{i\}and distinct edgeseie\_\{i\}withvi,vi\+1∈eiv\_\{i\},v\_\{i\+1\}\\in e\_\{i\}for everyii, indices read modulokk\. An*rr\-uniform tree*is a connectedrr\-uniform hypergraph containing no Berge cycle\. Equivalently\([Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4), Theorem 2\.1\), it is a hypergraph that can be assembled starting from a single edge, each subsequent edge meeting the union of the previous ones in exactly one vertex; in particular anrr\-uniform tree onmmvertices has exactly\(m−1\)/\(r−1\)\(m\-1\)/\(r\-1\)edges\.

A*weak coloring*ofHHis a coloring ofV⁡\(H\)V\(H\)in which no edge is monochromatic, the*weak chromatic number*χw​\(H\)\\chi\_\{w\}\(H\)is the least number of colors in a weak coloring, and the*chromatic surplus*s⁡\(H\)s\(H\)is the least size of a color class over all weak colorings ofHHwith exactlyχw​\(H\)\\chi\_\{w\}\(H\)colors\.[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)proved that every connectedrr\-uniform hypergraphGGof orderm≥rm\\geq rsatisfies

R⁡\(G,Kn\(r\),r\)≥\(m−1\)​\(χw​\(Kn\(r\)\)−1\)\+s⁡\(Kn\(r\)\),R\\bigl\(G,K\_\{n\}^\{\(r\)\};r\\bigr\)\\ \\geq\\ \(m\-1\)\\bigl\(\\chi\_\{w\}\(K\_\{n\}^\{\(r\)\}\)\-1\\bigr\)\+s\\bigl\(K\_\{n\}^\{\(r\)\}\\bigr\),\(6\)and they callGG*nn\-good*when equality holds; their notation for the chromatic surplus ist⁡\(⋅\)t\(\\cdot\), and we follow the laters⁡\(⋅\)s\(\\cdot\)of[Budden & Clifton 2022](https://arxiv.org/html/2608.16977#bibb.bib3)\. For an arbitraryrr\-uniform targetHHwiths⁡\(H\)≤ms\(H\)\\leq mthe same theorem givesR⁡\(G,H,r\)≥\(m−1\)​\(χw​\(H\)−1\)\+s⁡\(H\)R\(G,H;r\)\\geq\(m\-1\)\(\\chi\_\{w\}\(H\)\-1\)\+s\(H\), andGGis*HH\-good*when equality holds;nn\-goodness is the caseH=Kn\(r\)H=K\_\{n\}^\{\(r\)\}\. Hereχw​\(Kn\(r\)\)=⌈n/\(r−1\)⌉\\chi\_\{w\}\(K\_\{n\}^\{\(r\)\}\)=\\lceil n/\(r\-1\)\\rceil, since a color class of a weak coloring ofKn\(r\)K\_\{n\}^\{\(r\)\}is exactly a set of at mostr−1r\-1vertices\. Forr=2r=2, equation[6](https://arxiv.org/html/2608.16977#A3.E6)readsR⁡\(G,Kn\)≥\(m−1\)​\(n−1\)\+1R\(G,K\_\{n\}\)\\geq\(m\-1\)\(n\-1\)\+1, and[Chvátal 1977](https://arxiv.org/html/2608.16977#bibb.bib8)proved that every tree attains it: every treeTmT\_\{m\}onmmvertices satisfiesR⁡\(Tm,Kn\)=\(m−1\)​\(n−1\)\+1R\(T\_\{m\},K\_\{n\}\)=\(m\-1\)\(n\-1\)\+1\. The termnn\-good goes back to[Burr et al\. 1980](https://arxiv.org/html/2608.16977#bibb.bib7), and the systematic study of goodness to[Burr & Erdös 1983](https://arxiv.org/html/2608.16977#bibb.bib6)\.

[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)conjectured that hypertrees are optimal in the same sense:

*Ifr≥2r\\geq 2andTTis anyrr\-uniform tree, thenTTisnn\-good*\[for everyn≥rn\\geq r\]\.

We disprove this\.

The evidence behind the conjecture is substantial\.[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)record the hypertree embedding bound of[Loh 2009](https://arxiv.org/html/2608.16977#bibb.bib9), namely

R⁡\(Tm\(r\),Kn\(r\),r\)≤\(m−1\)​\(n−1\)r−1\+1R\\bigl\(T\_\{m\}^\{\(r\)\},K\_\{n\}^\{\(r\)\};r\\bigr\)\\ \\leq\\ \\frac\{\(m\-1\)\(n\-1\)\}\{r\-1\}\+1for everyrr\-uniform treeTm\(r\)T\_\{m\}^\{\(r\)\}onmmvertices, and deduce that the conjecture holds whenever\(r−1\)\|\(n−1\)\(r\-1\)\\mid\(n\-1\), because the two bounds then agree\. Forr=3r=3the divisibility condition says exactly thatnnis odd, and for the smallest nontrivial tree[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)push past it:R⁡\(T5\(3\),Kn\(3\),3\)=2​n−2R\(T\_\{5\}^\{\(3\)\},K\_\{n\}^\{\(3\)\};3\)=2n\-2for evennnas well, soT5\(3\)T\_\{5\}^\{\(3\)\}isnn\-good for everyn≥3n\\geq 3\. BeyondT5\(3\)T\_\{5\}^\{\(3\)\}their bounds for evennnleave a range of possible values rather than a single one, and they report finding no counterexample\.[Budden & Clifton 2022](https://arxiv.org/html/2608.16977#bibb.bib3)reduce the conjecture to the residuesn≡0\(modr−1\)n\\equiv 0\\pmod\{r\-1\}: a tree that isnn\-good for every multiplennofr−1r\-1isnn\-good for everyn≥rn\\geq r\. In the asymptotic direction,[Boyadzhiyska & Lo 2025](https://arxiv.org/html/2608.16977#bibb.bib2)prove that for everyr≥3r\\geq 3and everyrr\-uniform hypergraphHHwiths⁡\(H\)≤2​r−1s\(H\)\\leq 2r\-1, all sufficiently longrr\-uniform loose paths areHH\-good, the threshold depending onHH\. Every complete target satisfies the hypothesis, sinces⁡\(Kn\(r\)\)≤r−1s\(K\_\{n\}^\{\(r\)\}\)\\leq r\-1, so for each fixednnall sufficiently long loose paths arenn\-good\. A loose path that is notnn\-good is therefore short, and ours is the shortest nontrivial one\.

Connected hypergraphs that are notnn\-good were known, but no tree was among them;[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)show that the33\-uniform loose cycleC4\(3\)C\_\{4\}^\{\(3\)\}, two edges meeting in two vertices, is not55\-good, and record thatK4\(3\)K\_\{4\}^\{\(3\)\},K5\(3\)K\_\{5\}^\{\(3\)\}andK6\(3\)K\_\{6\}^\{\(3\)\}are not44\-good and thatK5\(4\)K\_\{5\}^\{\(4\)\}is not55\-good\. Two nearby negative results are of a different kind\. Theℓ\\ell\-paths withℓ≥2\\ell\\geq 2that[Boyadzhiyska & Lo 2025](https://arxiv.org/html/2608.16977#bibb.bib2)rule out are not trees, since consecutive edges meet inℓ≥2\\ell\\geq 2vertices and hence span a Berge cycle of length22\. The loose paths they rule out are trees, but the targets there are incomplete hypergraphsHHwiths⁡\(H\)\>2​r−1s\(H\)\>2r\-1, outside the hypothesis above\. Radziszowski’s dynamic survey describes the subsequent literature as “further results towards the conjecture that allrr\-uniform trees arenn\-good”\([Radziszowski 1994](https://arxiv.org/html/2608.16977#bibb.bib10)\)\. \(A corrigendum\([Budden & Penland 2019](https://arxiv.org/html/2608.16977#bibb.bib5)\)amends Theorem 3\.10 of the original paper, on disjoint unions ofnn\-good hypergraphs; Conjecture 6\.2 is unaffected\.\)

The counterexample is the smallest44\-uniform tree other than a single edge, equivalently the44\-uniform loose path with two edges, and it is taken againstK5\(4\)K\_\{5\}^\{\(4\)\}, a pair of parameters that the divisibility criterion does not cover\. LetTTbe the44\-uniform hypergraph on the vertex set\[7\]\[7\]with the two edges

e1=\{1,2,3,4\},e2=\{4,5,6,7\}\.e\_\{1\}=\\\{1,2,3,4\\\},\\qquad e\_\{2\}=\\\{4,5,6,7\\\}\.\(7\)It is connected, and it has no Berge cycle: a Berge cycle uses at least two distinct edges, so with onlye1e\_\{1\}ande2e\_\{2\}available it would need two distinct vertices ine1∩e2e\_\{1\}\\cap e\_\{2\}, whereas\|e1∩e2\|=1\|e\_\{1\}\\cap e\_\{2\}\|=1\. HenceTTis a44\-uniform tree, and it is the unique one of order77up to isomorphism, since a44\-uniform tree on77vertices has exactly\(7−1\)/\(4−1\)=2\(7\-1\)/\(4\-1\)=2edges and those edges meet in exactly one vertex\.

###### Theorem C\.2\.1\.

LetTTbe the44\-uniform tree of equation[7](https://arxiv.org/html/2608.16977#A3.E7)\. Then

R⁡\(T,K5\(4\),4\)=9,R\\bigl\(T,K\_\{5\}^\{\(4\)\};4\\bigr\)=9,while the right\-hand side of equation[6](https://arxiv.org/html/2608.16977#A3.E6)equals88forG=TG=T,r=4r=4andn=5n=5\. In particularTTis not55\-good, and Conjecture 6\.2 of[Budden & Penland 2017](https://arxiv.org/html/2608.16977#bibb.bib4)is false\.

The lower bound is a single explicit coloring\. The key point is that the affine planes of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}, regarded as44\-element subsets, meet one another in00or22points and never in11, while the two edges ofTTmeet in exactly11point; at the same time the planes are numerous enough that every55points contain one\. The matching upper bound is Loh’s hypertree embedding theorem, which at\(r,m,n\)=\(4,7,5\)\(r,m,n\)=\(4,7,5\)gives exactly99\.

#### C\.2\.2The affine\-plane coloring

Take the vertex set ofK8\(4\)K\_\{8\}^\{\(4\)\}to beV=𝔽23V=\\mathbb\{F\}\_\{2\}^\{3\}\. Fora∈𝔽23∖\{0\}a\\in\\mathbb\{F\}\_\{2\}^\{3\}\\setminus\\\{0\\\}andb∈𝔽2b\\in\\mathbb\{F\}\_\{2\}put

P⁡\(a,b\):=\{x∈𝔽23:a⋅x=b\},P\(a,b\):=\\\{x\\in\\mathbb\{F\}\_\{2\}^\{3\}:\\ a\\cdot x=b\\\},wherea⋅xa\\cdot xdenotes the standard bilinear form; such a set is an*affine plane*, that is, a coset of a two\-dimensional subspace\. Color a44\-subset ofVVred if it is an affine plane, and blue otherwise\. Figure[7](https://arxiv.org/html/2608.16977#A3.F7)depicts the two intersection patterns at issue\.

11223344556677e1e\_\{1\}e2e\_\{2\}the treeTT000000100100010010110110001001101101011011111111P⁡\(001,0\)P\(001,0\)andP⁡\(010,0\)P\(010,0\)Figure 7:On the left, the unique44\-uniform treeTTof order77: its two edges meet in the single vertex44\. On the right, two of the fourteen affine planes of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}, drawn as faces of the cube; they meet in the two points000000and100100, marked by the thick segment\. Lemma[C\.2\.4](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem4)says that two red edges can never meet the waye1e\_\{1\}ande2e\_\{2\}do\.###### Lemma C\.2\.3\.

There are exactly1414affine planes in𝔽23\\mathbb\{F\}\_\{2\}^\{3\}, and a44\-subsetS⊆𝔽23S\\subseteq\\mathbb\{F\}\_\{2\}^\{3\}is an affine plane if and only if∑x∈Sx=0\\sum\_\{x\\in S\}x=0\.

###### Proof\.

The mapa↦P⁡\(a,0\)a\\mapsto P\(a,0\)is a bijection from𝔽23∖\{0\}\\mathbb\{F\}\_\{2\}^\{3\}\\setminus\\\{0\\\}onto the set of two\-dimensional subspaces of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}, and each such subspace has exactly the two cosetsP⁡\(a,0\)P\(a,0\)andP⁡\(a,1\)P\(a,1\)\. Hence there are exactly1414affine planes, each of size44\.

A two\-dimensional subspace has the form\{0,p,q,p\+q\}\\\{0,p,q,p\+q\\\}and therefore sums to00, and translating a44\-set byxxchanges its sum by4​x=04x=0; so every affine plane sums to00\. Conversely, letS=\{x1,x2,x3,x4\}S=\\\{x\_\{1\},x\_\{2\},x\_\{3\},x\_\{4\}\\\}satisfy∑ixi=0\\sum\_\{i\}x\_\{i\}=0\. ThenS\+x4=\{x1\+x4,x2\+x4,x3\+x4,0\}S\+x\_\{4\}=\\\{x\_\{1\}\+x\_\{4\},\\,x\_\{2\}\+x\_\{4\},\\,x\_\{3\}\+x\_\{4\},\\,0\\\}consists of00together with three distinct nonzero elementsp,q,up,q,uwhose sum is00\. Thusu=p\+qu=p\+q, andp,qp,qare distinct nonzero vectors and hence linearly independent, soS\+x4=\{0,p,q,p\+q\}S\+x\_\{4\}=\\\{0,p,q,p\+q\\\}is a two\-dimensional subspace\. ThereforeSSis one of its cosets\. ∎

#### C\.2\.3The lower bound

We now check that the coloring contains neither of the two forbidden configurations\.

###### Lemma C\.2\.4\.

Any two distinct red edges meet in00or22vertices\. Consequently the coloring contains no red copy ofTT\.

###### Proof\.

LetP⁡\(a,b\)≠P⁡\(c,d\)P\(a,b\)\\neq P\(c,d\)be red edges\. Ifa=ca=cthenb≠db\\neq d, and the two sets are disjoint\. Ifa≠ca\\neq cthenaaandccare distinct nonzero vectors, hence linearly independent over𝔽2\\mathbb\{F\}\_\{2\}, so the linear mapx↦\(a⋅x,c⋅x\)x\\mapsto\(a\\cdot x,\\,c\\cdot x\)from𝔽23\\mathbb\{F\}\_\{2\}^\{3\}to𝔽22\\mathbb\{F\}\_\{2\}^\{2\}is surjective with kernel of dimension11\. Every one of its fibers, in particularP⁡\(a,b\)∩P⁡\(c,d\)P\(a,b\)\\cap P\(c,d\), therefore has exactly22elements\.

A red copy ofTTwould consist of two red edges meeting in exactly one vertex, which the above excludes\. ∎

###### Lemma C\.2\.5\.

Every55\-subset of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}contains a red44\-subset\. Consequently the coloring contains no blueK5\(4\)K\_\{5\}^\{\(4\)\}\.

###### Proof\.

LetX⊆𝔽23X\\subseteq\\mathbb\{F\}\_\{2\}^\{3\}with\|X\|=5\|X\|=5, writeY=𝔽23∖X=\{u,v,w\}Y=\\mathbb\{F\}\_\{2\}^\{3\}\\setminus X=\\\{u,v,w\\\}, and putσ=u\+v\+w\\sigma=u\+v\+w\. Ifσ∈Y\\sigma\\in Y, sayσ=u\\sigma=u, thenv\+w=0v\+w=0and hencev=wv=w, a contradiction; soσ∈X\\sigma\\in X\. In each coordinate exactly four of the eight vectors of𝔽23\\mathbb\{F\}\_\{2\}^\{3\}have entry11, so∑x∈𝔽23x=0\\sum\_\{x\\in\\mathbb\{F\}\_\{2\}^\{3\}\}x=0and therefore

∑x∈Xx=∑y∈Yy=σ\.\\sum\_\{x\\in X\}x=\\sum\_\{y\\in Y\}y=\\sigma\.It follows that∑x∈X∖\{σ\}x=0\\sum\_\{x\\in X\\setminus\\\{\\sigma\\\}\}x=0, soX∖\{σ\}X\\setminus\\\{\\sigma\\\}is red by Lemma[C\.2\.3](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem3)\. A blueK5\(4\)K\_\{5\}^\{\(4\)\}would be a55\-set all of whose44\-subsets are blue, which the above excludes\. ∎

Lemmas[C\.2\.4](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem4)and[C\.2\.5](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem5)exhibit a red/blue coloring ofK8\(4\)K\_\{8\}^\{\(4\)\}with no redTTand no blueK5\(4\)K\_\{5\}^\{\(4\)\}, so

R⁡\(T,K5\(4\),4\)≥9\.R\\bigl\(T,K\_\{5\}^\{\(4\)\};4\\bigr\)\\ \\geq\\ 9\.\(8\)

#### C\.2\.4Proof of the main theorem

The remaining input is the hypertree embedding theorem of[Loh 2009](https://arxiv.org/html/2608.16977#bibb.bib9), which answered a question of[Bohman et al\. 2010](https://arxiv.org/html/2608.16977#bibb.bib1)by removing all dependence on the uniformityrr\.

###### Theorem C\.2\.6\.

Everyrr\-uniform hypergraph with weak chromatic number greater thankkcontains a copy of everyrr\-uniform tree withkkedges\.

###### Proof of Theorem[C\.2\.1](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem1)\.

For the upper bound, consider any red/blue coloring ofE⁡\(K9\(4\)\)E\(K\_\{9\}^\{\(4\)\}\)and letHRH\_\{R\}be the spanning subhypergraph formed by the red edges\. The treeTThas exactly22edges\. Ifχw​\(HR\)\>2\\chi\_\{w\}\(H\_\{R\}\)\>2, then Theorem[C\.2\.6](https://arxiv.org/html/2608.16977#A3.SS2.Thmtheorem6)produces a red copy ofTT\. Otherwise fix a weak coloring ofHRH\_\{R\}using at most two colors\. Some color classCCsatisfies\|C\|≥⌈9/2⌉=5\|C\|\\geq\\lceil 9/2\\rceil=5, and no red edge lies inside a color class, so all44\-subsets of any five vertices ofCCare blue\. That is a blueK5\(4\)K\_\{5\}^\{\(4\)\}\. HenceR⁡\(T,K5\(4\),4\)≤9R\(T,K\_\{5\}^\{\(4\)\};4\)\\leq 9, and with equation[8](https://arxiv.org/html/2608.16977#A3.E8)we obtainR⁡\(T,K5\(4\),4\)=9R\(T,K\_\{5\}^\{\(4\)\};4\)=9\.

It remains to evaluate the right\-hand side of equation[6](https://arxiv.org/html/2608.16977#A3.E6)\. A weak coloring ofK5\(4\)K\_\{5\}^\{\(4\)\}is exactly a partition of a55\-set into classes of size at most33, so two classes are needed and3\+23\+2is the only partition into two such classes\. Henceχw​\(K5\(4\)\)=2\\chi\_\{w\}\(K\_\{5\}^\{\(4\)\}\)=2ands⁡\(K5\(4\)\)=2s\(K\_\{5\}^\{\(4\)\}\)=2, and since\|V⁡\(T\)\|=7\|V\(T\)\|=7the right\-hand side of equation[6](https://arxiv.org/html/2608.16977#A3.E6)equals\(7−1\)​\(2−1\)\+2=8\(7\-1\)\(2\-1\)\+2=8\. As9≠89\\neq 8, the treeTTis not55\-good\. ∎

## References\.

- Bohman et al\. \(2010\)Tom Bohman, Alan Frieze, and Dhruv Mubayi\.Coloring h\-free hypergraphs\.*Random Structures & Algorithms*, 36\(1\):11–25, 2010\.
- Boyadzhiyska & Lo \(2025\)Simona Boyadzhiyska and Allan Lo\.Ramsey goodness of k\-uniform paths, or the lack thereof\.*European Journal of Combinatorics*, 129:104021, 2025\.
- Budden & Clifton \(2022\)Mark Budden and Justin Clifton\.Hypergraph ramsey numbers involving trees, stars, and complete hypergraphs\.*Integers: Electronic Journal of Combinatorial Number Theory*, 22:1, 2022\.
- Budden & Penland \(2017\)Mark Budden and Andrew Penland\.Trees andnn\-good hypergraphs\.*arXiv preprint arXiv:1710\.05731*, 2017\.
- Budden & Penland \(2019\)Mark Budden and Andrew Penland\.Corrigendum: Trees and n\-good hypergraphs\.*Australas\. J Comb\.*, 75:171–173, 2019\.
- Burr & Erdös \(1983\)Stefan A Burr and Paul Erdös\.Generalizations of a ramsey\-theoretic result of chvátal\.*Journal of Graph Theory*, 7\(1\):39–51, 1983\.
- Burr et al\. \(1980\)Stefan A Burr, P Erdős, Ralph J Faudree, CC Rousseau, and RH Schelp\.An extremal problem in generalized ramsey theory\.*Ars Combinatoria*, 10:193–203, 1980\.
- Chvátal \(1977\)Vasek Chvátal\.Tree\-complete graph ramsey numbers\.*Journal of Graph Theory*, 1\(1\):93–93, 1977\.
- Loh \(2009\)Po\-Shen Loh\.A note on embedding hypertrees\.*arXiv preprint arXiv:0901\.2988*, 2009\.
- Radziszowski \(1994\)Stanisław P\. Radziszowski\.Small Ramsey numbers\.*The Electronic Journal of Combinatorics*, 1994\.doi:10\.37236/21\.URL[https://www\.combinatorics\.org/ojs/index\.php/eljc/article/view/DS1](https://www.combinatorics.org/ojs/index.php/eljc/article/view/DS1)\.Dynamic Survey DS1, revision 18, 24 April 2026\.

### C\.3Maximum versus average independent set size in triangle\-free graphs

This problem asks by how much the largest independent set of a triangle\-free graph exceeds a typical one\.[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)conjectured that a factor2−od​\(1\)2\-o\_\{d\}\(1\)is forced once the minimum degreeddis large\. We disprove this\. The Cartesian productsC5​□​Km,mC\_\{5\}\\,\\square\\,K\_\{m,m\}are triangle\-free and\(m\+2\)\(m\+2\)\-regular, and for them the ratio of the maximum to the average size of an independent set tends to24/13=1\.846​…24/13=1\.846\\ldots, which is bounded away from22\. The mechanism is that the two sides ofKm,mK\_\{m,m\}must draw their independent sets from disjoint parts of the pentagon\.

#### C\.3\.1Introduction

For a finite graphGGletℐ⁡\(G\)\\mathcal\{I\}\(G\)be its family of independent sets andα⁡\(G\)=maxI∈ℐ⁡\(G\)⁡\|I\|\\alpha\(G\)=\\max\_\{I\\in\\mathcal\{I\}\(G\)\}\|I\|its independence number\. The*hard\-core model*onGGat fugacityλ\>0\\lambda\>0is the probability distribution onℐ⁡\(G\)\\mathcal\{I\}\(G\)giving eachIImass proportional toλ\|I\|\\lambda^\{\|I\|\}, and

α¯G​\(λ\)=∑I∈ℐ⁡\(G\)\|I\|​λ\|I\|∑I∈ℐ⁡\(G\)λ\|I\|\\overline\{\\alpha\}\_\{G\}\(\\lambda\)=\\frac\{\\sum\_\{I\\in\\mathcal\{I\}\(G\)\}\|I\|\\,\\lambda^\{\|I\|\}\}\{\\sum\_\{I\\in\\mathcal\{I\}\(G\)\}\\lambda^\{\|I\|\}\}is the expected size of a set drawn from it\. Atλ=1\\lambda=1the distribution is uniform onℐ⁡\(G\)\\mathcal\{I\}\(G\), soα¯G​\(1\)\\overline\{\\alpha\}\_\{G\}\(1\)is the average size of an independent set ofGG; it lies between00andα⁡\(G\)\\alpha\(G\)and is not normalized by\|V⁡\(G\)\|\|V\(G\)\|\.

[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)proved that a triangle\-free graphGGonnnvertices withΔ⁡\(G\)≤d\\Delta\(G\)\\leq dsatisfiesα¯G​\(1\)≥\(1\+od​\(1\)\)​log⁡dd​n\\overline\{\\alpha\}\_\{G\}\(1\)\\geq\(1\+o\_\{d\}\(1\)\)\\frac\{\\log d\}\{d\}\\,n, so that the average independent set is already as large as the lower bound Shearer’s theorem\([Shearer 1983](https://arxiv.org/html/2608.16977#bibc.bib8)\)gives for the maximum one; in particular this gives a second proof ofR⁡\(3,k\)≤\(1\+o⁡\(1\)\)​k2/log⁡kR\(3,k\)\\leq\(1\+o\(1\)\)k^\{2\}/\\log k\. In Section 5 of the same paper they ask by how much the maximum exceeds the average and record four conjectures: three on triangle\-free graphs and a fourth, Conjecture 4, forKrK\_\{r\}\-free graphs\. The second of them,[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3), reads as follows:

*For every triangle\-free graphGGof minimum degreedd,α⁡\(G\)/α¯G​\(1\)≥2−od​\(1\)\\alpha\(G\)/\\overline\{\\alpha\}\_\{G\}\(1\)\\geq 2\-o\_\{d\}\(1\)\.*

[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)deduce from it thatR⁡\(3,k\)≤\(12\+o⁡\(1\)\)​k2/log⁡kR\(3,k\)\\leq\(\\tfrac\{1\}\{2\}\+o\(1\)\)k^\{2\}/\\log k, halving the constant in Shearer’s bound, which is still the best known\.

The general lower bound available for all graphs is[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3), which states thatα⁡\(G\)/α¯G​\(λ\)≥1\+α⁡\(G\)/\(λ​n\)\\alpha\(G\)/\\overline\{\\alpha\}\_\{G\}\(\\lambda\)\\geq 1\+\\alpha\(G\)/\(\\lambda n\)for every graphGGonnnvertices, with equality for a disjoint union of copies of a singleKrK\_\{r\}\. That bound degenerates as soon asα⁡\(G\)=o⁡\(n\)\\alpha\(G\)=o\(n\)\. The constant4/34/3of Conjecture 1 is the exact value ofα​\(G\)/α¯G​\(1\)\\alpha\(G\)/\\overline\{\\alpha\}\_\{G\}\(1\)atG=K3G=K\_\{3\}, and it also falls below197136/137585=1\.43283​…197136/137585=1\.43283\\ldots, the smallest ratio[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)report, attained by the cyclic triangle\-free graph witnessingR⁡\(3,9\)≥36R\(3,9\)\\geq 36\([Grinstead & Roberts 1982](https://arxiv.org/html/2608.16977#bibc.bib6)\)\.

The evidence offered for Conjecture 2 is the expectation that graphs produced by the triangle\-free process have ratio tending to22\. The conjecture remained open:[Davies & Kang 2025](https://arxiv.org/html/2608.16977#bibc.bib2)record it as Conjecture B in the open problems section of their survey of the hard\-core model in graph theory, and[Morris 2026](https://arxiv.org/html/2608.16977#bibc.bib7)restates it in his ICM survey of Ramsey theory, noting that any lower bound better than1\+o⁡\(1\)1\+o\(1\)would be a very significant breakthrough\. The one route toward it proposed in[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)is the identity

α⁡\(G\)=α¯G​\(1\)\+∫1∞Varλ​\(\|I\|\)λ​𝑑λ,\\alpha\(G\)=\\overline\{\\alpha\}\_\{G\}\(1\)\+\\int\_\{1\}^\{\\infty\}\\frac\{\\mathrm\{Var\}\_\{\\lambda\}\(\|I\|\)\}\{\\lambda\}\\,d\\lambda,which reduces the conjecture to lower bounds on the variance of the hard\-core model;[Davies et al\. 2025](https://arxiv.org/html/2608.16977#bibc.bib4)prove such bounds, but only for fugacities that are small in terms of the number of vertices\. In a neighboring circle of hard\-core conjectures,[Cambie & Jooken 2023](https://arxiv.org/html/2608.16977#bibc.bib1)disproved five conjectures on the extremal graphs for the occupancy fraction and the independence polynomial among regular graphs of given girth, by computer search over small graphs\.

###### Theorem C\.3\.1\.

Form≥1m\\geq 1letGm=C5​□​Km,mG\_\{m\}=C\_\{5\}\\,\\square\\,K\_\{m,m\}\. ThenGmG\_\{m\}is triangle\-free and\(m\+2\)\(m\+2\)\-regular on10​m10mvertices,α⁡\(Gm\)=4​m\\alpha\(G\_\{m\}\)=4m, and

α¯Gm​\(1\)=136​m​\(1\+O⁡\(\(2/3\)m\)\),so thatα⁡\(Gm\)α¯Gm​\(1\)⟶2413=1\.846​…<2\.\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)=\\frac\{13\}\{6\}\\,m\\Bigl\(1\+O\\bigl\(\(2/3\)^\{m\}\\bigr\)\\Bigr\),\\qquad\\text\{so that\}\\qquad\\frac\{\\alpha\(G\_\{m\}\)\}\{\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)\}\\longrightarrow\\frac\{24\}\{13\}=1\.846\\ldots<2\.

###### Corollary C\.3\.2\.

Conjecture 2 of[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)is false\. IndeedGmG\_\{m\}is triangle\-free of minimum degreed=m\+2→∞d=m\+2\\to\\inftywhileα⁡\(Gm\)/α¯Gm​\(1\)→24/13\\alpha\(G\_\{m\}\)/\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)\\to 24/13, so no bound of the form2−od​\(1\)2\-o\_\{d\}\(1\)can hold\. Since a triangle\-free graph isKrK\_\{r\}\-free for everyr≥3r\\geq 3, the same graphs refute the minimum\-degree assertion of[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)for every fixedr≥3r\\geq 3\.

The product structure is essential:Km,mK\_\{m,m\}alone is triangle\-free andmm\-regular withα=m\\alpha=m,\|ℐ⁡\(Km,m\)\|=2m\+1−1\|\\mathcal\{I\}\(K\_\{m,m\}\)\|=2^\{m\+1\}\-1and∑I\|I\|=m​2m\\sum\_\{I\}\|I\|=m2^\{m\}, so its ratio is exactly2−2−m2\-2^\{\-m\}\. Conjecture 1 of[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3), thatα⁡\(G\)/α¯G​\(1\)≥4/3\\alpha\(G\)/\\overline\{\\alpha\}\_\{G\}\(1\)\\geq 4/3for every triangle\-free graphGG, is untouched by these examples, since both24/1324/13and the value32/1932/19obtained below exceed4/34/3; so is the first assertion of Conjecture 4, which asks only for1\+1/r1\+1/rin theKrK\_\{r\}\-free case\. Conjecture 3 of[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3), the version at general fugacity, is also untouched: we compute only atλ=1\\lambda=1, whereas that conjecture is free to chooseλ\\lambdasmall, and by the same deduction it still impliesR⁡\(3,k\)≤\(12\+o⁡\(1\)\)​k2/log⁡kR\(3,k\)\\leq\(\\tfrac\{1\}\{2\}\+o\(1\)\)k^\{2\}/\\log k\. Finally,α⁡\(Gm\)/\|V⁡\(Gm\)\|=2/5\\alpha\(G\_\{m\}\)/\|V\(G\_\{m\}\)\|=2/5, so these graphs lie in the range where the general bound of[Davies et al\. 2018](https://arxiv.org/html/2608.16977#bibc.bib3)already forces the ratio to be at least7/57/5; they say nothing about triangle\-free graphs withα⁡\(G\)=o⁡\(\|V⁡\(G\)\|\)\\alpha\(G\)=o\(\|V\(G\)\|\), which is the range relevant toR⁡\(3,k\)R\(3,k\)\.

We now summarize the computation\. An independent set ofC5​□​Km,mC\_\{5\}\\,\\square\\,K\_\{m,m\}is a family of independent sets ofC5C\_\{5\}, one over each vertex ofKm,mK\_\{m,m\}, subject only to the requirement that the union of those over one side be disjoint from the union of those over the other\. Classifying such a family by its pair of unions and inverting over the Boolean lattice2ℤ52^\{\\mathbb\{Z\}\_\{5\}\}writes the independence polynomial ofGmG\_\{m\}as a signed sum of353^\{5\}terms\(FX​FY\)m\(F\_\{X\}F\_\{Y\}\)^\{m\}indexed by the disjoint pairsX,Y⊆ℤ5X,Y\\subseteq\\mathbb\{Z\}\_\{5\}, whereFXF\_\{X\}is the independence polynomial of the subgraph ofC5C\_\{5\}induced onXX\. The key point is thatFX​\(1\)​FY​\(1\)F\_\{X\}\(1\)F\_\{Y\}\(1\)is maximized only when one ofX,YX,Yis a nonadjacent pair ofC5C\_\{5\}and the other is its complement\. The mean32\+23=136\\tfrac\{3\}\{2\}\+\\tfrac\{2\}\{3\}=\\tfrac\{13\}\{6\}attached to that configuration is the average number of chosen vertices per pair of opposite fibers, which givesα¯Gm​\(1\)≈136​m\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)\\approx\\tfrac\{13\}\{6\}m\.

#### C\.3\.2The construction

In the*Cartesian product*G​□​HG\\,\\square\\,Hthe vertex set isV⁡\(G\)×V⁡\(H\)V\(G\)\\times V\(H\), and\(g,h\)\(g,h\)is adjacent to\(g′,h′\)\(g^\{\\prime\},h^\{\\prime\}\)exactly wheng=g′g=g^\{\\prime\}andh​h′∈E⁡\(H\)hh^\{\\prime\}\\in E\(H\), orh=h′h=h^\{\\prime\}andg​g′∈E⁡\(G\)gg^\{\\prime\}\\in E\(G\)\. Degrees add, soGm=C5​□​Km,mG\_\{m\}=C\_\{5\}\\,\\square\\,K\_\{m,m\}is\(m\+2\)\(m\+2\)\-regular on5⋅2​m=10​m5\\cdot 2m=10mvertices, and in particularδ⁡\(Gm\)=m\+2\\delta\(G\_\{m\}\)=m\+2\.

###### Lemma C\.3\.3\.

The Cartesian product of two triangle\-free graphs is triangle\-free\. In particularGmG\_\{m\}is triangle\-free\.

###### Proof\.

Every edge ofG​□​HG\\,\\square\\,Hchanges exactly one coordinate\. Letu,v,wu,v,wspan a triangle\. By the pigeonhole principle two of its three edges change the same coordinate, say the first, and these two edges share a vertex, sayu​vuvandv​wvw\. Thenu,v,wu,v,wall have the same second coordinate, so the third edgeu​wuwalso changes only the first coordinate\. Hence the first coordinates ofu,v,wu,v,ware pairwise distinct and pairwise adjacent, giving a triangle inGG\. The same argument with the coordinates exchanged gives a triangle inHHwhen the repeated coordinate is the second\. SinceC5C\_\{5\}andKm,mK\_\{m,m\}are triangle\-free, so isGmG\_\{m\}\. ∎

Throughout,ℤ5=\{0,1,2,3,4\}\\mathbb\{Z\}\_\{5\}=\\\{0,1,2,3,4\\\}is the vertex set ofC5C\_\{5\}, withiiadjacent toi±1i\\pm 1, andLLandRRare the two sides ofKm,mK\_\{m,m\}, each of sizemm\. ForX⊆ℤ5X\\subseteq\\mathbb\{Z\}\_\{5\}we writeℐ⁡\(X\)\\mathcal\{I\}\(X\)for the family of independent sets of the induced subgraphC5​\[X\]C\_\{5\}\[X\]and put

FX​\(y\)=∑S∈ℐ⁡\(X\)y\|S\|,f⁡\(X\)=FX​\(1\)=\|ℐ⁡\(X\)\|,F\_\{X\}\(y\)=\\sum\_\{S\\in\\mathcal\{I\}\(X\)\}y^\{\|S\|\},\\qquad f\(X\)=F\_\{X\}\(1\)=\|\\mathcal\{I\}\(X\)\|,so thatFXF\_\{X\}is the*independence polynomial*ofC5​\[X\]C\_\{5\}\[X\]\. Observe thatffis monotone:X⊆X′X\\subseteq X^\{\\prime\}impliesf⁡\(X\)≤f⁡\(X′\)f\(X\)\\leq f\(X^\{\\prime\}\)\.

The*fiber*ℤ5×\{x\}\\mathbb\{Z\}\_\{5\}\\times\\\{x\\\}over a vertexxxofKm,mK\_\{m,m\}induces a copy ofC5C\_\{5\}, and the two fibers over adjacentx,x′x,x^\{\\prime\}are joined by a perfect matching that preserves theℤ5\\mathbb\{Z\}\_\{5\}\-coordinate, as in Figure[8](https://arxiv.org/html/2608.16977#A3.F8)\. Hence an independent setIIofGmG\_\{m\}is the same thing as a family\(Sx\)x∈L∪R\(S\_\{x\}\)\_\{x\\in L\\cup R\}with

Sx∈ℐ⁡\(ℤ5\)​for every​x,Sx∩Sx′=∅​whenever​x​x′∈E⁡\(Km,m\),S\_\{x\}\\in\\mathcal\{I\}\(\\mathbb\{Z\}\_\{5\}\)\\ \\text\{for every \}x,\\qquad S\_\{x\}\\cap S\_\{x^\{\\prime\}\}=\\varnothing\\ \\text\{whenever \}xx^\{\\prime\}\\in E\(K\_\{m,m\}\),and then\|I\|=∑x\|Sx\|\|I\|=\\sum\_\{x\}\|S\_\{x\}\|\. SinceKm,mK\_\{m,m\}is complete bipartite, the disjointness constraints say precisely that

\(⋃x∈LSx\)∩\(⋃x∈RSx\)=∅\.\\Bigl\(\\bigcup\_\{x\\in L\}S\_\{x\}\\Bigr\)\\cap\\Bigl\(\\bigcup\_\{x\\in R\}S\_\{x\}\\Bigr\)=\\varnothing\.\(9\)
00112233440011223344x∈Lx\\in Lx′∈Rx^\{\\prime\}\\in RFigure 8:Two fibers ofGm=C5​□​Km,mG\_\{m\}=C\_\{5\}\\,\\square\\,K\_\{m,m\}, over adjacent verticesx∈Lx\\in Landx′∈Rx^\{\\prime\}\\in R\. Each fiber induces a copy ofC5C\_\{5\}, and the gray matching between them preserves theℤ5\\mathbb\{Z\}\_\{5\}\-coordinate, so the independent sets chosen in fibers on opposite sides must be disjoint\. The gray halos mark a dominant splitting from Lemma[C\.3\.6](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem6): every fiber overLLtakes its independent set inside the nonadjacent pairX=\{0,2\}X=\\\{0,2\\\}and every fiber overRRinside the complementY=\{1,3,4\}Y=\\\{1,3,4\\\}, and this is the splitting ofℤ5\\mathbb\{Z\}\_\{5\}that maximizesf⁡\(X\)​f​\(Y\)f\(X\)f\(Y\)\. The filled vertices show one admissible choice inside them, namely\{0,2\}\\\{0,2\\\}overxxand\{1,3\}\\\{1,3\\\}overx′x^\{\\prime\}\.###### Lemma C\.3\.4\.

α⁡\(Gm\)=4​m\\alpha\(G\_\{m\}\)=4m\.

###### Proof\.

EachSxS\_\{x\}is an independent set ofC5C\_\{5\}, hence\|Sx\|≤2\|S\_\{x\}\|\\leq 2, and summing over the2​m2mfibers givesα⁡\(Gm\)≤4​m\\alpha\(G\_\{m\}\)\\leq 4m\. For the lower bound takeSx=\{0,2\}S\_\{x\}=\\\{0,2\\\}for everyx∈Lx\\in LandSx=\{1,3\}S\_\{x\}=\\\{1,3\\\}for everyx∈Rx\\in R\. Both sets are independent inC5C\_\{5\}and they are disjoint, so equation[9](https://arxiv.org/html/2608.16977#A3.E9)holds and the resulting independent set has size4​m4m\. ∎

#### C\.3\.3Exact counting by Möbius inversion

Let

Zm​\(y\)=∑I∈ℐ⁡\(Gm\)y\|I\|Z\_\{m\}\(y\)=\\sum\_\{I\\in\\mathcal\{I\}\(G\_\{m\}\)\}y^\{\|I\|\}be the independence polynomial ofGmG\_\{m\}, so that\|ℐ⁡\(Gm\)\|=Zm​\(1\)\|\\mathcal\{I\}\(G\_\{m\}\)\|=Z\_\{m\}\(1\)andα¯Gm​\(1\)=Zm′​\(1\)/Zm​\(1\)\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)=Z\_\{m\}^\{\\prime\}\(1\)/Z\_\{m\}\(1\)\.

One is tempted to sum\(FX​\(y\)​FY​\(y\)\)m\\bigl\(F\_\{X\}\(y\)F\_\{Y\}\(y\)\\bigr\)^\{m\}over the disjoint pairs\(X,Y\)\(X,Y\), readingXXandYYas the two unions in equation[9](https://arxiv.org/html/2608.16977#A3.E9)\. This overcounts: the quantity\(FX​\(y\)​FY​\(y\)\)m\\bigl\(F\_\{X\}\(y\)F\_\{Y\}\(y\)\\bigr\)^\{m\}records the families withSx⊆XS\_\{x\}\\subseteq Xforx∈Lx\\in LandSx⊆YS\_\{x\}\\subseteq Yforx∈Rx\\in R, so a single independent set is counted once for every pair\(X,Y\)\(X,Y\)whose two parts contain its two unions\. The remedy is to force the unions to be attained exactly\. ForX⊆ℤ5X\\subseteq\\mathbb\{Z\}\_\{5\}define

gX​\(y\)=∑\(S1,…,Sm\)∈ℐ​\(ℤ5\)mS1∪⋯∪Sm=Xy\|S1\|\+⋯\+\|Sm\|,g\_\{X\}\(y\)=\\sum\_\{\\begin\{subarray\}\{c\}\(S\_\{1\},\\dots,S\_\{m\}\)\\in\\mathcal\{I\}\(\\mathbb\{Z\}\_\{5\}\)^\{m\}\\\\ S\_\{1\}\\cup\\dots\\cup S\_\{m\}=X\\end\{subarray\}\}y^\{\|S\_\{1\}\|\+\\dots\+\|S\_\{m\}\|\},the generating function of themm\-tuples of independent sets ofC5C\_\{5\}whose union is exactlyXX\. A tuple has union contained inXXif and only if everySiS\_\{i\}lies inℐ⁡\(X\)\\mathcal\{I\}\(X\), so

∑X′⊆XgX′​\(y\)=FX​\(y\)m\.\\sum\_\{X^\{\\prime\}\\subseteq X\}g\_\{X^\{\\prime\}\}\(y\)=F\_\{X\}\(y\)^\{m\}\.Möbius inversion in the Boolean lattice2ℤ52^\{\\mathbb\{Z\}\_\{5\}\}therefore gives

gX​\(y\)=∑X′⊆X\(−1\)\|X∖X′\|​FX′​\(y\)m\.g\_\{X\}\(y\)=\\sum\_\{X^\{\\prime\}\\subseteq X\}\(\-1\)^\{\|X\\setminus X^\{\\prime\}\|\}F\_\{X^\{\\prime\}\}\(y\)^\{m\}\.\(10\)Classifying an independent set ofGmG\_\{m\}by the ordered pair of unions in equation[9](https://arxiv.org/html/2608.16977#A3.E9)partitionsℐ⁡\(Gm\)\\mathcal\{I\}\(G\_\{m\}\), and yields the exact identity

Zm​\(y\)=∑X,Y⊆ℤ5X∩Y=∅gX​\(y\)​gY​\(y\)\.Z\_\{m\}\(y\)=\\sum\_\{\\begin\{subarray\}\{c\}X,Y\\subseteq\\mathbb\{Z\}\_\{5\}\\\\ X\\cap Y=\\varnothing\\end\{subarray\}\}g\_\{X\}\(y\)\\,g\_\{Y\}\(y\)\.\(11\)Substituting equation[10](https://arxiv.org/html/2608.16977#A3.E10)into equation[11](https://arxiv.org/html/2608.16977#A3.E11)collapses to a sum over the same index set with explicit signs\.

###### Lemma C\.3\.5\.

For everym≥1m\\geq 1,

Zm​\(y\)=∑X,Y⊆ℤ5X∩Y=∅\(−1\)5−\|X\|−\|Y\|​\(FX​\(y\)​FY​\(y\)\)m\.Z\_\{m\}\(y\)=\\sum\_\{\\begin\{subarray\}\{c\}X,Y\\subseteq\\mathbb\{Z\}\_\{5\}\\\\ X\\cap Y=\\varnothing\\end\{subarray\}\}\(\-1\)^\{\\,5\-\|X\|\-\|Y\|\}\\bigl\(F\_\{X\}\(y\)F\_\{Y\}\(y\)\\bigr\)^\{m\}\.

###### Proof\.

IfX∩Y=∅X\\cap Y=\\varnothingandX′⊆XX^\{\\prime\}\\subseteq X,Y′⊆YY^\{\\prime\}\\subseteq Y, thenX′∩Y′=∅X^\{\\prime\}\\cap Y^\{\\prime\}=\\varnothing, so expanding equation[10](https://arxiv.org/html/2608.16977#A3.E10)in equation[11](https://arxiv.org/html/2608.16977#A3.E11)produces a linear combination of the terms\(FX′​FY′\)m\\bigl\(F\_\{X^\{\\prime\}\}F\_\{Y^\{\\prime\}\}\\bigr\)^\{m\}indexed by disjoint pairs\(X′,Y′\)\(X^\{\\prime\},Y^\{\\prime\}\)\. Fix such a pair and setW=ℤ5∖\(X′∪Y′\)W=\\mathbb\{Z\}\_\{5\}\\setminus\(X^\{\\prime\}\\cup Y^\{\\prime\}\)\. The pairs\(X,Y\)\(X,Y\)contributing to it are exactlyX=X′∪AX=X^\{\\prime\}\\cup AandY=Y′∪BY=Y^\{\\prime\}\\cup BwithA,BA,Bdisjoint subsets ofWW, and each contributes the sign\(−1\)\|A\|\+\|B\|\(\-1\)^\{\|A\|\+\|B\|\}\. Assigning eachw∈Ww\\in Windependently toAA, toBB, or to neither, with respective weights−1,−1,\+1\-1,\-1,\+1, the coefficient equals

∑A,B⊆WA∩B=∅\(−1\)\|A\|\+\|B\|=∏w∈W\(1−1−1\)=\(−1\)\|W\|=\(−1\)5−\|X′\|−\|Y′\|\.∎\\sum\_\{\\begin\{subarray\}\{c\}A,B\\subseteq W\\\\ A\\cap B=\\varnothing\\end\{subarray\}\}\(\-1\)^\{\|A\|\+\|B\|\}=\\prod\_\{w\\in W\}\(1\-1\-1\)=\(\-1\)^\{\|W\|\}=\(\-1\)^\{\\,5\-\|X^\{\\prime\}\|\-\|Y^\{\\prime\}\|\}\.\\qed

#### C\.3\.4The dominant pairs

By Lemma[C\.3\.5](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem5)the exponential growth ofZm​\(1\)Z\_\{m\}\(1\)is governed bymax⁡f⁡\(X\)​f​\(Y\)\\max f\(X\)f\(Y\)over disjoint pairs\. We determine both the maximum and the next value, the latter being what controls the error term\.

###### Lemma C\.3\.6\.

LetX,Y⊆ℤ5X,Y\\subseteq\\mathbb\{Z\}\_\{5\}be disjoint\. Thenf⁡\(X\)​f​\(Y\)≤24f\(X\)f\(Y\)\\leq 24, with equality exactly when one ofX,YX,Yis a pair of nonadjacent vertices ofC5C\_\{5\}and the other is its complement, which happens for1010ordered pairs\. Moreover, iff⁡\(X\)​f​\(Y\)≠24f\(X\)f\(Y\)\\neq 24thenf⁡\(X\)​f​\(Y\)≤16f\(X\)f\(Y\)\\leq 16\.

###### Proof\.

A direct count gives the value offfon every induced subgraph ofC5C\_\{5\}: it is11on∅\\varnothing,22on a single vertex,33on an edge and44on a nonedge,55onP3P\_\{3\}and66onK2∪K1K\_\{2\}\\cup K\_\{1\},88onP4P\_\{4\}, and1111onC5C\_\{5\}itself\. Sinceffis monotone andX∩Y=∅X\\cap Y=\\varnothing, replacingYYbyℤ5∖X\\mathbb\{Z\}\_\{5\}\\setminus Xcan only increasef⁡\(X\)​f​\(Y\)f\(X\)f\(Y\)\. It therefore suffices to tabulate the complementary pairs\(X,ℤ5∖X\)\(X,\\mathbb\{Z\}\_\{5\}\\setminus X\), up to the rotational symmetry ofC5C\_\{5\}and up to swapping the two parts:

\|X\|C5​\[X\]C5​\[ℤ5∖X\]f⁡\(X\)​f​\(ℤ5∖X\)0∅C51⋅11=111K1P42⋅8=162,X​an edgeK2P33⋅5=152,X​a nonedgeK2¯K2∪K14⋅6=24\\begin\{array\}\[\]\{c\|c\|c\|c\}\|X\|&C\_\{5\}\[X\]&C\_\{5\}\[\\mathbb\{Z\}\_\{5\}\\setminus X\]&f\(X\)f\(\\mathbb\{Z\}\_\{5\}\\setminus X\)\\\\ \\hline\\cr 0&\\varnothing&C\_\{5\}&1\\cdot 11=11\\\\ 1&K\_\{1\}&P\_\{4\}&2\\cdot 8=16\\\\ 2,\\ X\\ \\text\{an edge\}&K\_\{2\}&P\_\{3\}&3\\cdot 5=15\\\\ 2,\\ X\\ \\text\{a nonedge\}&\\overline\{K\_\{2\}\}&K\_\{2\}\\cup K\_\{1\}&4\\cdot 6=24\\end\{array\}The cases\|X\|≥3\|X\|\\geq 3are obtained by swapping the two parts\. Hencemax⁡f⁡\(X\)​f​\(Y\)=24\\max f\(X\)f\(Y\)=24over all disjoint pairs, attained on complementary pairs only in the stated configuration; there are55nonadjacent pairs inC5C\_\{5\}, giving1010ordered pairs\.

Suppose nowf⁡\(X\)​f​\(Y\)\>16f\(X\)f\(Y\)\>16\. EnlargingYYtoℤ5∖X\\mathbb\{Z\}\_\{5\}\\setminus Xgivesf⁡\(X\)​f​\(ℤ5∖X\)\>16f\(X\)f\(\\mathbb\{Z\}\_\{5\}\\setminus X\)\>16, so by the tableXXis either a nonadjacent pair or the complement of one\. IfX=\{a,a\+2\}X=\\\{a,a\+2\\\}is the nonadjacent pair thenf⁡\(X\)=4f\(X\)=4, sof⁡\(Y\)\>4f\(Y\)\>4; butYYis contained inℤ5∖X\\mathbb\{Z\}\_\{5\}\\setminus X, an edge plus an isolated vertex, whose proper subsets all havef≤4f\\leq 4, soY=ℤ5∖XY=\\mathbb\{Z\}\_\{5\}\\setminus Xandf⁡\(X\)​f​\(Y\)=24f\(X\)f\(Y\)=24\. If insteadf⁡\(X\)=6f\(X\)=6thenf⁡\(Y\)\>16/6f\(Y\)\>16/6withYYcontained in a nonadjacent pair, whencef⁡\(Y\)∈\{1,2,4\}f\(Y\)\\in\\\{1,2,4\\\}and onlyf⁡\(Y\)=4f\(Y\)=4survives, again giving2424\. Therefore every value other than2424is at most1616\. ∎

For a dominant pair, sayX=\{0,2\}X=\\\{0,2\\\}andY=\{1,3,4\}Y=\\\{1,3,4\\\}, the induced subgraphs areK2¯\\overline\{K\_\{2\}\}andK2∪K1K\_\{2\}\\cup K\_\{1\}, so

FX​\(y\)=\(1\+y\)2,FY​\(y\)=\(1\+y\)​\(1\+2​y\),P⁡\(y\):=FX​\(y\)​FY​\(y\)=\(1\+y\)3​\(1\+2​y\)\.F\_\{X\}\(y\)=\(1\+y\)^\{2\},\\qquad F\_\{Y\}\(y\)=\(1\+y\)\(1\+2y\),\\qquad P\(y\):=F\_\{X\}\(y\)F\_\{Y\}\(y\)=\(1\+y\)^\{3\}\(1\+2y\)\.ThusP⁡\(1\)=8⋅3=24P\(1\)=8\\cdot 3=24and

P′​\(1\)P⁡\(1\)=31\+y\|y=1\+21\+2​y\|y=1=32\+23=136\.\\frac\{P^\{\\prime\}\(1\)\}\{P\(1\)\}=\\Bigl\.\\frac\{3\}\{1\+y\}\\Bigr\|\_\{y=1\}\+\\Bigl\.\\frac\{2\}\{1\+2y\}\\Bigr\|\_\{y=1\}=\\frac\{3\}\{2\}\+\\frac\{2\}\{3\}=\\frac\{13\}\{6\}\.\(12\)All five rotations of the pair\(X,Y\)\(X,Y\), in both orders, give the same polynomialPP\. SinceP′​\(1\)/P​\(1\)P^\{\\prime\}\(1\)/P\(1\)is the sum of the mean sizes of a uniformly random element ofℐ⁡\(X\)\\mathcal\{I\}\(X\)and ofℐ⁡\(Y\)\\mathcal\{I\}\(Y\), the constant13/613/6has a per\-vertex reading: a vertex ofLLcontributes on average11, the mean of\|S\|\|S\|over the four independent subsets of a nonadjacent pair, and a vertex ofRRcontributes7/67/6, the mean of\|S\|\|S\|over the six independent subsets of an edge plus an isolated vertex\.

#### C\.3\.5Proof of the main theorem

###### Proof of Theorem[C\.3\.1](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem1)\.

The graph\-theoretic assertions are Lemmas[C\.3\.3](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem3)and[C\.3\.4](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem4)together with the degree count above\. Split the sum of Lemma[C\.3\.5](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem5)into the1010dominant pairs, each of which contributes\+P​\(y\)m\+P\(y\)^\{m\}by Lemma[C\.3\.6](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem6)and the sign\(−1\)5−2−3=\+1\(\-1\)^\{5\-2\-3\}=\+1, and a remainder:

Zm​\(y\)=10​P​\(y\)m\+Em​\(y\),Em​\(y\)=∑\(X,Y\)​disjointnot dominant\(−1\)5−\|X\|−\|Y\|​\(FX​\(y\)​FY​\(y\)\)m\.Z\_\{m\}\(y\)=10\\,P\(y\)^\{m\}\+E\_\{m\}\(y\),\\qquad E\_\{m\}\(y\)=\\sum\_\{\\begin\{subarray\}\{c\}\(X,Y\)\\ \\text\{disjoint\}\\\\ \\text\{not dominant\}\\end\{subarray\}\}\(\-1\)^\{5\-\|X\|\-\|Y\|\}\\bigl\(F\_\{X\}\(y\)F\_\{Y\}\(y\)\\bigr\)^\{m\}\.Assigning each element ofℤ5\\mathbb\{Z\}\_\{5\}toXX, toYY, or to neither shows that there are35=2433^\{5\}=243ordered disjoint pairs in all, and by Lemma[C\.3\.6](https://arxiv.org/html/2608.16977#A3.SS3.Thmtheorem6)every nondominant one hasFX​\(1\)​FY​\(1\)≤16F\_\{X\}\(1\)F\_\{Y\}\(1\)\\leq 16\. Hence

\|Em​\(1\)\|≤243⋅16m\.\|E\_\{m\}\(1\)\|\\leq 243\\cdot 16^\{m\}\.For the derivative, each nondominant term satisfies

\|dd​y​\(FX​\(y\)​FY​\(y\)\)m\|y=1=m​\(FX​\(1\)​FY​\(1\)\)m⋅\(FX​FY\)′​\(1\)FX​\(1\)​FY​\(1\)≤4​m⋅16m,\\Bigl\|\\tfrac\{d\}\{dy\}\\bigl\(F\_\{X\}\(y\)F\_\{Y\}\(y\)\\bigr\)^\{m\}\\Bigr\|\_\{y=1\}=m\\bigl\(F\_\{X\}\(1\)F\_\{Y\}\(1\)\\bigr\)^\{m\}\\cdot\\frac\{\(F\_\{X\}F\_\{Y\}\)^\{\\prime\}\(1\)\}\{F\_\{X\}\(1\)F\_\{Y\}\(1\)\}\\leq 4m\\cdot 16^\{m\},because\(FX​FY\)′​\(1\)/\(FX​FY\)​\(1\)\(F\_\{X\}F\_\{Y\}\)^\{\\prime\}\(1\)/\(F\_\{X\}F\_\{Y\}\)\(1\)is a sum of two mean independent set sizes inC5C\_\{5\}, each at most22\. Hence\|Em′​\(1\)\|≤972​m​16m\|E\_\{m\}^\{\\prime\}\(1\)\|\\leq 972\\,m\\,16^\{m\}\. UsingP⁡\(1\)=24P\(1\)=24and equation[12](https://arxiv.org/html/2608.16977#A3.E12),

Zm​\(1\)=10⋅24m​\(1\+O⁡\(\(2/3\)m\)\),Zm′​\(1\)=136​m⋅10⋅24m​\(1\+O⁡\(\(2/3\)m\)\),Z\_\{m\}\(1\)=10\\cdot 24^\{m\}\\Bigl\(1\+O\\bigl\(\(2/3\)^\{m\}\\bigr\)\\Bigr\),\\qquad Z\_\{m\}^\{\\prime\}\(1\)=\\frac\{13\}\{6\}\\,m\\cdot 10\\cdot 24^\{m\}\\Bigl\(1\+O\\bigl\(\(2/3\)^\{m\}\\bigr\)\\Bigr\),since972​m​16m972\\,m\\,16^\{m\}divided by136​m⋅10⋅24m\\tfrac\{13\}\{6\}m\\cdot 10\\cdot 24^\{m\}isO⁡\(\(2/3\)m\)O\\bigl\(\(2/3\)^\{m\}\\bigr\)\. Dividing,

α¯Gm​\(1\)=Zm′​\(1\)Zm​\(1\)=136​m​\(1\+O⁡\(\(2/3\)m\)\),\\overline\{\\alpha\}\_\{G\_\{m\}\}\(1\)=\\frac\{Z\_\{m\}^\{\\prime\}\(1\)\}\{Z\_\{m\}\(1\)\}=\\frac\{13\}\{6\}\\,m\\Bigl\(1\+O\\bigl\(\(2/3\)^\{m\}\\bigr\)\\Bigr\),and withα⁡\(Gm\)=4​m\\alpha\(G\_\{m\}\)=4mthe ratio tends to4/\(13/6\)=24/134/\(13/6\)=24/13, completing the proof\. ∎

## References\.

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- Davies & Kang \(2025\)Ewan Davies and Ross J Kang\.The hard\-core model in graph theory\.*arXiv preprint arXiv:2501\.03379*, 2025\.
- Davies et al\. \(2018\)Ewan Davies, Matthew Jenssen, Will Perkins, and Barnaby Roberts\.On the average size of independent sets in triangle\-free graphs\.*Proceedings of the American Mathematical Society*, 146\(1\):111–124, 2018\.
- Davies et al\. \(2025\)Ewan Davies, Juspreet Singh Sandhu, and Brian Tan\.On expectations and variances in the hard\-core model on bounded degree graphs\.*arXiv preprint arXiv:2505\.13396*, 2025\.
- Greenwood & Gleason \(1955\)Robert E Greenwood and Andrew Mattei Gleason\.Combinatorial relations and chromatic graphs\.*Canadian Journal of Mathematics*, 7:1–7, 1955\.
- Grinstead & Roberts \(1982\)Charles M Grinstead and Sam M Roberts\.On the ramsey numbers r \(3, 8\) and r \(3, 9\)\.*Journal of Combinatorial Theory, Series B*, 33\(1\):27–51, 1982\.
- Morris \(2026\)Robert Morris\.Some recent results in ramsey theory\.In*International Congress of Mathematicians 2026*, pp\. 210–239\. SIAM, 2026\.
- Shearer \(1983\)James B Shearer\.A note on the independence number of triangle\-free graphs\.*Discrete Mathematics*, 46\(1\):83–87, 1983\.

### C\.4Sets with no large divisor difference

This problem asks how large a setA⊆\[n\]A\\subseteq\[n\]can be if no differenceb−a≥tb\-a\\geq tbetween two of its elements dividesbb\. The odd numbers form such a set of size⌈n/2⌉\\lceil n/2\\rceil, and Erdős asked whether\|A\|≤\(12\+ot​\(1\)\)​n\\lvert A\\rvert\\leq\\bigl\(\\tfrac\{1\}\{2\}\+o\_\{t\}\(1\)\\bigr\)nmust hold\. We prove that it does, with the explicit error term3​n/log⁡log⁡n3n/\\sqrt\{\\log\\log n\}\. The proof weights each integer by the number of its prime factors drawn from the odd primes in\(t,n1/3\]\(t,n^\{1/3\}\], compares the even elements ofAAwith the odd integers omitted from it through the injectiona↦a−pa\\mapsto a\-p, and converts the resulting weighted inequality into a bound on\|A\|\\lvert A\\rvertby Cauchy–Schwarz\.

#### C\.4\.1Introduction

Erdős asked, in a letter to Ruzsa written around 1980, how dense a set of integers can be if no sufficiently large difference between two of its elements divides one of them\. The question is recorded in the miscellany of Erdős problems of[Guy 1983](https://arxiv.org/html/2608.16977#bibd.bib3)and in Ruzsa’s survey of Erdős’s work on the integers\([Ruzsa 1999](https://arxiv.org/html/2608.16977#bibd.bib6)\), and is Problem 635 on the Erdős problems website\([Bloom 2026](https://arxiv.org/html/2608.16977#bibd.bib1)\), where it reads as follows\.

*Lett≥1t\\geq 1andA⊆\{1,…,N\}A\\subseteq\\\{1,\\ldots,N\\\}be such that whenevera,b∈Aa,b\\in Awithb−a≥tb\-a\\geq twe haveb−a∤bb\-a\\nmid b\. How large can\|A\|\\lvert A\\rvertbe? Is it true that\|A\|≤\(12\+ot​\(1\)\)​N\\lvert A\\rvert\\leq\\left\(\\frac\{1\}\{2\}\+o\_\{t\}\(1\)\\right\)N?*

We writennthroughout for the integer calledNNthere\. We answer the second question affirmatively, with an explicit rate\.

Fix an integert≥1t\\geq 1\. Call a setA⊆\[n\]A\\subseteq\[n\]*tt\-admissible*if no two elementsa<ba<bofAAsatisfyb−a≥tb\-a\\geq tandb−a\|bb\-a\\mid b, and writeF⁡\(n,t\)F\(n;t\)for the largest size of att\-admissible subset of\[n\]\[n\]\. Sinceb=a\+\(b−a\)b=a\+\(b\-a\), the divisibilityb−a\|bb\-a\\mid bis equivalent tob−a\|ab\-a\\mid a, so the forbidden configuration is symmetric in the two elements; we call such a difference a*divisor difference*of the pair\. Enlargingttconstrains fewer pairs, soF⁡\(n,t\)F\(n;t\)is nondecreasing intt\.

Two cases are immediate\. A11\-admissible set contains no two consecutive integers, since a difference of11is at leastt=1t=1and divides everything, soF⁡\(n,1\)≤⌈n/2⌉F\(n;1\)\\leq\\lceil n/2\\rceil\. The odd numbers in\[n\]\[n\]are11\-admissible, because the difference of two odd numbers is even and an even number cannot divide an odd one, so in factF⁡\(n,1\)=⌈n/2⌉F\(n;1\)=\\lceil n/2\\rceil, andF⁡\(n,t\)≥⌈n/2⌉F\(n;t\)\\geq\\lceil n/2\\rceilfor everytt\. Oncen≥2n\\geq 2, the odd numbers stop being optimal as soon ast≥2t\\geq 2\. Erdős observed that the set

A=\{m≤n:m​odd\}∪\{2k≤n:k​odd\}A=\\\{m\\leq n:m\\text\{ odd\}\\\}\\cup\\\{2^\{k\}\\leq n:k\\text\{ odd\}\\\}is22\-admissible\([Bloom 2026](https://arxiv.org/html/2608.16977#bibd.bib1)\)\. Indeed, pairs of odd elements are handled by the previous paragraph\. Two powers2k<2j2^\{k\}<2^\{j\}havej−k≥2j\-k\\geq 2, sincejjandkkare distinct odd numbers, so their difference2k​\(2j−k−1\)2^\{k\}\(2^\{j\-k\}\-1\)has odd part2j−k−1\>12^\{j\-k\}\-1\>1and therefore does not divide2j2^\{j\}\. If exactly one ofa<ba<bis an added power, thenb−ab\-ais odd, while the equivalent divisibilitiesb−a\|ab\-a\\mid aandb−a\|bb\-a\\mid bmakeb−ab\-aa divisor of that power of two, forcingb−a=1<2b\-a=1<2\. Since the two sets above are disjoint, this gives

F⁡\(n,t\)≥⌈n2⌉\+log2⁡n−12for every​t≥2,F\(n;t\)\\ \\geq\\ \\Bigl\\lceil\\frac\{n\}\{2\}\\Bigr\\rceil\+\\frac\{\\log\_\{2\}n\-1\}\{2\}\\qquad\\text\{for every \}t\\geq 2,so the inequality asked for cannot hold withot​\(1\)o\_\{t\}\(1\)replaced by00\.

The second question was answered affirmatively in January 2026 by Liam Price with ChatGPT\-5\.2, as recorded on the problem’s page\([Bloom 2026](https://arxiv.org/html/2608.16977#bibd.bib1)\); the first question, which asks for the exact value ofF⁡\(n,t\)F\(n;t\), remains open\. Tao observed that an affirmative answer also follows quickly from an inequality of[Elliott 2012](https://arxiv.org/html/2608.16977#bibd.bib2), which bounds a weighted mean square, over the small primespp, of the difference between the average of an arbitrary function on an intervalIIand its average along the multiples ofppinII, by the mean square of that function onII\. Elliott’s inequality can be substituted for Lemma[C\.4\.4](https://arxiv.org/html/2608.16977#A3.SS4.Thmtheorem4)below\. Tao also pointed out that the graph in whichaaandbbare joined wheneverb−ab\-adividesbb, onceb−ab\-ais restricted to primes or almost primes, is closely related to the divisibility graphs through which pair correlations of multiplicative functions are studied\([Matomäki et al\. 2016](https://arxiv.org/html/2608.16977#bibd.bib5);[Helfgott & Radziwiłł 2021](https://arxiv.org/html/2608.16977#bibd.bib4)\); that literature is concerned with connectivity and expansion rather than with independent sets, which is what Erdős’s question asks about\([Bloom 2026](https://arxiv.org/html/2608.16977#bibd.bib1)\)\.

###### Theorem C\.4\.1\.

For every integert≥1t\\geq 1there is ann0​\(t\)n\_\{0\}\(t\)such that

⌈n2⌉≤F⁡\(n,t\)≤n2\+3​nlog⁡log⁡n\\Bigl\\lceil\\frac\{n\}\{2\}\\Bigr\\rceil\\ \\leq\\ F\(n;t\)\\ \\leq\\ \\frac\{n\}\{2\}\+\\frac\{3n\}\{\\sqrt\{\\log\\log n\}\}for alln≥n0​\(t\)n\\geq n\_\{0\}\(t\)\. In particularF⁡\(n,t\)=\(12\+ot​\(1\)\)​nF\(n;t\)=\\bigl\(\\tfrac\{1\}\{2\}\+o\_\{t\}\(1\)\\bigr\)nfor each fixedtt\.

The upper bound rests on a single injection\. Letppbe an odd prime withp\>tp\>t\. Ifaais an even multiple ofpp, thena−pa\-pis an odd multiple ofppat distancep≥tp\\geq tfromaa, andppdividesaa; soaaanda−pa\-pcannot both lie in att\-admissible set\. Summing this over a setPPof odd primes larger thanttweights each even element of att\-admissible setAA, and each odd element excluded fromAA, by its number of prime factors drawn fromPP\. That weight has meanκ=∑p∈P1/p\\kappa=\\sum\_\{p\\in P\}1/pon each parity class, and its total square deviation there is at mostn​κ/2\+O⁡\(\|P\|2\)n\\kappa/2\+O\(\\lvert P\\rvert^\{2\}\), so Cauchy–Schwarz converts the weighted inequality into a bound of sizen/κn/\\sqrt\{\\kappa\}for the excess of even elements ofAAover missing odd elements\. TakingPPto be the odd primes in\(t,n1/3\]\(t,n^\{1/3\}\]makesκ\\kappaas large aslog⁡log⁡n\\log\\log n\.

#### C\.4\.2A second moment estimate

The following elementary estimate will be used\.

###### Lemma C\.4\.4\.

LetPPbe a finite set of odd primes, and put

κ=∑p∈P1p,ωP\(m\)=\#\{p∈P:p∣m\}\.\\kappa=\\sum\_\{p\\in P\}\\frac\{1\}\{p\},\\qquad\\omega\_\{P\}\(m\)=\\\#\\\{p\\in P:p\\mid m\\\}\.Then for everyn≥1n\\geq 1,

∑m≤nm​even\(ωP​\(m\)−κ\)2≤n​κ2\+2​\|P\|2and∑m≤nm​odd\(ωP​\(m\)−κ\)2≤n​κ2\+2​\|P\|2\.\\sum\_\{\\begin\{subarray\}\{c\}m\\leq n\\\\ m\\text\{ even\}\\end\{subarray\}\}\\bigl\(\\omega\_\{P\}\(m\)\-\\kappa\\bigr\)^\{2\}\\leq\\frac\{n\\kappa\}\{2\}\+2\\lvert P\\rvert^\{2\}\\qquad\\text\{and\}\\qquad\\sum\_\{\\begin\{subarray\}\{c\}m\\leq n\\\\ m\\text\{ odd\}\\end\{subarray\}\}\\bigl\(\\omega\_\{P\}\(m\)\-\\kappa\\bigr\)^\{2\}\\leq\\frac\{n\\kappa\}\{2\}\+2\\lvert P\\rvert^\{2\}\.

###### Proof\.

Letℳ\\mathcal\{M\}be either the set of even integers in\[n\]\[n\]or the set of odd ones, and for an odd squarefreeddputN\(d\)=\#\{m∈ℳ:d∣m\}N\(d\)=\\\#\\\{m\\in\\mathcal\{M\}:d\\mid m\\\}\. Sinceddis odd,N⁡\(d\)=⌊n/2​d⌋N\(d\)=\\lfloor n/2d\\rfloorin the even case andN⁡\(d\)=⌊n/d⌋−⌊n/2​d⌋N\(d\)=\\lfloor n/d\\rfloor\-\\lfloor n/2d\\rfloorin the odd case\. In both cases

N⁡\(d\)=n2​d\+δd,\|δd\|≤1\.N\(d\)=\\frac\{n\}\{2d\}\+\\delta\_\{d\},\\qquad\\lvert\\delta\_\{d\}\\rvert\\leq 1\.For distinctp,q∈Pp,q\\in Pwe have𝟏\{p∣m\}​𝟏\{q∣m\}=𝟏\{p​q∣m\}\\mathbf\{1\}\_\{\\\{p\\mid m\\\}\}\\mathbf\{1\}\_\{\\\{q\\mid m\\\}\}=\\mathbf\{1\}\_\{\\\{pq\\mid m\\\}\}, so expanding the square gives

∑m∈ℳ\(ωP​\(m\)−κ\)2\\displaystyle\\sum\_\{m\\in\\mathcal\{M\}\}\\bigl\(\\omega\_\{P\}\(m\)\-\\kappa\\bigr\)^\{2\}=∑p,q∈Pp≠qN⁡\(p​q\)\+\(1−2​κ\)​∑p∈PN⁡\(p\)\+κ2​N​\(1\)\\displaystyle=\\sum\_\{\\begin\{subarray\}\{c\}p,q\\in P\\\\ p\\neq q\\end\{subarray\}\}N\(pq\)\+\(1\-2\\kappa\)\\sum\_\{p\\in P\}N\(p\)\+\\kappa^\{2\}N\(1\)=n2​\(κ2−∑p∈P1p2\)\+n​κ2−n​κ2\+n​κ22\+ℰ\\displaystyle=\\frac\{n\}\{2\}\\Bigl\(\\kappa^\{2\}\-\\sum\_\{p\\in P\}\\frac\{1\}\{p^\{2\}\}\\Bigr\)\+\\frac\{n\\kappa\}\{2\}\-n\\kappa^\{2\}\+\\frac\{n\\kappa^\{2\}\}\{2\}\+\\mathcal\{E\}=n​κ2−n2​∑p∈P1p2\+ℰ,\\displaystyle=\\frac\{n\\kappa\}\{2\}\-\\frac\{n\}\{2\}\\sum\_\{p\\in P\}\\frac\{1\}\{p^\{2\}\}\+\\mathcal\{E\},where

ℰ=∑p,q∈Pp≠qδp​q\+\(1−2​κ\)​∑p∈Pδp\+κ2​δ1\.\\mathcal\{E\}=\\sum\_\{\\begin\{subarray\}\{c\}p,q\\in P\\\\ p\\neq q\\end\{subarray\}\}\\delta\_\{pq\}\+\(1\-2\\kappa\)\\sum\_\{p\\in P\}\\delta\_\{p\}\+\\kappa^\{2\}\\delta\_\{1\}\.Everyp∈Pp\\in Pis odd, soκ≤\|P\|/3\\kappa\\leq\\lvert P\\rvert/3and therefore

\|ℰ\|≤\(\|P\|2−\|P\|\)\+\(1\+2​κ\)​\|P\|\+κ2≤\(1\+23\+19\)​\|P\|2≤2​\|P\|2\.\\lvert\\mathcal\{E\}\\rvert\\leq\\bigl\(\\lvert P\\rvert^\{2\}\-\\lvert P\\rvert\\bigr\)\+\(1\+2\\kappa\)\\lvert P\\rvert\+\\kappa^\{2\}\\leq\\Bigl\(1\+\\frac\{2\}\{3\}\+\\frac\{1\}\{9\}\\Bigr\)\\lvert P\\rvert^\{2\}\\leq 2\\lvert P\\rvert^\{2\}\.Discarding the negative term−n2∑p∈Pp−2\-\\frac\{n\}\{2\}\\sum\_\{p\\in P\}p^\{\-2\}completes the proof\. ∎

#### C\.4\.3Proof of the main theorem

###### Proof of Theorem[C\.4\.1](https://arxiv.org/html/2608.16977#A3.SS4.Thmtheorem1)\.

The lower bound is the set of odd numbers in\[n\]\[n\], as noted above\.

For the upper bound, fixt≥1t\\geq 1and set

z=n1/3,P=\{p​an odd prime:t<p≤z\},κ=∑p∈P1p\.z=n^\{1/3\},\\qquad P=\\\{p\\text\{ an odd prime\}:t<p\\leq z\\\},\\qquad\\kappa=\\sum\_\{p\\in P\}\\frac\{1\}\{p\}\.By Mertens’ theorem∑p≤z1/p=log⁡log⁡z\+O⁡\(1\)\\sum\_\{p\\leq z\}1/p=\\log\\log z\+O\(1\), andlog⁡log⁡z=log⁡log⁡n−log⁡3\\log\\log z=\\log\\log n\-\\log 3, soκ=log⁡log⁡n\+Ot​\(1\)\\kappa=\\log\\log n\+O\_\{t\}\(1\)\. Choosen0​\(t\)n\_\{0\}\(t\)so thatκ≥max⁡\(1,12​log⁡log⁡n\)\\kappa\\geq\\max\\bigl\(1,\\tfrac\{1\}\{2\}\\log\\log n\\bigr\),n1/3≥4n^\{1/3\}\\geq 4, andlog⁡log⁡n≤n\\sqrt\{\\log\\log n\}\\leq nfor alln≥n0​\(t\)n\\geq n\_\{0\}\(t\)\.

LetA⊆\[n\]A\\subseteq\[n\]bett\-admissible and putB=\[n\]∖AB=\[n\]\\setminus A\. We claim that for eachp∈Pp\\in P,

\#\{a∈A:aeven,p∣a\}≤\#\{b∈B:bodd,p∣b\}\.\\\#\\\{a\\in A:a\\text\{ even\},\\ p\\mid a\\\}\\ \\leq\\ \\\#\\\{b\\in B:b\\text\{ odd\},\\ p\\mid b\\\}\.Indeed, leta∈Aa\\in Abe even withp\|ap\\mid a\. Asppis odd,a=2​r​pa=2rpfor somer≥1r\\geq 1, anda−p=\(2​r−1\)​pa\-p=\(2r\-1\)pis an odd multiple ofpplying in\[1,n\]\[1,n\]\. The paira−p<aa\-p<ahas differencep\>tp\>tandp\|ap\\mid a, soa−p∉Aa\-p\\notin A, that is,a−p∈Ba\-p\\in B\. The mapa↦a−pa\\mapsto a\-pis injective, which proves the claim\.

Summing the claim overp∈Pp\\in Pgives

∑a∈Aa​evenωP​\(a\)≤∑b∈Bb​oddωP​\(b\)\.\\sum\_\{\\begin\{subarray\}\{c\}a\\in A\\\\ a\\text\{ even\}\\end\{subarray\}\}\\omega\_\{P\}\(a\)\\ \\leq\\ \\sum\_\{\\begin\{subarray\}\{c\}b\\in B\\\\ b\\text\{ odd\}\\end\{subarray\}\}\\omega\_\{P\}\(b\)\.\(13\)Write

x=\#⁡\{a∈A:a​even\},y=\#⁡\{b∈B:b​odd\}\.x=\\\#\\\{a\\in A:a\\text\{ even\}\\\},\\qquad y=\\\#\\\{b\\in B:b\\text\{ odd\}\\\}\.Subtractingκ\\kappafrom every summand in equation[13](https://arxiv.org/html/2608.16977#A3.E13)and rearranging,

κ⁡\(x−y\)≤∑b∈Bb​odd\(ωP​\(b\)−κ\)−∑a∈Aa​even\(ωP​\(a\)−κ\)\.\\kappa\(x\-y\)\\ \\leq\\ \\sum\_\{\\begin\{subarray\}\{c\}b\\in B\\\\ b\\text\{ odd\}\\end\{subarray\}\}\\bigl\(\\omega\_\{P\}\(b\)\-\\kappa\\bigr\)\-\\sum\_\{\\begin\{subarray\}\{c\}a\\in A\\\\ a\\text\{ even\}\\end\{subarray\}\}\\bigl\(\\omega\_\{P\}\(a\)\-\\kappa\\bigr\)\.Apply Cauchy–Schwarz to each of the two sums, extend the resulting sums of squares to all integers of the relevant parity in\[n\]\[n\], and invoke Lemma[C\.4\.4](https://arxiv.org/html/2608.16977#A3.SS4.Thmtheorem4):

κ⁡\(x−y\)≤\(x\+y\)​\(n​κ2\+2​\|P\|2\)1/2\.\\kappa\(x\-y\)\\ \\leq\\ \\bigl\(\\sqrt\{x\}\+\\sqrt\{y\}\\bigr\)\\Bigl\(\\frac\{n\\kappa\}\{2\}\+2\\lvert P\\rvert^\{2\}\\Bigr\)^\{1/2\}\.Now\|P\|≤z=n1/3\\lvert P\\rvert\\leq z=n^\{1/3\}, andn1/3≥4≥4/κn^\{1/3\}\\geq 4\\geq 4/\\kappa, so2​\|P\|2≤2​n2/3≤n​κ/22\\lvert P\\rvert^\{2\}\\leq 2n^\{2/3\}\\leq n\\kappa/2\. Moreoverx≤n/2x\\leq n/2andy≤\(n\+1\)/2y\\leq\(n\+1\)/2, whencex\+y≤2​\(n\+1\)≤3​n\\sqrt\{x\}\+\\sqrt\{y\}\\leq\\sqrt\{2\(n\+1\)\}\\leq\\sqrt\{3n\}\. Thereforeκ⁡\(x−y\)≤3​n⋅n​κ\\kappa\(x\-y\)\\leq\\sqrt\{3n\}\\cdot\\sqrt\{n\\kappa\}, that is,

x−y≤n​3κ\.x\-y\\leq n\\sqrt\{\\frac\{3\}\{\\kappa\}\}\.The setAAhas exactly⌈n/2⌉−y\\lceil n/2\\rceil\-yodd elements andxxeven elements, so

\|A\|=⌈n2⌉−y\+x≤⌈n2⌉\+n​3κ≤n\+12\+n​6log⁡log⁡n≤n2\+3​nlog⁡log⁡n,\\lvert A\\rvert=\\Bigl\\lceil\\frac\{n\}\{2\}\\Bigr\\rceil\-y\+x\\leq\\Bigl\\lceil\\frac\{n\}\{2\}\\Bigr\\rceil\+n\\sqrt\{\\frac\{3\}\{\\kappa\}\}\\leq\\frac\{n\+1\}\{2\}\+n\\sqrt\{\\frac\{6\}\{\\log\\log n\}\}\\leq\\frac\{n\}\{2\}\+\\frac\{3n\}\{\\sqrt\{\\log\\log n\}\},the last step becauselog⁡log⁡n≤n\\sqrt\{\\log\\log n\}\\leq nand3−6\>123\-\\sqrt\{6\}\>\\tfrac\{1\}\{2\}\. SinceAAwas an arbitrarytt\-admissible subset of\[n\]\[n\], this is the desired bound onF⁡\(n,t\)F\(n;t\)\. ∎

## References\.

- Bloom \(2026\)T\. F\. Bloom\.Erdős problem \#635\.[https://www\.erdosproblems\.com/635](https://www.erdosproblems.com/635), 2026\.Accessed 12 August 2026\.
- Elliott \(2012\)Peter DTA Elliott\.*Probabilistic number theory I: Mean\-value theorems*\.Springer Science & Business Media, 2012\.
- Guy \(1983\)Richard K\. Guy\.A miscellany of Erdős problems\.*The American Mathematical Monthly*, 90\(2\):118–120, 1983\.ISSN 00029890, 19300972\.URL[http://www\.jstor\.org/stable/2975810](http://www.jstor.org/stable/2975810)\.
- Helfgott & Radziwiłł \(2021\)Harald Andres Helfgott and Maksym Radziwiłł\.Expansion, divisibility and parity\.*arXiv preprint arXiv:2103\.06853*, 2021\.
- Matomäki et al\. \(2016\)Kaisa Matomäki, Maksym Radziwiłł, and Terence Tao\.Sign patterns of the liouville and möbius functions\.In*Forum of Mathematics, Sigma*, volume 4, pp\. e14\. Cambridge University Press, 2016\.
- Ruzsa \(1999\)Imre Z Ruzsa\.Erdős and the integers\.*Journal of Number Theory*, 79\(1\):115–163, 1999\.

### C\.5Divisibility among binomial coefficients

This problem asks for the density of the set ofmmfor which some admissiblekkmakes a product ofkkconsecutive integers starting just abovenndivide the product ofkkconsecutive integers starting just abovemm\.[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2)asked whether all, or almost all, largemmadmit such akk, and if not what the density of those that do is; we show that the density exists and equals11for every fixedn≥2n\\geq 2\. The proof produces, for eachBB, a singlekkfor which\(n\+kn\)\\binom\{n\+k\}\{n\}has no prime factor belowBB, so that the required divisibility is forced by one congruence condition onmmmodulo each prime divisor of that coefficient\.

#### C\.5\.1Introduction

For positive integersnnandkkwriteA⁡\(n,k\)=\(n\+k\)\!/n\!A\(n,k\)=\(n\+k\)\!/n\!for the product of thekkconsecutive integersn\+1,…,n\+kn\+1,\\dots,n\+k\.[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2)study the divisibility relationA⁡\(n,k\)\|A⁡\(m,k\)A\(n,k\)\\mid A\(m,k\)form\>nm\>n\. Since

A⁡\(m,k\)A⁡\(n,k\)=\(m\+kk\)/\(n\+kk\),\\frac\{A\(m,k\)\}\{A\(n,k\)\}=\\binom\{m\+k\}\{k\}\\Bigm/\\binom\{n\+k\}\{k\},that relation is exactly the divisibility

\(n\+kn\)\|\(m\+kk\)\\binom\{n\+k\}\{n\}\\ \\Bigm\|\\ \\binom\{m\+k\}\{k\}\(14\)between two binomial coefficients\. Boundingkkis essential here\. Indeed, fork\>m−nk\>m\-none has

A⁡\(m,k\)A⁡\(n,k\)=A⁡\(n\+k,m−n\)A⁡\(n,m−n\),\\frac\{A\(m,k\)\}\{A\(n,k\)\}=\\frac\{A\(n\+k,m\-n\)\}\{A\(n,m\-n\)\},and[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2)note that this quotient is an integer fork=A⁡\(n,m−n\)−mk=A\(n,m\-n\)\-m, a value exceedingm−nm\-nas soon asm≥n\+2m\\geq n\+2\. Form=n\+1m=n\+1that value is00; there one may takek=n\+1k=n\+1instead, with quotient22\. Without a restriction onkkthe problem is therefore vacuous\. With the natural restriction1≤k≤m−n1\\leq k\\leq m\-nin place, they ask the following question in their §1, in which the displayed divisibility is their \(1\.4\)\.

*Givenn\>1n\>1is it true that for all \(almost all\) largemmthere exists akk,1≤k≤m−n1\\leq k\\leq m\-nso that\(k\+nn\)\|\(m\+kk\)\\binom\{k\+n\}\{n\}\\mid\\binom\{m\+k\}\{k\}? If not, what is the densityd∗​\(n\)d^\{\*\}\(n\)of integersmmfor which\(1\.4\)has a solution with1≤k≤m−n1\\leq k\\leq m\-n?*

For a fixed integern≥1n\\geq 1let

Gn:=\{m∈ℕ:m\>n,∃k∈ℕ,1≤k≤m−n,\(n\+kn\)\|\(m\+kk\)\},G\_\{n\}:=\\Bigl\\\{m\\in\\mathbb\{N\}:\\ m\>n,\\ \\exists\\,k\\in\\mathbb\{N\},\\ 1\\leq k\\leq m\-n,\\ \\binom\{n\+k\}\{n\}\\Bigm\|\\binom\{m\+k\}\{k\}\\Bigr\\\},and for a setS⊆ℕS\\subseteq\\mathbb\{N\}write

d⁡\(S\):=limx→∞\|S∩\[1,x\]\|xd\(S\):=\\lim\_\{x\\to\\infty\}\\frac\{\|S\\cap\[1,x\]\|\}\{x\}for its natural density, when the limit exists, so thatd∗​\(n\)=d⁡\(Gn\)d^\{\*\}\(n\)=d\(G\_\{n\}\)\. We prove thatd∗​\(n\)d^\{\*\}\(n\)exists and equals11for every fixedn≥2n\\geq 2, so that the “almost all” alternative holds\.

The casen=1n=1, excluded from the question above, is settled outright in[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2), and its proof is the carry argument used below\. For a primepp, takingk=p−1k=p\-1turns equation[14](https://arxiv.org/html/2608.16977#A3.E14)into the assertionp\|\(m\+p−1p−1\)p\\mid\\binom\{m\+p\-1\}\{p\-1\}\. The base\-ppdigits ofp−1p\-1are\(p−1,0,0,…\)\(p\-1,0,0,\\dots\), so addingmmandp−1p\-1in baseppproduces a carry out of the units place as soon asp∤mp\\nmid m, and Kummer’s theorem then givesp\|\(m\+p−1p−1\)p\\mid\\binom\{m\+p\-1\}\{p\-1\}\. Nom\>2m\>2is divisible by every primep≤mp\\leq m, and any primep≤mp\\leq mwithp∤mp\\nmid msupplies an admissiblek=p−1k=p\-1in\[1,m−1\]\[1,m\-1\], so everym\>2m\>2lies inG1G\_\{1\}\. Erdős and Straus write that already the next case,n=2n=2, “seems much more difficult to decide”\.

Erdős and Straus themselves prove two results about equation[14](https://arxiv.org/html/2608.16977#A3.E14), both in the regime wherem/nm/nis bounded\.[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2)show that for fixedc\>0c\>0andΛ\>1\\Lambda\>1only finitely many triplesn,k,mn,k,mwithk≥c​nk\\geq cnandn\+k≤m≤Λ​nn\+k\\leq m\\leq\\Lambda nsatisfy it, and their Theorem 3\.3 removes the hypothesisk≥c​nk\\geq cnat the cost of requiringk≥2k\\geq 2and a prime in\[n\+1,n\+k\]\[n\+1,n\+k\]\. Both leave the range studied here untouched, since we fixnnand letm→∞m\\to\\infty, so thatm/n→∞m/n\\to\\infty\. Apart from these, work on equation[14](https://arxiv.org/html/2608.16977#A3.E14)has concentrated on the boundary casek=m−nk=m\-n, that is, on the question whether for each fixednnsomekksatisfies\(n\+kn\)\|\(n\+2​kk\)\\binom\{n\+k\}\{n\}\\mid\\binom\{n\+2k\}\{k\}\. That case is raised in[Erdős & Straus 1977](https://arxiv.org/html/2608.16977#bibe.bib2), recorded by[Erdös & Graham 1980](https://arxiv.org/html/2608.16977#bibe.bib3), and appears as Problem 389 on the Erdős problems website\([Bloom 2026](https://arxiv.org/html/2608.16977#bibe.bib1)\), where it is listed as open;[Ulas & Schinzel 2013](https://arxiv.org/html/2608.16977#bibe.bib7)gives computational results for it and for a companion question of Erdős and Graham, verifying that a suitablekkexists for everyn≤20n\\leq 20, and settling the companion Erdős–Graham question for everyn≤9n\\leq 9\.

[Pomerance 2015](https://arxiv.org/html/2608.16977#bibe.bib5)proves that for each fixedk≥1k\\geq 1the set ofMMwithM\+k\|\(2​MM\)M\+k\\mid\\binom\{2M\}\{M\}has asymptotic density11, and notes that the same proof yields density one for the divisibility\(M\+1\)\(M\+2\)⋯\(M\+k\)∣\(2​MM\)\(M\+1\)\(M\+2\)\\cdots\(M\+k\)\\mid\\binom\{2M\}\{M\}\.[Pomerance 2026](https://arxiv.org/html/2608.16977#bibe.bib6)allowskkto grow withMM: the product\(M\+1\)⋯\(M\+k\)\(M\+1\)\\cdots\(M\+k\)divides\(2​MM\)\\binom\{2M\}\{M\}for allk≤η​log⁡Mk\\leq\\eta\\log M, for any fixedη<1/log⁡4\\eta<1/\\log 4, and\(M\+kk\)\\binom\{M\+k\}\{k\}divides\(2​MM\)\\binom\{2M\}\{M\}for allk≤e0\.8​log⁡Mk\\leq e^\{0\.8\\sqrt\{\\log M\}\}, in both cases for a set ofMMof density11\. The mechanism there is the one we use: a high power ofppdividingM\+kM\+kforces the low base\-ppdigits ofMMto be large, and Kummer’s theorem converts this into carries\. None of these statements implies our theorem: in each of them the dividend is the central binomial coefficient\(2​MM\)\\binom\{2M\}\{M\}andkkis small compared withMM, whereas in equation[14](https://arxiv.org/html/2608.16977#A3.E14)the dividend is\(m\+kk\)\\binom\{m\+k\}\{k\}and the divisor is\(n\+kn\)\\binom\{n\+k\}\{n\}withnnfixed andkkunbounded\. In a similar vein[Harborth 1979](https://arxiv.org/html/2608.16977#bibe.bib4)showed that for each fixedkkalmost all entries\(Mj\)\\binom\{M\}\{j\}of Pascal’s triangle are divisible byM\(M−1\)⋯\(M−k\+1\)M\(M\-1\)\\cdots\(M\-k\+1\); see the discussion in[Pomerance 2015](https://arxiv.org/html/2608.16977#bibe.bib5)\. We have found no prior source asserting thatGnG\_\{n\}has positive lower density, let alone density one\.

Throughout,ppandqqdenote primes,vp​\(⋅\)v\_\{p\}\(\\cdot\)is thepp\-adic valuation,π⁡\(B\)=∑p≤B1\\pi\(B\)=\\sum\_\{p\\leq B\}1is the prime\-counting function,ϑ⁡\(B\)=∑p≤Blog⁡p\\vartheta\(B\)=\\sum\_\{p\\leq B\}\\log pis Chebyshev’s function, andω⁡\(N\)\\omega\(N\)is the number of distinct prime divisors ofNN\.

###### Theorem C\.5\.1\.

For every fixed integern≥2n\\geq 2the natural densityd∗​\(n\)d^\{\*\}\(n\)exists and

SinceGn⊆ℕG\_\{n\}\\subseteq\\mathbb\{N\}, its upper density is at most11, so the entire content of the theorem is a lower bound on the lower density\.

We now summarize the proof\. For each integer parameterBBwe exhibit one valuek=kBk=k\_\{B\}of the free variable in equation[14](https://arxiv.org/html/2608.16977#A3.E14)that works simultaneously for everymmin an explicit periodic setTBT\_\{B\}, and we show that the density ofTBT\_\{B\}tends to11asB→∞B\\to\\infty\. The key point is to choosekBk\_\{B\}so thatNB=\(n\+kBn\)N\_\{B\}=\\binom\{n\+k\_\{B\}\}\{n\}has no prime factor belowBB\. Every primeq\|NBq\\mid N\_\{B\}then divides exactly one of thennintegerskB\+1,…,kB\+nk\_\{B\}\+1,\\dots,k\_\{B\}\+n, saykB\+i⁡\(q\)k\_\{B\}\+i\(q\), and does so to the full multiplicityvq​\(NB\)v\_\{q\}\(N\_\{B\}\)\. The single congruence conditionmmodq≥i⁡\(q\)m\\bmod q\\geq i\(q\)is then enough to forceqvq​\(NB\)\|\(m\+kBkB\)q^\{v\_\{q\}\(N\_\{B\}\)\}\\mid\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}, becausekBk\_\{B\}has residueqs−i⁡\(q\)q^\{s\}\-i\(q\)moduloqsq^\{s\}for everys≤vq​\(NB\)s\\leq v\_\{q\}\(N\_\{B\}\)and each suchsstherefore contributes a carry\. Only Kummer’s theorem, Legendre’s formula, the Chinese remainder theorem and Chebyshev’s elementary bounds onπ\\piandϑ\\varthetaare needed\.

#### C\.5\.2A binomial coefficient free of small primes

Fixn≥2n\\geq 2\. For an integer parameterB\>nB\>ndefine

kB:=∏p≤Bpep,ep:=min⁡\{e≥1:pe\>n\},k\_\{B\}:=\\prod\_\{p\\leq B\}p^\{e\_\{p\}\},\\qquad e\_\{p\}:=\\min\\\{e\\geq 1:\\ p^\{e\}\>n\\\},\(15\)and set

NB:=\(n\+kBn\)\.N\_\{B\}:=\\binom\{n\+k\_\{B\}\}\{n\}\.
###### Lemma C\.5\.3\.

Every prime divisor ofNBN\_\{B\}is larger thanBB\.

###### Proof\.

Letp≤Bp\\leq B\. By equation[15](https://arxiv.org/html/2608.16977#A3.E15)we havepep\|kBp^\{e\_\{p\}\}\\mid k\_\{B\}, so theepe\_\{p\}lowest base\-ppdigits ofkBk\_\{B\}vanish, whilen<pepn<p^\{e\_\{p\}\}, so every base\-ppdigit ofnnof indexepe\_\{p\}or higher vanishes\. AddingnnandkBk\_\{B\}in basepptherefore produces no carry at all: in each of the positions0,…,ep−10,\\dots,e\_\{p\}\-1the digit ofkBk\_\{B\}is00, and in each higher position the digit ofnnis00, so no position ever sums toppor more\. By Kummer’s theoremvp​\(n\+kBn\)v\_\{p\}\\binom\{n\+k\_\{B\}\}\{n\}equals the number of carries in this addition, whence

vp​\(NB\)=0\.v\_\{p\}\(N\_\{B\}\)=0\.Thus no primep≤Bp\\leq BdividesNBN\_\{B\}\. ∎

###### Lemma C\.5\.4\.

Letqqbe a prime withq\|NBq\\mid N\_\{B\}and putaq:=vq​\(NB\)a\_\{q\}:=v\_\{q\}\(N\_\{B\}\)\. Then there is a unique indexi⁡\(q\)∈\{1,…,n\}i\(q\)\\in\\\{1,\\dots,n\\\}withq\|kB\+i⁡\(q\)q\\mid k\_\{B\}\+i\(q\), and for this index

qaq\|kB\+i⁡\(q\),q^\{a\_\{q\}\}\\ \\big\\\|\\ k\_\{B\}\+i\(q\),that is,aq=vq​\(kB\+i⁡\(q\)\)a\_\{q\}=v\_\{q\}\\bigl\(k\_\{B\}\+i\(q\)\\bigr\)\.

###### Proof\.

By Lemma[C\.5\.3](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem3)we haveq\>B\>nq\>B\>n, soq∤n\!q\\nmid n\!and hence

vq​\(NB\)=vq​\(∏i=1n\(kB\+i\)\)−vq​\(n\!\)=∑i=1nvq​\(kB\+i\)\.v\_\{q\}\(N\_\{B\}\)=v\_\{q\}\\Bigl\(\\prod\_\{i=1\}^\{n\}\(k\_\{B\}\+i\)\\Bigr\)\-v\_\{q\}\(n\!\)=\\sum\_\{i=1\}^\{n\}v\_\{q\}\(k\_\{B\}\+i\)\.The left\-hand side is positive, so at least one summand is positive\. Two distinct indices1≤i<j≤n1\\leq i<j\\leq nwithq\|kB\+iq\\mid k\_\{B\}\+iandq\|kB\+jq\\mid k\_\{B\}\+jwould giveq\|j−iq\\mid j\-iwith0<j−i<n<q0<j\-i<n<q, which is impossible\. Hence exactly one indexi⁡\(q\)i\(q\)contributes, and the displayed identity reduces toaq=vq​\(kB\+i⁡\(q\)\)a\_\{q\}=v\_\{q\}\(k\_\{B\}\+i\(q\)\)\. ∎

#### C\.5\.3A congruence forcing the divisibility

###### Lemma C\.5\.5\.

Letq\|NBq\\mid N\_\{B\}, writei=i⁡\(q\)i=i\(q\)anda=aqa=a\_\{q\}\. Ifmmis a nonnegative integer with

mmodq∈\{i,i\+1,…,q−1\},m\\bmod q\\in\\\{i,i\+1,\\dots,q\-1\\\},then

vq​\(m\+kBkB\)≥a\.v\_\{q\}\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}\\ \\geq\\ a\.

###### Proof\.

Fixsswith1≤s≤a1\\leq s\\leq a\. By Lemma[C\.5\.4](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem4)we haveqa\|kB\+iq^\{a\}\\mid k\_\{B\}\+i, henceqs\|kB\+iq^\{s\}\\mid k\_\{B\}\+i, sokB≡−i\(modqs\)k\_\{B\}\\equiv\-i\\pmod\{q^\{s\}\}\. Since1≤i≤n<q≤qs1\\leq i\\leq n<q\\leq q^\{s\}, the residue ofkBk\_\{B\}moduloqsq^\{s\}is exactlyqs−iq^\{s\}\-i, that is,

kB=us​qs\+\(qs−i\)for some integer​us=⌊kB/qs⌋≥0\.k\_\{B\}=u\_\{s\}q^\{s\}\+\(q^\{s\}\-i\)\\qquad\\text\{for some integer \}u\_\{s\}=\\lfloor k\_\{B\}/q^\{s\}\\rfloor\\geq 0\.Writem=vs​qs\+tsm=v\_\{s\}q^\{s\}\+t\_\{s\}with0≤ts<qs0\\leq t\_\{s\}<q^\{s\}\. Sincets≡m\(modq\)t\_\{s\}\\equiv m\\pmod\{q\}andts≥0t\_\{s\}\\geq 0, we havets≥tsmodq=mmodq≥it\_\{s\}\\geq t\_\{s\}\\bmod q=m\\bmod q\\geq i\. Consequently

i≤ts<qs,soqs≤ts\+qs−i<2​qs,i\\leq t\_\{s\}<q^\{s\},\\qquad\\text\{so\}\\qquad q^\{s\}\\ \\leq\\ t\_\{s\}\+q^\{s\}\-i\\ <\\ 2q^\{s\},and therefore

⌊m\+kBqs⌋−⌊mqs⌋−⌊kBqs⌋=\(vs\+us\)\+⌊ts\+qs−iqs⌋−vs−us=⌊ts\+qs−iqs⌋=1\.\\Bigl\\lfloor\\frac\{m\+k\_\{B\}\}\{q^\{s\}\}\\Bigr\\rfloor\-\\Bigl\\lfloor\\frac\{m\}\{q^\{s\}\}\\Bigr\\rfloor\-\\Bigl\\lfloor\\frac\{k\_\{B\}\}\{q^\{s\}\}\\Bigr\\rfloor=\(v\_\{s\}\+u\_\{s\}\)\+\\Bigl\\lfloor\\frac\{t\_\{s\}\+q^\{s\}\-i\}\{q^\{s\}\}\\Bigr\\rfloor\-v\_\{s\}\-u\_\{s\}=\\Bigl\\lfloor\\frac\{t\_\{s\}\+q^\{s\}\-i\}\{q^\{s\}\}\\Bigr\\rfloor=1\.By Legendre’s formula applied to\(m\+kB\)\!\(m\+k\_\{B\}\)\!,m\!m\!andkB\!k\_\{B\}\!,

vq​\(m\+kBkB\)=∑s≥1\(⌊m\+kBqs⌋−⌊mqs⌋−⌊kBqs⌋\)\.v\_\{q\}\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}=\\sum\_\{s\\geq 1\}\\Bigl\(\\Bigl\\lfloor\\frac\{m\+k\_\{B\}\}\{q^\{s\}\}\\Bigr\\rfloor\-\\Bigl\\lfloor\\frac\{m\}\{q^\{s\}\}\\Bigr\\rfloor\-\\Bigl\\lfloor\\frac\{k\_\{B\}\}\{q^\{s\}\}\\Bigr\\rfloor\\Bigr\)\.Each summand is nonnegative, since⌊α\+β⌋≥⌊α⌋\+⌊β⌋\\lfloor\\alpha\+\\beta\\rfloor\\geq\\lfloor\\alpha\\rfloor\+\\lfloor\\beta\\rfloor, and theaasummands with1≤s≤a1\\leq s\\leq aeach equal11\. Hence the sum is at leastaa\. ∎

Observe that the hypothesis of Lemma[C\.5\.5](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem5)is a condition moduloqqalone, even though its conclusion concerns the prime powerqaq^\{a\}\. This is what makes the Chinese remainder theorem cheap to apply\. Define

TB:=\{m∈ℕ:mmodq∈\{i\(q\),i\(q\)\+1,…,q−1\}for every primeq∣NB\}\.T\_\{B\}:=\\Bigl\\\{m\\in\\mathbb\{N\}:\\ m\\bmod q\\in\\\{i\(q\),i\(q\)\+1,\\dots,q\-1\\\}\\ \\text\{ for every prime \}q\\mid N\_\{B\}\\Bigr\\\}\.\(16\)
###### Corollary C\.5\.6\.

For everym∈TBm\\in T\_\{B\}we haveNB\|\(m\+kBkB\)N\_\{B\}\\mid\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}\. Consequently

TB∩\[n\+kB,∞\)⊆Gn\.T\_\{B\}\\cap\[\\,n\+k\_\{B\},\\infty\)\\ \\subseteq\\ G\_\{n\}\.

###### Proof\.

Letm∈TBm\\in T\_\{B\}\. For each primeq\|NBq\\mid N\_\{B\}, Lemma[C\.5\.5](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem5)givesvq​\(m\+kBkB\)≥aq=vq​\(NB\)v\_\{q\}\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}\\geq a\_\{q\}=v\_\{q\}\(N\_\{B\}\); henceNB\|\(m\+kBkB\)N\_\{B\}\\mid\\binom\{m\+k\_\{B\}\}\{k\_\{B\}\}\. If moreoverm≥n\+kBm\\geq n\+k\_\{B\}, thenk:=kBk:=k\_\{B\}satisfies1≤k≤m−n1\\leq k\\leq m\-nand\(n\+kn\)=NB\\binom\{n\+k\}\{n\}=N\_\{B\}divides\(m\+kk\)\\binom\{m\+k\}\{k\}, som∈Gnm\\in G\_\{n\}\. ∎

###### Example C\.5\.7\.

Taken=2n=2andB=5B=5\. Thene2=2e\_\{2\}=2ande3=e5=1e\_\{3\}=e\_\{5\}=1, sok5=4⋅3⋅5=60k\_\{5\}=4\\cdot 3\\cdot 5=60and

N5=\(622\)=1891=31⋅61,N\_\{5\}=\\binom\{62\}\{2\}=1891=31\\cdot 61,whose prime factors indeed exceed55\. Here31\|62=k5\+231\\mid 62=k\_\{5\}\+2and61\|61=k5\+161\\mid 61=k\_\{5\}\+1, soi⁡\(31\)=2i\(31\)=2andi⁡\(61\)=1i\(61\)=1, both withaq=1a\_\{q\}=1\. Corollary[C\.5\.6](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem6)therefore says that

\(622\)\|\(m\+6060\)whenever​m≥62,m≢0,1\(mod31\),m≢0\(mod61\),\\binom\{62\}\{2\}\\ \\Bigm\|\\ \\binom\{m\+60\}\{60\}\\qquad\\text\{whenever \}m\\geq 62,\\quad m\\not\\equiv 0,1\\pmod\{31\},\\quad m\\not\\equiv 0\\pmod\{61\},a set of density2931⋅6061=17401891\>0\.92\\tfrac\{29\}\{31\}\\cdot\\tfrac\{60\}\{61\}=\\tfrac\{1740\}\{1891\}\>0\.92\. The congruence conditions are sufficient but not necessary: for instancem=930m=930is excluded by the condition at3131, yet\(622\)\|\(99060\)\\binom\{62\}\{2\}\\mid\\binom\{990\}\{60\}\. Indeed930930and6060have base\-3131digits\(0,30\)\(0,30\)and\(29,1\)\(29,1\), written from least to most significant, so their addition still carries out of the second place and31\|\(99060\)31\\mid\\binom\{990\}\{60\}, while930mod61=15≥i⁡\(61\)930\\bmod 61=15\\geq i\(61\)leaves930930inside the condition at6161\.

#### C\.5\.4The density of the congruence set

The definition equation[16](https://arxiv.org/html/2608.16977#A3.E16)imposes congruence conditions modulo the finitely many distinct primesq\|NBq\\mid N\_\{B\}, soTBT\_\{B\}is a union of residue classes modulo∏q\|NBq\\prod\_\{q\\mid N\_\{B\}\}qand its natural densityd⁡\(TB\)d\(T\_\{B\}\)exists\. Sincei⁡\(q\)≤n<qi\(q\)\\leq n<q, the condition atqqexcludes exactly thei⁡\(q\)i\(q\)residues0,1,…,i⁡\(q\)−10,1,\\dots,i\(q\)\-1, so by the Chinese remainder theorem

d⁡\(TB\)=∏q\|NB\(1−i⁡\(q\)q\)≥∏q\|NB\(1−nq\)\>0\.d\(T\_\{B\}\)=\\prod\_\{q\\mid N\_\{B\}\}\\Bigl\(1\-\\frac\{i\(q\)\}\{q\}\\Bigr\)\\ \\geq\\ \\prod\_\{q\\mid N\_\{B\}\}\\Bigl\(1\-\\frac\{n\}\{q\}\\Bigr\)\\ \>\\ 0\.\(17\)
###### Lemma C\.5\.8\.

For every integerB\>nB\>nwe havelog⁡kB=On​\(B\)\\log k\_\{B\}=O\_\{n\}\(B\)and

ω⁡\(NB\)=On​\(Blog⁡B\)\.\\omega\(N\_\{B\}\)=O\_\{n\}\\\!\\Bigl\(\\frac\{B\}\{\\log B\}\\Bigr\)\.

###### Proof\.

By the minimality ofepe\_\{p\}in equation[15](https://arxiv.org/html/2608.16977#A3.E15)we havepep−1≤np^\{e\_\{p\}\-1\}\\leq n, whenceep​log⁡p≤log⁡n\+log⁡pe\_\{p\}\\log p\\leq\\log n\+\\log p, and therefore

log⁡kB=∑p≤Bep​log⁡p≤π⁡\(B\)​log⁡n\+ϑ⁡\(B\)\.\\log k\_\{B\}=\\sum\_\{p\\leq B\}e\_\{p\}\\log p\\leq\\pi\(B\)\\log n\+\\vartheta\(B\)\.Chebyshev’s boundsπ⁡\(B\)=O⁡\(B/log⁡B\)\\pi\(B\)=O\(B/\\log B\)andϑ⁡\(B\)=O⁡\(B\)\\vartheta\(B\)=O\(B\)givelog⁡kB=On​\(B\)\\log k\_\{B\}=O\_\{n\}\(B\)\.

For the second assertion, note thatNB⋅n\!=∏i=1n\(kB\+i\)N\_\{B\}\\cdot n\!=\\prod\_\{i=1\}^\{n\}\(k\_\{B\}\+i\), so every prime divisor ofNBN\_\{B\}divides somekB\+ik\_\{B\}\+iwith1≤i≤n1\\leq i\\leq n\. Fix such aniiand letttbe the number of distinct primes exceedingBBthat dividekB\+ik\_\{B\}\+i\. Each such prime is at leastB\+1B\+1becauseBBis an integer, so their product divideskB\+ik\_\{B\}\+iand is at least\(B\+1\)t\(B\+1\)^\{t\}, whence

t≤log⁡\(kB\+n\)log⁡\(B\+1\)\.t\\leq\\frac\{\\log\(k\_\{B\}\+n\)\}\{\\log\(B\+1\)\}\.By Lemma[C\.5\.3](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem3)every prime divisor ofNBN\_\{B\}exceedsBB, so summing over thennvalues ofiigives

ω⁡\(NB\)≤n​log⁡\(kB\+n\)log⁡\(B\+1\)=On​\(Blog⁡B\),\\omega\(N\_\{B\}\)\\leq n\\,\\frac\{\\log\(k\_\{B\}\+n\)\}\{\\log\(B\+1\)\}=O\_\{n\}\\\!\\Bigl\(\\frac\{B\}\{\\log B\}\\Bigr\),where the last step useslog⁡kB=On​\(B\)\\log k\_\{B\}=O\_\{n\}\(B\)\. ∎

###### Lemma C\.5\.9\.

We haved⁡\(TB\)→1d\(T\_\{B\}\)\\to 1asB→∞B\\to\\infty\.

###### Proof\.

Assume thatB≥2​nB\\geq 2n\. Every primeq\|NBq\\mid N\_\{B\}satisfiesq≥B\+1\>2​nq\\geq B\+1\>2nby Lemma[C\.5\.3](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem3)and the integrality ofBB, son/q<1/2n/q<1/2\. For0≤x≤1/20\\leq x\\leq 1/2one haslog⁡\(1−x\)≥−2​x\\log\(1\-x\)\\geq\-2x; indeedh⁡\(x\):=log⁡\(1−x\)\+2​xh\(x\):=\\log\(1\-x\)\+2xsatisfiesh⁡\(0\)=0h\(0\)=0andh′​\(x\)=2−11−x≥0h^\{\\prime\}\(x\)=2\-\\tfrac\{1\}\{1\-x\}\\geq 0on\[0,1/2\]\[0,1/2\], soh≥0h\\geq 0there\. Applying this to each factor of equation[17](https://arxiv.org/html/2608.16977#A3.E17),

logd\(TB\)≥∑q\|NBlog\(1−nq\)≥−2∑q\|NBnq≥−2​n​ω​\(NB\)B\+1,\\log d\(T\_\{B\}\)\\ \\geq\\ \\sum\_\{q\\mid N\_\{B\}\}\\log\\Bigl\(1\-\\frac\{n\}\{q\}\\Bigr\)\\ \\geq\\ \-2\\sum\_\{q\\mid N\_\{B\}\}\\frac\{n\}\{q\}\\ \\geq\\ \-\\frac\{2n\\,\\omega\(N\_\{B\}\)\}\{B\+1\},where the last step usesq≥B\+1q\\geq B\+1for every suchqq\. By Lemma[C\.5\.8](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem8)the right\-hand side isOn​\(1/log⁡B\)O\_\{n\}\(1/\\log B\)in absolute value, hencelog⁡d⁡\(TB\)→0\\log d\(T\_\{B\}\)\\to 0andd⁡\(TB\)→1d\(T\_\{B\}\)\\to 1\. ∎

Note thatd⁡\(TB\)d\(T\_\{B\}\)need not increase withBB\. Continuing Example[C\.5\.7](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem7), we havek13=4⋅3⋅5⋅7⋅11⋅13=60060k\_\{13\}=4\\cdot 3\\cdot 5\\cdot 7\\cdot 11\\cdot 13=60060and

N13=\(600622\)=17⋅59⋅509⋅3533,N\_\{13\}=\\binom\{60062\}\{2\}=17\\cdot 59\\cdot 509\\cdot 3533,withi⁡\(17\)=i⁡\(3533\)=1i\(17\)=i\(3533\)=1andi⁡\(59\)=i⁡\(509\)=2i\(59\)=i\(509\)=2, so that equation[17](https://arxiv.org/html/2608.16977#A3.E17)gives

d⁡\(T13\)=1617⋅5759⋅507509⋅35323533=0\.9054​…<0\.9201​…=d⁡\(T5\)\.d\(T\_\{13\}\)=\\frac\{16\}\{17\}\\cdot\\frac\{57\}\{59\}\\cdot\\frac\{507\}\{509\}\\cdot\\frac\{3532\}\{3533\}=0\.9054\\ldots\\ <\\ 0\.9201\\ldots=d\(T\_\{5\}\)\.RaisingBBfrom55to1313has replaced two prime factors by four, one of them barely aboveBB\. Only the limit is asserted\.

#### C\.5\.5Proof of the main theorem

###### Proof of Theorem[C\.5\.1](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem1)\.

Letε\>0\\varepsilon\>0\. By Lemma[C\.5\.9](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem9)choose an integerB≥2​nB\\geq 2nwithd⁡\(TB\)\>1−εd\(T\_\{B\}\)\>1\-\\varepsilon\. By Corollary[C\.5\.6](https://arxiv.org/html/2608.16977#A3.SS5.Thmtheorem6)every element ofTBT\_\{B\}that is at leastn\+kBn\+k\_\{B\}lies inGnG\_\{n\}, so for everyx≥n\+kBx\\geq n\+k\_\{B\},

\|Gn∩\[1,x\]\|≥\|TB∩\[n\+kB,x\]\|≥\|TB∩\[1,x\]\|−\(n\+kB\)\.\|G\_\{n\}\\cap\[1,x\]\|\\ \\geq\\ \\bigl\|T\_\{B\}\\cap\[n\+k\_\{B\},x\]\\bigr\|\\ \\geq\\ \|T\_\{B\}\\cap\[1,x\]\|\-\(n\+k\_\{B\}\)\.The subtracted quantity is independent ofxx, so dividing byxxand lettingx→∞x\\to\\inftygives

lim infx→∞\|Gn∩\[1,x\]\|x≥limx→∞\|TB∩\[1,x\]\|x=d⁡\(TB\)\>1−ε\.\\liminf\_\{x\\to\\infty\}\\frac\{\|G\_\{n\}\\cap\[1,x\]\|\}\{x\}\\ \\geq\\ \\lim\_\{x\\to\\infty\}\\frac\{\|T\_\{B\}\\cap\[1,x\]\|\}\{x\}=d\(T\_\{B\}\)\>1\-\\varepsilon\.Sinceε\>0\\varepsilon\>0was arbitrary, the lower density ofGnG\_\{n\}is11, while its upper density is trivially at most11\. Hence the natural density exists andd∗​\(n\)=1d^\{\*\}\(n\)=1, completing the proof\. ∎

## References\.

- Bloom \(2026\)T\. F\. Bloom\.Erdős problem \#389\.[https://www\.erdosproblems\.com/389](https://www.erdosproblems.com/389), 2026\.Accessed 12 August 2026\.
- Erdős & Straus \(1977\)P\. Erdős and E\. G\. Straus\.On products of consecutive integers\.In*Number Theory and Algebra*, pp\. 63–70\. Academic Press, New York, 1977\.
- Erdös & Graham \(1980\)Paul Erdös and Ronald L Graham\.*Old and new problems and results in combinatorial number theory*, volume 28\.L’Enseignement Mathematiques Un\. Geneve, 1980\.
- Harborth \(1979\)Heiko Harborth\.Divisibility of\.*The American Mathematical Monthly*, 86\(2\):115–117, 1979\.
- Pomerance \(2015\)Carl Pomerance\.Divisors of the middle binomial coefficient\.*The American Mathematical Monthly*, 122\(7\):636–644, 2015\.
- Pomerance \(2026\)Carl Pomerance\.Remarks on the middle binomial coefficient\.*Integers*, 26, 2026\.
- Ulas & Schinzel \(2013\)Maciej Ulas and Andrzej Schinzel\.A note on erdős–straus and erdős–graham divisibility problems \(with an appendix by andrzej schinzel\)\.*International Journal of Number Theory*, 9\(03\):583–599, 2013\.

### C\.6Many\-one𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-completeness for binary symmetric group characters

This problem concerns the complexity of evaluating an irreducible character of the symmetric group when both partitions are written in binary\.[Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)proved that this evaluation is𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under Turing reductions and conjectured that many\-one reductions already suffice; we prove the conjecture, in the sharper form that the partition indexing the character may always be taken to have at most two parts\. The reduction encodes an arbitrary𝖦𝖺𝗉𝖯\\mathsf\{GapP\}function as a subset\-sum count in a large base and reads the value off a two\-row character, which by a classical identity is a difference of two such counts at consecutive targets\.

#### C\.6\.1Introduction

For a partitionλ⊢n\\lambda\\vdash nletχλ\\chi^\{\\lambda\}denote the irreducible character of the symmetric group𝔖n\\mathfrak\{S\}\_\{n\}indexed byλ\\lambda\. Its value at a permutation depends only on the cycle type of that permutation, so forμ⊢n\\mu\\vdash nwe writeχλ​\(μ\)\\chi^\{\\lambda\}\(\\mu\)for the common value ofχλ\\chi^\{\\lambda\}on the conjugacy class of cycle typeμ\\mu\. The problemComputeCharBinarytakes as input two partitionsλ,μ⊢n\\lambda,\\mu\\vdash n, each presented as a list of parts written in binary, and outputs the integerχλ​\(μ\)\\chi^\{\\lambda\}\(\\mu\)\. Because a character value can be negative, the natural home for this function is not the counting class\#​𝖯\\\#\\mathsf\{P\}of[Valiant 1979](https://arxiv.org/html/2608.16977#bibf.bib13)but the gap class𝖦𝖺𝗉𝖯=\#​𝖯−\#​𝖯\\mathsf\{GapP\}=\\\#\\mathsf\{P\}\-\\\#\\mathsf\{P\}of differences of two\#​𝖯\\\#\\mathsf\{P\}functions, introduced by[Fenner et al\. 1994](https://arxiv.org/html/2608.16977#bibf.bib2)\. A functionFFis*𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-hard under many\-one reductions*if for everyf∈𝖦𝖺𝗉𝖯f\\in\\mathsf\{GapP\}there is a polynomial\-time computable mapRRwithf⁡\(x\)=F⁡\(R⁡\(x\)\)f\(x\)=F\(R\(x\)\)for allxx, and*𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete*if moreoverF∈𝖦𝖺𝗉𝖯F\\in\\mathsf\{GapP\}\. This is stronger than hardness under Turing reductions, whereffis only required to be computable in polynomial time given an oracle forFF; we follow[Papadimitriou 2003](https://arxiv.org/html/2608.16977#bibf.bib10)for the standard conventions\.

[Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)proved that decidingχλ​\(μ\)=0\\chi^\{\\lambda\}\(\\mu\)=0is𝖢=​𝖯\\mathsf\{C\}\_\{=\}\\mathsf\{P\}\-complete and that decidingχλ​\(μ\)≥0\\chi^\{\\lambda\}\(\\mu\)\\geq 0is𝖯𝖯\\mathsf\{PP\}\-complete, both under many\-one reductions, and deduced that neither\|χλ​\(μ\)\|\|\\chi^\{\\lambda\}\(\\mu\)\|norχλ​\(μ\)2\\chi^\{\\lambda\}\(\\mu\)^\{2\}lies in\#​𝖯\\\#\\mathsf\{P\}unless the polynomial hierarchy collapses to its second level\. As a byproduct of the same reduction they obtained thatComputeCharBinaryis𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under Turing reductions\([Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)\), and they noted that their route cannot be pushed as far as a parsimonious reduction\. They then stated the following\([Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6), Conjecture 5\.2\)\.

*The problemComputeCharBinaryis𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under many\-one reductions\.*

We prove this conjecture, in the sharper form that the reduction may always be taken to output a partitionλ\\lambdawith at most two parts\.

The closest previous result is the Turing\-reduction completeness stated above\. Before that,[Hepler 1994](https://arxiv.org/html/2608.16977#bibf.bib4)proved that computingχλ​\(μ\)\\chi^\{\\lambda\}\(\\mu\)is\#​𝖯\\\#\\mathsf\{P\}\-hard under many\-one reductions already for unary input, hence also for binary input;\#​𝖯\\\#\\mathsf\{P\}\-hardness constrains only the nonnegative part of the character and does not by itself give𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-hardness\.[Pak & Panova 2017](https://arxiv.org/html/2608.16977#bibf.bib8)showed that the positivity of a Kronecker coefficient can be decided in timeO⁡\(log⁡N\)O\(\\log N\)for partitions with a bounded number of parts and largest partNN, while[Ikenmeyer et al\. 2017](https://arxiv.org/html/2608.16977#bibf.bib5)showed that deciding positivity of a Kronecker coefficient is NP\-hard;[Panova 2023](https://arxiv.org/html/2608.16977#bibf.bib9)surveys this circle of questions\. On the algorithmic side,[Bravyi et al\. 2025](https://arxiv.org/html/2608.16977#bibf.bib1)give an algorithm computing a matrix product state that encodes the column\(χλ​\(μ\)\)λ⊢n\(\\chi^\{\\lambda\}\(\\mu\)\)\_\{\\lambda\\vdash n\}of the character table, and record the worst\-case\#​𝖯\\\#\\mathsf\{P\}\-hardness of a single entry as the obstruction to a general polynomial\-time algorithm\. We are not aware of a previous many\-one𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-hardness result for this function\.

###### Theorem C\.6\.1\.

The functionComputeCharBinaryis𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-complete under polynomial\-time many\-one reductions\. More precisely, for everyf∈𝖦𝖺𝗉𝖯f\\in\\mathsf\{GapP\}there is a polynomial\-time computable mapx↦\(λx,μx\)x\\mapsto\(\\lambda\_\{x\},\\mu\_\{x\}\), whose values are pairs of partitions of a common integernxn\_\{x\}withλx\\lambda\_\{x\}having at most two parts, such that

f⁡\(x\)=χλx​\(μx\)f\(x\)=\\chi^\{\\lambda\_\{x\}\}\(\\mu\_\{x\}\)for every inputxx\.

The proof rests on one classical identity and one gadget\. For a two\-row shape the character value is a difference of two subset\-sum counts at consecutive targets,

χ\(n−s,s\)​\(μ\)=Nμ​\(s\)−Nμ​\(s−1\),\\chi^\{\(n\-s,s\)\}\(\\mu\)=N\_\{\\mu\}\(s\)\-N\_\{\\mu\}\(s\-1\),whereNμ​\(t\)N\_\{\\mu\}\(t\)is the number of subsets of the parts ofμ\\muwith total sizett\. It therefore suffices to manufacture, out of two given counting problems, a single multiset of binary integers whose subset sums realize the first count atssand the second ats−1s\-1\. The key point is that the two targets differ by exactly11, so the two problems have to be separated inside a single units digit\. We take the two problems to be counts of exact covers of a finite set, write every part in a large baseQQ, give the two instances disjoint blocks of base\-QQdigits, and adjoin two selector parts that differ in the units digit and each of which pre\-fills the digit block of the other instance\.

#### C\.6\.2Characters of two\-row shape

Throughout, a partitionμ=\(μ1,…,μm\)⊢n\\mu=\(\\mu\_\{1\},\\ldots,\\mu\_\{m\}\)\\vdash nhas positive parts, and parts of equal size are regarded as distinct, indexed by\[m\]\[m\]\. Fort∈ℤt\\in\\mathbb\{Z\}set

Nμ​\(t\):=\#⁡\{I⊆\[m\]:∑i∈Iμi=t\},N\_\{\\mu\}\(t\):=\\\#\\Big\\\{I\\subseteq\[m\]:\\sum\_\{i\\in I\}\\mu\_\{i\}=t\\Big\\\},so thatNμ​\(t\)=0N\_\{\\mu\}\(t\)=0fort<0t<0andNμ​\(0\)=1N\_\{\\mu\}\(0\)=1\.

###### Lemma C\.6\.3\.

Letμ⊢n\\mu\\vdash nand let0≤s≤n/20\\leq s\\leq n/2\. Then

χ\(n−s,s\)​\(μ\)=Nμ​\(s\)−Nμ​\(s−1\)\.\\chi^\{\(n\-s,s\)\}\(\\mu\)=N\_\{\\mu\}\(s\)\-N\_\{\\mu\}\(s\-1\)\.

###### Proof\.

For an integer vectorα=\(α1,α2,…\)\\alpha=\(\\alpha\_\{1\},\\alpha\_\{2\},\\ldots\)with entries summing tonnletϕα\\phi^\{\\alpha\}be the character of the representation of𝔖n\\mathfrak\{S\}\_\{n\}induced from the trivial representation of the Young subgroup𝔖α1×𝔖α2×⋯\\mathfrak\{S\}\_\{\\alpha\_\{1\}\}\\times\\mathfrak\{S\}\_\{\\alpha\_\{2\}\}\\times\\cdots, with the conventionϕα=0\\phi^\{\\alpha\}=0when someαi<0\\alpha\_\{i\}<0\. Equivalently,ϕα\\phi^\{\\alpha\}is the character of the action of𝔖n\\mathfrak\{S\}\_\{n\}on the words containing exactlyαi\\alpha\_\{i\}letters equal toii\. Such a word is fixed by a permutationπ\\piprecisely when it is constant on each cycle ofπ\\pi, so forπ\\piof cycle typeμ\\muthe valueϕα​\(μ\)\\phi^\{\\alpha\}\(\\mu\)is the number of ways to label themmcycles by letters so that the cycles labelediihave total lengthαi\\alpha\_\{i\}\([Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)\)\. In particular, for a vector with two entries,

ϕ\(n−t,t\)​\(μ\)=Nμ​\(t\)\.\\phi^\{\(n\-t,t\)\}\(\\mu\)=N\_\{\\mu\}\(t\)\.The Frobenius character formula\([James 2006](https://arxiv.org/html/2608.16977#bibf.bib7), Eq\. 2\.3\.8\), equivalently Young’s rule\([Sagan 2001](https://arxiv.org/html/2608.16977#bibf.bib11)\), gives

χ\(n−s,s\)=ϕ\(n−s,s\)−ϕ\(n−s\+1,s−1\),\\chi^\{\(n\-s,s\)\}=\\phi^\{\(n\-s,s\)\}\-\\phi^\{\(n\-s\+1,s\-1\)\},and evaluating at cycle typeμ\\muyields the identity\. ∎

#### C\.6\.3A difference of two exact cover counts

An instance of\#​ExactCover\\\#\\textnormal\{\{ExactCover\}\}is a pair\(X,C\)\(X,C\)in whichX=\[k\]X=\[k\]withk≥0k\\geq 0andC=\(S1,…,Sr\)C=\(S\_\{1\},\\ldots,S\_\{r\}\)is a list of nonempty subsets ofXX, members of equal content being regarded as distinct and indexed by\[r\]\[r\]\. A*selection*is a subsetF⊆\[r\]F\\subseteq\[r\], and it is an*exact cover*if everyu∈Xu\\in Xlies inSiS\_\{i\}for exactly onei∈Fi\\in F\. The value of the instance is the number

ec⁡\(X,C\):=\#⁡\{F⊆\[r\]:F​is an exact cover\}\\operatorname\{ec\}\(X,C\):=\\\#\\\{F\\subseteq\[r\]:F\\text\{ is an exact cover\}\\\}of exact covers, so thatec⁡\(∅,\(\)\)=1\\operatorname\{ec\}\(\\varnothing,\(\)\)=1, the empty selection covering the empty ground set\. Deciding whether an exact cover exists is NP\-complete already when every member ofCChas three elements\([Garey & Johnson 2002](https://arxiv.org/html/2608.16977#bibf.bib3)\); we put no bound on the sizes of the members, which the gadget below does not need\. An element ofXXlying in no member ofCCis covered by no selection, so an instance containing such an element has value00\. Replacing every such instance by the fixed instance\(\[3\],\(\{1,2\},\{2,3\}\)\)\\big\(\[3\],\(\\\{1,2\\\},\\\{2,3\\\}\)\\big\), whose value is also00, we may and do assume

k≤∑S∈C\|S\|,k\\leq\\sum\_\{S\\in C\}\|S\|,so thatkkis bounded by the length of the instance\.

###### Lemma C\.6\.4\.

There is a polynomial\-time computable map sending a conjunctive normal form formulaφ\\varphiwith clauses of between one and three literals to an instance\(Xφ,Cφ\)\(X\_\{\\varphi\},C\_\{\\varphi\}\)of\#​ExactCover\\\#\\textnormal\{\{ExactCover\}\}withec⁡\(Xφ,Cφ\)\\operatorname\{ec\}\(X\_\{\\varphi\},C\_\{\\varphi\}\)equal to the number of satisfying assignments ofφ\\varphi\.

###### Proof\.

Letφ\\varphihave variablesx1,…,xNx\_\{1\},\\ldots,x\_\{N\}and clausesc1,…,cMc\_\{1\},\\ldots,c\_\{M\}, and letcjc\_\{j\}be the disjunction of the literalsℓj,1,…,ℓj,wj\\ell\_\{j,1\},\\ldots,\\ell\_\{j,w\_\{j\}\}with1≤wj≤31\\leq w\_\{j\}\\leq 3; repeated and complementary literals inside a clause are allowed\. Take the ground set

Xφ:=\{vi:i∈\[N\]\}∪\{zj:j∈\[M\]\}∪\{pj,q:j∈\[M\],q∈\[wj\]\},X\_\{\\varphi\}:=\\\{v\_\{i\}:i\\in\[N\]\\\}\\cup\\\{z\_\{j\}:j\\in\[M\]\\\}\\cup\\\{p\_\{j,q\}:j\\in\[M\],\\ q\\in\[w\_\{j\}\]\\\},one element for each variable, one for each clause and one for each position inside a clause, and letCφC\_\{\\varphi\}consist of the sets

Vi,t:=\{vi\}∪\{pj,q:ℓj,q∈\{xi,¬xi\}​is falsified by​xi=t\}\(i∈\[N\],t∈\{0,1\}\)V\_\{i,t\}:=\\\{v\_\{i\}\\\}\\cup\\\{p\_\{j,q\}:\\ell\_\{j,q\}\\in\\\{x\_\{i\},\\neg x\_\{i\}\\\}\\text\{ is falsified by \}x\_\{i\}=t\\\}\\qquad\(i\\in\[N\],\\ t\\in\\\{0,1\\\}\)together with the sets

Wj,T:=\{zj\}∪\{pj,q:q∈T\}\(j∈\[M\],∅≠T⊆\[wj\]\)\.W\_\{j,T\}:=\\\{z\_\{j\}\\\}\\cup\\\{p\_\{j,q\}:q\\in T\\\}\\qquad\(j\\in\[M\],\\ \\varnothing\\neq T\\subseteq\[w\_\{j\}\]\)\.All of them are nonempty, there are2​N\+∑j\(2wj−1\)≤2​N\+7​M2N\+\\sum\_\{j\}\(2^\{w\_\{j\}\}\-1\)\\leq 2N\+7Mof them, and the list is written down in time linear in the length ofφ\\varphi\.

LetFFbe an exact cover\. The elementviv\_\{i\}lies only inVi,0V\_\{i,0\}andVi,1V\_\{i,1\}, so exactly one of the two is selected; letα⁡\(i\)\\alpha\(i\)be the corresponding value ofxix\_\{i\}\. The elementzjz\_\{j\}lies only in the setsWj,TW\_\{j,T\}, so for eachjjexactly oneT=TjT=T\_\{j\}is selected\. The selected setsVi,α⁡\(i\)V\_\{i,\\alpha\(i\)\}cover between them precisely thosepj,qp\_\{j,q\}whose literal is false underα\\alpha, so exactness at the elementspj,qp\_\{j,q\}forcesTjT\_\{j\}to be the set of positionsqqwithℓj,q\\ell\_\{j,q\}true underα\\alpha\. AsTjT\_\{j\}is nonempty,α\\alphasatisfies every clause\. Conversely, letα\\alphasatisfyφ\\varphiand letTjT\_\{j\}be the set of positions ofcjc\_\{j\}holding a literal true underα\\alpha, which is nonempty\. The setsVi,α⁡\(i\)V\_\{i,\\alpha\(i\)\}withi∈\[N\]i\\in\[N\]andWj,TjW\_\{j,T\_\{j\}\}withj∈\[M\]j\\in\[M\]then cover each element ofXφX\_\{\\varphi\}exactly once\. The two constructions are mutually inverse, so the exact covers correspond bijectively to the satisfying assignments ofφ\\varphi\. ∎

Given two instances of\#​ExactCover\\\#\\textnormal\{\{ExactCover\}\}, write

Diff​\#​ExactCover​\(\(X,C\),\(X′,C′\)\):=ec⁡\(X,C\)−ec⁡\(X′,C′\)\.\\textnormal\{\{Diff\}\}\\\#\\textnormal\{\{ExactCover\}\}\\big\(\(X,C\),\(X^\{\\prime\},C^\{\\prime\}\)\\big\):=\\operatorname\{ec\}\(X,C\)\-\\operatorname\{ec\}\(X^\{\\prime\},C^\{\\prime\}\)\.
###### Lemma C\.6\.5\.

The functionDiff​\#​ExactCover\\textnormal\{\{Diff\}\}\\\#\\textnormal\{\{ExactCover\}\}is𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\-hard under polynomial\-time many\-one reductions\.

###### Proof\.

Letf∈𝖦𝖺𝗉𝖯f\\in\\mathsf\{GapP\}and writef=g−hf=g\-hwithg,h∈\#​𝖯g,h\\in\\\#\\mathsf\{P\}\. Ifg⁡\(x\)=\#⁡\{w∈\{0,1\}q⁡\(\|x\|\):V⁡\(x,w\)=1\}g\(x\)=\\\#\\\{w\\in\\\{0,1\\\}^\{q\(\|x\|\)\}:V\(x,w\)=1\\\}for a polynomialqqand a polynomial\-time predicateVV, then the standard simulation of a polynomial\-time machine by a Boolean circuit\([Papadimitriou 2003](https://arxiv.org/html/2608.16977#bibf.bib10)\)produces in polynomial time a circuitΓx\\Gamma\_\{x\}with gates of fan\-in at most two and with\#​CircuitSat​\(Γx\)=g⁡\(x\)\\\#\\textnormal\{\{CircuitSat\}\}\(\\Gamma\_\{x\}\)=g\(x\); hence\#​CircuitSat\\\#\\textnormal\{\{CircuitSat\}\}is\#​𝖯\\\#\\mathsf\{P\}\-complete under parsimonious reductions\. The Tseitin transformation\([Tseitin 1983](https://arxiv.org/html/2608.16977#bibf.bib12)\)turns a circuit into a formula in conjunctive normal form with clauses of between one and three literals: it introduces one variable for each gate, adds the clauses forcing that variable to equal the value of the gate, and adds a unit clause forcing the output gate to take the value11\. The gate variables are determined by the input variables, so every satisfying assignment of the circuit extends to exactly one satisfying assignment of the formula, and the transformation is parsimonious\. Composing it with Lemma[C\.6\.4](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem4)and normalizing as above, we obtain polynomial\-time computable instances with

ec⁡\(Xx,Cx\)=g⁡\(x\),ec⁡\(Xx′,Cx′\)=h⁡\(x\),\\operatorname\{ec\}\(X\_\{x\},C\_\{x\}\)=g\(x\),\\qquad\\operatorname\{ec\}\(X^\{\\prime\}\_\{x\},C^\{\\prime\}\_\{x\}\)=h\(x\),and thereforef⁡\(x\)=Diff​\#​ExactCover​\(\(Xx,Cx\),\(Xx′,Cx′\)\)f\(x\)=\\textnormal\{\{Diff\}\}\\\#\\textnormal\{\{ExactCover\}\}\(\(X\_\{x\},C\_\{x\}\),\(X^\{\\prime\}\_\{x\},C^\{\\prime\}\_\{x\}\)\)\. ∎

#### C\.6\.4The digit gadget

The construction is a subset\-sum encoding in a large base, in the style of the strong NP\-hardness proof of44\-Partition\([Garey & Johnson 2002](https://arxiv.org/html/2608.16977#bibf.bib3)\)\. Fix two instances\(X,C\)\(X,C\)and\(X′,C′\)\(X^\{\\prime\},C^\{\\prime\}\)as above, withX=\[k\]X=\[k\],X′=\[k′\]X^\{\\prime\}=\[k^\{\\prime\}\],C=\(S1,…,Sr\)C=\(S\_\{1\},\\ldots,S\_\{r\}\)andC′=\(S1′,…,Sr′′\)C^\{\\prime\}=\(S^\{\\prime\}\_\{1\},\\ldots,S^\{\\prime\}\_\{r^\{\\prime\}\}\), and set

Q:=r\+r′\+4\.Q:=r\+r^\{\\prime\}\+4\.We index base\-QQdigit positions by0,1,…,k\+k′\+10,1,\\ldots,k\+k^\{\\prime\}\+1, and call position00the*units digit*, positions1,…,k1,\\ldots,kthe*XX\-block*, positionsk\+1,…,k\+k′k\+1,\\ldots,k\+k^\{\\prime\}the*X′X^\{\\prime\}\-block*, and positionk\+k′\+1k\+k^\{\\prime\}\+1the*selector digit*\. ForS∈CS\\in CandS′∈C′S^\{\\prime\}\\in C^\{\\prime\}put

aS:=∑u∈SQu,bS′:=∑u∈S′Qk\+u,a\_\{S\}:=\\sum\_\{u\\in S\}Q^\{u\},\\qquad b\_\{S^\{\\prime\}\}:=\\sum\_\{u\\in S^\{\\prime\}\}Q^\{k\+u\},so thataSa\_\{S\}marks the elements covered bySSinside theXX\-block, and likewise forbS′b\_\{S^\{\\prime\}\}\. Let

A:=∑u∈XQu,B:=∑u∈X′Qk\+uA:=\\sum\_\{u\\in X\}Q^\{u\},\\qquad B:=\\sum\_\{u\\in X^\{\\prime\}\}Q^\{k\+u\}be the all\-ones patterns on the two blocks, and letD:=Qk\+k′\+1D:=Q^\{k\+k^\{\\prime\}\+1\}\. Introduce the two*selectors*and the*guard*

cA:=B\+D\+1,cB:=A\+D,H:=A\+B\+D\+2,c\_\{A\}:=B\+D\+1,\\qquad c\_\{B\}:=A\+D,\\qquad H:=A\+B\+D\+2,and set

so thatH=s\+1H=s\+1\. Letμ\\mube the partition obtained by sorting the multiset

\{aS:S∈C\}∪\{bS′:S′∈C′\}∪\{cA,cB,H\}\\\{a\_\{S\}:S\\in C\\\}\\cup\\\{b\_\{S^\{\\prime\}\}:S^\{\\prime\}\\in C^\{\\prime\}\\\}\\cup\\\{c\_\{A\},c\_\{B\},H\\\}into weakly decreasing order, letn:=\|μ\|n:=\|\\mu\|, and put

λ:=\(n−s,s\)\.\\lambda:=\(n\-s,s\)\.Table[5](https://arxiv.org/html/2608.16977#A3.T5)displays the base\-QQdigits of every quantity involved\.

Table 5:The base\-QQdigits of the parts ofμ\\muand of the two targetsssands−1s\-1\. Here𝟏\\mathbf\{1\}and𝟎\\mathbf\{0\}denote the all\-ones and all\-zeros patterns on the block indicated, and𝟏S\\mathbf\{1\}\_\{S\}denotes the indicator pattern ofSSon theXX\-block, with𝟏S′\\mathbf\{1\}\_\{S^\{\\prime\}\}defined analogously on theX′X^\{\\prime\}\-block\. No column can reachQQ, so subset sums may be compared digit by digit: a subset of parts summing tossmust takecAc\_\{A\}, hence miss theX′X^\{\\prime\}\-block entirely and pick out an exact cover of\(X,C\)\(X,C\), while a subset summing tos−1s\-1must takecBc\_\{B\}and pick out an exact cover of\(X′,C′\)\(X^\{\\prime\},C^\{\\prime\}\)\.###### Lemma C\.6\.6\.

Letm=r\+r′\+3m=r\+r^\{\\prime\}\+3be the number of parts ofμ\\mu\. For everyI⊆\[m\]I\\subseteq\[m\]the base\-QQdigits of∑i∈Iμi\\sum\_\{i\\in I\}\\mu\_\{i\}are obtained by adding the digit patterns of the parts indexed byII, without carrying\. Consequently, fort∈\{s−1,s\}t\\in\\\{s\-1,s\\\}one has∑i∈Iμi=t\\sum\_\{i\\in I\}\\mu\_\{i\}=tif and only if those digit patterns sum to the digit pattern oftt\.

###### Proof\.

By Table[5](https://arxiv.org/html/2608.16977#A3.T5), every part ofμ\\muhas all its base\-QQdigits in\{0,1,2\}\\\{0,1,2\\\}\. At a position of theXX\-block the parts with a nonzero digit there are among therrpartsaSa\_\{S\}, the selectorcBc\_\{B\}and the guardHH, each contributing11, so the total over anyIIis at mostr\+2r\+2\. Symmetrically a position of theX′X^\{\\prime\}\-block receives at mostr′\+2r^\{\\prime\}\+2\. The selector digit receives at most33, fromcAc\_\{A\},cBc\_\{B\}andHH, and the units digit receives at most33, namely11fromcAc\_\{A\}and22fromHH\. Every other position receives00\. All of these totals are smaller thanQ=r\+r′\+4Q=r\+r^\{\\prime\}\+4, so no carrying occurs\. Finallysshas digit11at the units digit, at every position of the two blocks, and at the selector digit, whiles−1s\-1has the same digits except00at the units digit; both patterns have entries belowQQ, so two such integers agree if and only if their digit patterns agree\. ∎

###### Lemma C\.6\.7\.

The pair\(λ,μ\)\(\\lambda,\\mu\)is a valid input toComputeCharBinary, withλ\\lambdaa partition ofnnhaving at most two parts and0≤s≤n/20\\leq s\\leq n/2, and it is computable from\(X,C\)\(X,C\)and\(X′,C′\)\(X^\{\\prime\},C^\{\\prime\}\)in polynomial time\.

###### Proof\.

SincecA\+cB\+H=2​A\+2​B\+3​D\+3c\_\{A\}\+c\_\{B\}\+H=2A\+2B\+3D\+3, we get

n=∑S∈CaS\+∑S′∈C′bS′\+2​A\+2​B\+3​D\+3,n=\\sum\_\{S\\in C\}a\_\{S\}\+\\sum\_\{S^\{\\prime\}\\in C^\{\\prime\}\}b\_\{S^\{\\prime\}\}\+2A\+2B\+3D\+3,and therefore

n−2​s=∑S∈CaS\+∑S′∈C′bS′\+D\+1\>0\.n\-2s=\\sum\_\{S\\in C\}a\_\{S\}\+\\sum\_\{S^\{\\prime\}\\in C^\{\\prime\}\}b\_\{S^\{\\prime\}\}\+D\+1\>0\.Hencen−s\>s≥0n\-s\>s\\geq 0, soλ=\(n−s,s\)\\lambda=\(n\-s,s\)is a partition ofnnwith at most two parts ands≤n/2s\\leq n/2\. For the running time, the normalization givesk≤∑S∈C\|S\|k\\leq\\sum\_\{S\\in C\}\|S\|andk′≤∑S′∈C′\|S′\|k^\{\\prime\}\\leq\\sum\_\{S^\{\\prime\}\\in C^\{\\prime\}\}\|S^\{\\prime\}\|, so the largest exponentk\+k′\+1k\+k^\{\\prime\}\+1is linear in the length of the input, whilelog2⁡Q=O⁡\(log⁡\(r\+r′\+4\)\)\\log\_\{2\}Q=O\(\\log\(r\+r^\{\\prime\}\+4\)\)\. Thus each of ther\+r′\+3r\+r^\{\\prime\}\+3parts ofμ\\mu, and each ofnnandss, is an integer ofO⁡\(\(k\+k′\+1\)​log⁡\(r\+r′\+4\)\)O\\big\(\(k\+k^\{\\prime\}\+1\)\\log\(r\+r^\{\\prime\}\+4\)\\big\)bits, and all of them are produced by polynomially many arithmetic operations on integers of that size\. ∎

###### Lemma C\.6\.8\.

For the partitionμ\\muconstructed above,

Nμ​\(s\)=ec⁡\(X,C\),Nμ​\(s−1\)=ec⁡\(X′,C′\)\.N\_\{\\mu\}\(s\)=\\operatorname\{ec\}\(X,C\),\\qquad N\_\{\\mu\}\(s\-1\)=\\operatorname\{ec\}\(X^\{\\prime\},C^\{\\prime\}\)\.

###### Proof\.

LetIIindex a subset of the parts with∑i∈Iμi∈\{s−1,s\}\\sum\_\{i\\in I\}\\mu\_\{i\}\\in\\\{s\-1,s\\\}, which by Lemma[C\.6\.6](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem6)we may analyze digit by digit\. SinceH=s\+1\>sH=s\+1\>s, the guard is not taken\. Bothssands−1s\-1have selector digit11, and apart fromHHonlycAc\_\{A\}andcBc\_\{B\}have a nonzero selector digit, each equal to11; hence exactly one ofcAc\_\{A\}andcBc\_\{B\}is taken\.

Suppose first that∑i∈Iμi=s\\sum\_\{i\\in I\}\\mu\_\{i\}=s\. The units digit ofssis11, and among the remaining parts onlycAc\_\{A\}has a nonzero units digit, socAc\_\{A\}is taken andcBc\_\{B\}is not\. NowcAc\_\{A\}already contributes11at every position of theX′X^\{\\prime\}\-block, which is exactly the digit ofssthere, so no partbS′b\_\{S^\{\\prime\}\}is taken\. The parts taken are therefore\{cA\}∪\{aSi:i∈F\}\\\{c\_\{A\}\\\}\\cup\\\{a\_\{S\_\{i\}\}:i\\in F\\\}for someF⊆\[r\]F\\subseteq\[r\], and the one remaining requirement is that∑i∈FaSi\\sum\_\{i\\in F\}a\_\{S\_\{i\}\}have digit11at every position of theXX\-block\. By the definition ofaSa\_\{S\}this says exactly that the selected sets cover each element of the ground set exactly once, that is, thatFFis an exact cover of\(X,C\)\(X,C\)\. Hence the subsets of parts summing tosscorrespond bijectively to the exact covers of\(X,C\)\(X,C\), andNμ​\(s\)=ec⁡\(X,C\)N\_\{\\mu\}\(s\)=\\operatorname\{ec\}\(X,C\)\.

Suppose now that∑i∈Iμi=s−1\\sum\_\{i\\in I\}\\mu\_\{i\}=s\-1\. The units digit ofs−1s\-1is00, socAc\_\{A\}is not taken, and thereforecBc\_\{B\}is\. ThencBc\_\{B\}contributes11at every position of theXX\-block, so no partaSa\_\{S\}is taken, and the parts taken are\{cB\}∪\{bSi′:i∈F′\}\\\{c\_\{B\}\\\}\\cup\\\{b\_\{S^\{\\prime\}\_\{i\}\}:i\\in F^\{\\prime\}\\\}withF′⊆\[r′\]F^\{\\prime\}\\subseteq\[r^\{\\prime\}\]subject to the same condition on theX′X^\{\\prime\}\-block\. HenceNμ​\(s−1\)=ec⁡\(X′,C′\)N\_\{\\mu\}\(s\-1\)=\\operatorname\{ec\}\(X^\{\\prime\},C^\{\\prime\}\)\. ∎

#### C\.6\.5Membership in𝖦𝖺𝗉𝖯\\mathsf\{GapP\}

###### Proposition C\.6\.9\.

ComputeCharBinarybelongs to𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\.

###### Proof\.

Let the input beλ=\(λ1,…,λp\)\\lambda=\(\\lambda\_\{1\},\\ldots,\\lambda\_\{p\}\)andμ=\(μ1,…,μm\)\\mu=\(\\mu\_\{1\},\\ldots,\\mu\_\{m\}\), both partitions ofnnwritten as binary lists of parts, so thatpp,mmand all the bit lengths involved are bounded by the length of the input; an input not of this form is recognized in polynomial time and given the value00\. For an integer vectorβ=\(β1,…,βp\)\\beta=\(\\beta\_\{1\},\\ldots,\\beta\_\{p\}\)letPμ​\(β\)P\_\{\\mu\}\(\\beta\)denote the number of ordered tuples\(K1,…,Kp\)\(K\_\{1\},\\ldots,K\_\{p\}\)of pairwise disjoint, possibly empty subsets of\[m\]\[m\]whose union is\[m\]\[m\], such that∑j∈Kiμj=βi\\sum\_\{j\\in K\_\{i\}\}\\mu\_\{j\}=\\beta\_\{i\}for everyii; this is00if someβi<0\\beta\_\{i\}<0, because the parts ofμ\\muare positive\. As in the proof of Lemma[C\.6\.3](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem3)one hasϕβ​\(μ\)=Pμ​\(β\)\\phi^\{\\beta\}\(\\mu\)=P\_\{\\mu\}\(\\beta\), so the Frobenius character formula\([James 2006](https://arxiv.org/html/2608.16977#bibf.bib7), Eq\. 2\.3\.8\)reads

χλ​\(μ\)=∑σ∈𝔖psgn⁡\(σ\)​Pμ​\(λ\+σ−id\),\\chi^\{\\lambda\}\(\\mu\)=\\sum\_\{\\sigma\\in\\mathfrak\{S\}\_\{p\}\}\\operatorname\{sgn\}\(\\sigma\)\\,P\_\{\\mu\}\(\\lambda\+\\sigma\-\\mathrm\{id\}\),whereλ\+σ−id\\lambda\+\\sigma\-\\mathrm\{id\}is the vector whoseiith entry isλi\+σ⁡\(i\)−i\\lambda\_\{i\}\+\\sigma\(i\)\-i; see also[Ikenmeyer et al\. 2024](https://arxiv.org/html/2608.16977#bibf.bib6)\. Letggandhhcount the pairs\(σ,K\)\(\\sigma,K\)in whichσ∈𝔖p\\sigma\\in\\mathfrak\{S\}\_\{p\}hassgn⁡\(σ\)=\+1\\operatorname\{sgn\}\(\\sigma\)=\+1andsgn⁡\(σ\)=−1\\operatorname\{sgn\}\(\\sigma\)=\-1respectively, andK=\(K1,…,Kp\)K=\(K\_\{1\},\\ldots,K\_\{p\}\)is a tuple as above whose blocks haveμ\\mu\-weights∑j∈Kiμj\\sum\_\{j\\in K\_\{i\}\}\\mu\_\{j\}equal to the entries ofλ\+σ−id\\lambda\+\\sigma\-\\mathrm\{id\}\. Such a pair is specified byO⁡\(\(p\+m\)​log⁡\(p\+1\)\)O\\big\(\(p\+m\)\\log\(p\+1\)\\big\)bits and is verified by computingsgn⁡\(σ\)\\operatorname\{sgn\}\(\\sigma\)and comparingppsums of binary integers ofO⁡\(log⁡n\)O\(\\log n\)bits each, sog,h∈\#​𝖯g,h\\in\\\#\\mathsf\{P\}\. Thereforeχλ​\(μ\)=g−h\\chi^\{\\lambda\}\(\\mu\)=g\-hlies in𝖦𝖺𝗉𝖯\\mathsf\{GapP\}\. ∎

#### C\.6\.6Proof of the main theorem

###### Proof of Theorem[C\.6\.1](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem1)\.

Membership in𝖦𝖺𝗉𝖯\\mathsf\{GapP\}is Proposition[C\.6\.9](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem9)\. For hardness, fixf∈𝖦𝖺𝗉𝖯f\\in\\mathsf\{GapP\}\. By Lemma[C\.6\.5](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem5)there is a polynomial\-time computable map sending an inputxxto two exact cover instances with

f⁡\(x\)=ec⁡\(Xx,Cx\)−ec⁡\(Xx′,Cx′\)\.f\(x\)=\\operatorname\{ec\}\(X\_\{x\},C\_\{x\}\)\-\\operatorname\{ec\}\(X^\{\\prime\}\_\{x\},C^\{\\prime\}\_\{x\}\)\.Apply the digit gadget to those two instances, and let\(λx,μx\)\(\\lambda\_\{x\},\\mu\_\{x\}\)be the resulting pair, withλx=\(nx−sx,sx\)\\lambda\_\{x\}=\(n\_\{x\}\-s\_\{x\},s\_\{x\}\)\. By Lemma[C\.6\.7](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem7)the mapx↦\(λx,μx\)x\\mapsto\(\\lambda\_\{x\},\\mu\_\{x\}\)is computable in polynomial time, its values are partitions ofnxn\_\{x\}, the shapeλx\\lambda\_\{x\}has at most two parts, and0≤sx≤nx/20\\leq s\_\{x\}\\leq n\_\{x\}/2\. By Lemma[C\.6\.3](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem3)and Lemma[C\.6\.8](https://arxiv.org/html/2608.16977#A3.SS6.Thmtheorem8),

χλx​\(μx\)=Nμx​\(sx\)−Nμx​\(sx−1\)=ec⁡\(Xx,Cx\)−ec⁡\(Xx′,Cx′\)=f⁡\(x\)\.\\chi^\{\\lambda\_\{x\}\}\(\\mu\_\{x\}\)=N\_\{\\mu\_\{x\}\}\(s\_\{x\}\)\-N\_\{\\mu\_\{x\}\}\(s\_\{x\}\-1\)=\\operatorname\{ec\}\(X\_\{x\},C\_\{x\}\)\-\\operatorname\{ec\}\(X^\{\\prime\}\_\{x\},C^\{\\prime\}\_\{x\}\)=f\(x\)\.Hence everyf∈𝖦𝖺𝗉𝖯f\\in\\mathsf\{GapP\}many\-one reduces toComputeCharBinary, completing the proof\. ∎

## References\.

- Bravyi et al\. \(2025\)Sergey Bravyi, David Gosset, Vojtech Havlicek, and Louis Schatzki\.Classical and quantum algorithms for characters of the symmetric group\.*PRX Quantum*, 6\(3\):030323, 2025\.
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- Ikenmeyer et al\. \(2024\)Christian Ikenmeyer, Igor Pak, and Greta Panova\.Positivity of the symmetric group characters is as hard as the polynomial time hierarchy\.*International Mathematics Research Notices*, 2024\(10\):8442–8458, 2024\.
- James \(2006\)Gordon Douglas James\.*The representation theory of the symmetric groups*\.Springer, 2006\.
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### C\.7Matching variance versus the residual matching number

Kahn’s normal law for matchings characterizes asymptotic normality of the size of a uniformly random matching through five graph statistics that are bounded or unbounded together, and asks how tightly two of them, the varianceσ2\\sigma^\{2\}and the residual matching numberλ\\lambda, are tied to one another\. We show that in one direction they are not tied at all: along cliques carrying private pendant leaves the ratioσ2/λ\\sigma^\{2\}/\\lambdagrows at least linearly in the number of leaves at each clique vertex, while both parameters tend to infinity\. The mechanism is that extra leaves inflate the fluctuation of the clique matching without inflating the residue\.

#### C\.7\.1Introduction

For a finite simple graphGGletℳ⁡\(G\)\\mathcal\{M\}\(G\)be the set of matchings ofGG, letMMbe drawn uniformly fromℳ⁡\(G\)\\mathcal\{M\}\(G\), and putξG=\|M\|\\xi\_\{G\}=\|M\|\. Writeμ⁡\(G\)=𝔼⁡\[ξG\]\\mu\(G\)=\\mathbb\{E\}\[\\xi\_\{G\}\]andσ2​\(G\)=Var⁡\[ξG\]\\sigma^\{2\}\(G\)=\\mathrm\{Var\}\[\\xi\_\{G\}\], and writeν⁡\(G\)\\nu\(G\)andτ⁡\(G\)\\tau\(G\)for the matching number and the vertex cover number ofGG\. Forx∈V⁡\(G\)x\\in V\(G\)letp⁡\(x\)p\(x\)denote the probability thatxxis not covered byMM\.

By a theorem of[Godsil 1981](https://arxiv.org/html/2608.16977#bibg.bib1), resting on the real\-rootedness of the matching polynomial of[Heilmann & Lieb 1972](https://arxiv.org/html/2608.16977#bibg.bib2), the distribution ofξGn\\xi\_\{G\_\{n\}\}along a sequence of graphs is asymptotically normal if and only ifσ⁡\(Gn\)→∞\\sigma\(G\_\{n\}\)\\to\\infty\.[Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3)identified four combinatorial statistics with exactly the same threshold behavior, two of which are introduced there for the purpose\. The first is the*cover defect*

κ⁡\(G\)=min⁡\{∑y∈Yp⁡\(y\):Y​a vertex cover of​G\},\\kappa\(G\)=\\min\\Bigl\\\{\\sum\_\{y\\in Y\}p\(y\):\\ Y\\text\{ a vertex cover of \}G\\Bigr\\\},and the second, writingFMF\_\{M\}for the set of edges ofGGmeeting no edge ofMM, is the*residual matching number*

λ⁡\(G\)=𝔼⁡\[ν⁡\(FM\)\]\.\\lambda\(G\)=\\mathbb\{E\}\\bigl\[\\nu\(F\_\{M\}\)\\bigr\]\.ThusFMF\_\{M\}is the edge set induced by the vertices left uncovered byMM, andλ\\lambdameasures how much of a matching still fits into the residue\. Kahn’s theorem states that for any sequence\(Gn\)\(G\_\{n\}\)with\|V⁡\(Gn\)\|→∞\|V\(G\_\{n\}\)\|\\to\\inftyandδ⁡\(Gn\)≥1\\delta\(G\_\{n\}\)\\geq 1, and withσn=σ⁡\(Gn\)\\sigma\_\{n\}=\\sigma\(G\_\{n\}\)and so on, the five conditions

σn=O⁡\(1\),νn−μn=O⁡\(1\),τn−μn=O⁡\(1\),κn=O⁡\(1\),λn=O⁡\(1\)\\sigma\_\{n\}=O\(1\),\\qquad\\nu\_\{n\}\-\\mu\_\{n\}=O\(1\),\\qquad\\tau\_\{n\}\-\\mu\_\{n\}=O\(1\),\\qquad\\kappa\_\{n\}=O\(1\),\\qquad\\lambda\_\{n\}=O\(1\)are equivalent\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3), Theorem 1\.10\)\. He then asks how tight the comparison between the first and the last of these is\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3), Question 7\.3\):

*How closely related areσ2\\sigma^\{2\}andλ\\lambda? In particular, is it true thatλ=Θ⁡\(σ2\)\\lambda=\\Theta\(\\sigma^\{2\}\)\(that is, are there bounds on the ratiosλ/σ2\\lambda/\\sigma^\{2\}andσ2/λ\\sigma^\{2\}/\\lambda\)?*

We show that the answer is no, by breaking the directionσ2=O⁡\(λ\)\\sigma^\{2\}=O\(\\lambda\)\.

What Kahn’s own inequalities give is the following\. He provesσ2,λ≤ν−μ≤τ−μ≤κ2/2\+O⁡\(κ\)\\sigma^\{2\},\\lambda\\leq\\nu\-\\mu\\leq\\tau\-\\mu\\leq\\kappa^\{2\}/2\+O\(\\kappa\), and alsoκ=O⁡\(λ2\+λ\)\\kappa=O\(\\lambda^\{2\}\+\\lambda\)andκ=O⁡\(σ4\+σ2\)\\kappa=O\(\\sigma^\{4\}\+\\sigma^\{2\}\)\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3)\)\. Combining these gives

σ2=O⁡\(λ4\)​whenever​λ=Ω⁡\(1\),λ≤ν−μ=O⁡\(σ8\)​whenever​σ=Ω⁡\(1\),\\sigma^\{2\}=O\(\\lambda^\{4\}\)\\ \\text\{ whenever \}\\lambda=\\Omega\(1\),\\qquad\\lambda\\leq\\nu\-\\mu=O\(\\sigma^\{8\}\)\\ \\text\{ whenever \}\\sigma=\\Omega\(1\),and Kahn conjectures that the second of these can be improved toν−μ=O⁡\(σ6\)\\nu\-\\mu=O\(\\sigma^\{6\}\), which would be best possible\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3)\)\. The examples he records all haveσ2≍λ\\sigma^\{2\}\\asymp\\lambda\. For his Example 7\.1, a cliqueKnK\_\{n\}with one private pendant leaf at each clique vertex, one hasσ2=12​n\+O⁡\(1\)\\sigma^\{2\}=\\tfrac\{1\}\{2\}\\sqrt\{n\}\+O\(1\)andλ=κ=n\+O⁡\(1\)\\lambda=\\kappa=\\sqrt\{n\}\+O\(1\), and for his Example 7\.2 bothσ2\\sigma^\{2\}andλ\\lambdaare of ordern1/3n^\{1/3\}\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3)\)\. For a disjoint union of copies ofK1,mK\_\{1,m\}Kahn computesλ=ν−μ\\lambda=\\nu\-\\muandσ2=m⁡\(ν−μ\)/\(m\+1\)\\sigma^\{2\}=m\(\\nu\-\\mu\)/\(m\+1\), so thatσ2/λ=m/\(m\+1\)<1\\sigma^\{2\}/\\lambda=m/\(m\+1\)<1\([Kahn 2000](https://arxiv.org/html/2608.16977#bibg.bib3)\); there the leaf count movesσ2\\sigma^\{2\}andλ\\lambdatogether, and in the construction below it is the clique that decouples them\. General criteria for central limit theorems of this kind, in terms of the location of the zeros of graph\-counting polynomials, were later given by[Lebowitz et al\. 2016](https://arxiv.org/html/2608.16977#bibg.bib4); they say nothing about the size ofλ\\lambda\. Apart from Kahn’s own examples we are aware of no work that bears on Question 7\.3\.

#### C\.7\.2The construction

Our graphs are the graphs of Kahn’s Example 7\.1 with more leaves\.

###### Definition C\.7\.1\.

For integersn≥0n\\geq 0anda≥2a\\geq 2letGn,aG\_\{n,a\}be the graph with vertex set

V\(Gn,a\)=\{v1,…,vn\}∪\{yi,t:i∈\[n\],t∈\[a−1\]\}V\(G\_\{n,a\}\)=\\\{v\_\{1\},\\dots,v\_\{n\}\\\}\\cup\\\{y\_\{i,t\}:i\\in\[n\],\\ t\\in\[a\-1\]\\\}and edge set

E\(Gn,a\)=\{\{vi,vj\}:1≤i<j≤n\}∪\{\{vi,yi,t\}:i∈\[n\],t∈\[a−1\]\}\.E\(G\_\{n,a\}\)=\\bigl\\\{\\\{v\_\{i\},v\_\{j\}\\\}:1\\leq i<j\\leq n\\bigr\\\}\\cup\\bigl\\\{\\\{v\_\{i\},y\_\{i,t\}\\\}:i\\in\[n\],\\ t\\in\[a\-1\]\\bigr\\\}\.That is,Gn,aG\_\{n,a\}is a cliqueKnK\_\{n\}carryinga−1a\-1private pendant leaves at each clique vertex\. We callv1,…,vnv\_\{1\},\\dots,v\_\{n\}the*clique vertices*\.

The casea=2a=2is Kahn’s Example 7\.1\. Forn≥1n\\geq 1every leaf has degree11and every clique vertex has degreen−1\+\(a−1\)≥1n\-1\+\(a\-1\)\\geq 1, soGn,aG\_\{n,a\}is a finite simple graph withδ⁡\(Gn,a\)≥1\\delta\(G\_\{n,a\}\)\\geq 1and the family satisfies the standing hypothesis of Kahn’s theorem\. The parameteraais the number of configurations available at a clique vertex that is not matched inside the clique: it may stay uncovered, or take any one of itsa−1a\-1leaves\. Figure[9](https://arxiv.org/html/2608.16977#A3.F9)shows a matching ofG5,3G\_\{5,3\}and the residual graph it leaves behind\.

v1v\_\{1\}v2v\_\{2\}v3v\_\{3\}v4v\_\{4\}v5v\_\{5\}y3,1y\_\{3,1\}Figure 9:The graphG5,3G\_\{5,3\}, with its ten clique edges and its two private leaves at each clique vertex drawn in gray, and with the vertices covered by the matching drawn solid\. The matchingMMconsists of the two heavy black edgesv1​v2v\_\{1\}v\_\{2\}andv3​y3,1v\_\{3\}y\_\{3,1\}, so the set of clique vertices missed by the clique part ofMMis\{v3,v4,v5\}\\\{v\_\{3\},v\_\{4\},v\_\{5\}\\\}and exactly one of them takes a leaf\. The residual graphFMF\_\{M\}, the set of edges meeting no edge ofMM, is drawn heavy and dashed: it is the copy ofG2,3G\_\{2,3\}carried by the two clique vertices that took no leaf, andν⁡\(FM\)=2\\nu\(F\_\{M\}\)=2\.###### Theorem C\.7\.2\.

There are absolute constantsn0n\_\{0\}andc\>0c\>0such that for all integersn≥n0n\\geq n\_\{0\}andaawith4≤a≤n1/44\\leq a\\leq n^\{1/4\},

14​n≤λ⁡\(Gn,a\)≤5​n,σ2​\(Gn,a\)≥c​a​n\.\\tfrac\{1\}\{4\}\\sqrt\{n\}\\ \\leq\\ \\lambda\(G\_\{n,a\}\)\\ \\leq\\ 5\\sqrt\{n\},\\qquad\\sigma^\{2\}\(G\_\{n,a\}\)\\ \\geq\\ c\\,a\\sqrt\{n\}\.In particular

σ2​\(Gn,a\)λ⁡\(Gn,a\)≥c5​a\.\\frac\{\\sigma^\{2\}\(G\_\{n,a\}\)\}\{\\lambda\(G\_\{n,a\}\)\}\\ \\geq\\ \\frac\{c\}\{5\}\\,a\.

###### Corollary C\.7\.3\.

Putan=⌊n1/4⌋a\_\{n\}=\\lfloor n^\{1/4\}\\rfloorandHn=Gn,anH\_\{n\}=G\_\{n,a\_\{n\}\}forn≥16n\\geq 16\. Thenλ⁡\(Hn\)→∞\\lambda\(H\_\{n\}\)\\to\\infty,σ2​\(Hn\)→∞\\sigma^\{2\}\(H\_\{n\}\)\\to\\infty, and

σ2​\(Hn\)λ⁡\(Hn\)=Ω⁡\(n1/4\)⟶∞\.\\frac\{\\sigma^\{2\}\(H\_\{n\}\)\}\{\\lambda\(H\_\{n\}\)\}=\\Omega\\bigl\(n^\{1/4\}\\bigr\)\\longrightarrow\\infty\.Hence there is no absolute constantCCwithσ2​\(G\)≤C​λ​\(G\)\\sigma^\{2\}\(G\)\\leq C\\lambda\(G\)for all finite simple graphsGG, andλ=Θ⁡\(σ2\)\\lambda=\\Theta\(\\sigma^\{2\}\)fails\.

Both parameters diverge along\(Hn\)\(H\_\{n\}\), so this is not the degenerate kind of counterexample in whichσ2\\sigma^\{2\}andλ\\lambdaare bothO⁡\(1\)O\(1\)\.

The proof is a direct computation with the exact law ofξGn,a\\xi\_\{G\_\{n,a\}\}\. A matching ofGn,aG\_\{n,a\}consists of a matching of the clique together with, at each clique vertex left uncovered by it, one ofaachoices: no leaf, or one of thea−1a\-1leaves\. WritingSSfor the number of clique vertices missed by the clique part of the matching andBBfor the number of those that do take a leaf, one hasξ=\(n−S\)/2\+B\\xi=\(n\-S\)/2\+Bandν⁡\(FM\)=S−B\\nu\(F\_\{M\}\)=S\-B, andBBis binomial with parametersSSand\(a−1\)/a\(a\-1\)/a\. Henceλ=𝔼⁡\[S\]/a\\lambda=\\mathbb\{E\}\[S\]/a, while fora≥4a\\geq 4the varianceσ2\\sigma^\{2\}retains a fixed fraction ofVar⁡\[S\]\\mathrm\{Var\}\[S\], and we show that𝔼⁡\[S\]≍a​n\\mathbb\{E\}\[S\]\\asymp a\\sqrt\{n\}andVar⁡\[S\]≫a​n\\mathrm\{Var\}\[S\]\\gg a\\sqrt\{n\}\. The key point is that the leaf countaadividesλ\\lambdabut notσ2\\sigma^\{2\}: multiplying the number of admissible configurations at a free clique vertex byaashifts the equilibrium of the clique matching, pushing𝔼⁡\[S\]\\mathbb\{E\}\[S\]up to ordera​na\\sqrt\{n\}andVar⁡\[S\]\\mathrm\{Var\}\[S\]up to at least that order, whereas the residue is governed by the free clique vertices that took*no*leaf, and their expected number𝔼⁡\[S\]/a\\mathbb\{E\}\[S\]/astays of ordern\\sqrt\{n\}\.

#### C\.7\.3The law of a uniform matching

Fixnnanda≥2a\\geq 2, writeG=Gn,aG=G\_\{n,a\}, and abbreviateξ=ξG=\|M\|\\xi=\\xi\_\{G\}=\|M\|\. Given a matchingMMofGG, let

SM=\{i∈\[n\]:vi​is not covered by an edge of​M∩E⁡\(Kn\)\},S=\|SM\|,S\_\{M\}=\\\{i\\in\[n\]:\\ v\_\{i\}\\text\{ is not covered by an edge of \}M\\cap E\(K\_\{n\}\)\\\},\\qquad S=\|S\_\{M\}\|,and letBBbe the number ofi∈SMi\\in S\_\{M\}for whichviv\_\{i\}is matched byMMto one of its leaves\. Setq=\(a−1\)/aq=\(a\-1\)/a\.

###### Lemma C\.7\.6\.

LetMMbe uniform onℳ⁡\(G\)\\mathcal\{M\}\(G\)\. ThenS≡n\(mod2\)S\\equiv n\\pmod\{2\}, and for0≤s≤n0\\leq s\\leq nwiths≡n\(mod2\)s\\equiv n\\pmod\{2\},

ℙ⁡\(S=s\)=wsW,ws:=n\!s\!​2k​k\!​as,k:=n−s2,W:=∑sws\.\\mathbb\{P\}\(S=s\)=\\frac\{w\_\{s\}\}\{W\},\\qquad w\_\{s\}:=\\frac\{n\!\}\{s\!\\,2^\{k\}\\,k\!\}\\,a^\{s\},\\qquad k:=\\frac\{n\-s\}\{2\},\\qquad W:=\\sum\_\{s\}w\_\{s\}\.\(18\)Conditionally onS=sS=s, the variableBBis binomialBin⁡\(s,q\)\\mathrm\{Bin\}\(s,q\)\. Moreover, for every matchingMM,

\|M\|=n−S2\+B,ν⁡\(FM\)=S−B\.\|M\|=\\frac\{n\-S\}\{2\}\+B,\\qquad\\nu\(F\_\{M\}\)=S\-B\.\(19\)

###### Proof\.

A matching ofGGis determined by two successive choices: a matchingM0⊆E⁡\(Kn\)M\_\{0\}\\subseteq E\(K\_\{n\}\)of the clique, and then, for each clique vertex left uncovered byM0M\_\{0\}, either nothing or one of itsa−1a\-1leaves\. Indeed leaf edges at distinct clique vertices are disjoint, and a leaf edge atviv\_\{i\}is compatible withM0M\_\{0\}exactly whenviv\_\{i\}is uncovered byM0M\_\{0\}\.

The number of matchingsM0M\_\{0\}ofKnK\_\{n\}leaving a prescribed set ofssclique vertices uncovered is the number of perfect matchings ofKn−sK\_\{n\-s\}, namely\(n−s\)\!/\(2k​k\!\)\(n\-s\)\!/\(2^\{k\}k\!\)withk=\(n−s\)/2k=\(n\-s\)/2; this forcess≡n\(mod2\)s\\equiv n\\pmod\{2\}\. Choosing the set of uncovered vertices in\(ns\)\\binom\{n\}\{s\}ways and then making the leaf choices inasa^\{s\}ways gives the number of matchingsMMwithS=sS=sas

\(ns\)​\(n−s\)\!2k​k\!​as=n\!s\!​2k​k\!​as=ws,\\binom\{n\}\{s\}\\frac\{\(n\-s\)\!\}\{2^\{k\}k\!\}\\,a^\{s\}=\\frac\{n\!\}\{s\!\\,2^\{k\}k\!\}\\,a^\{s\}=w\_\{s\},which is equation[18](https://arxiv.org/html/2608.16977#A3.E18)\. Since thessleaf choices are made independently and uniformly amongaaoptions, of whicha−1a\-1produce a leaf edge,BBisBin⁡\(s,q\)\\mathrm\{Bin\}\(s,q\)givenS=sS=s\.

For the first identity in equation[19](https://arxiv.org/html/2608.16977#A3.E19), note thatMMconsists of\(n−S\)/2\(n\-S\)/2clique edges andBBleaf edges\. For the second, a clique vertexviv\_\{i\}is uncovered byMMprecisely wheni∈SMi\\in S\_\{M\}andviv\_\{i\}took no leaf, which happens for exactlyS−BS\-Bindices, while a leafyi,ty\_\{i,t\}is uncovered precisely when it was not chosen\. An edge lies inFMF\_\{M\}if and only if both its ends are uncovered, soFMF\_\{M\}consists of all clique edges between theS−BS\-Buncovered clique vertices together with alla−1a\-1leaf edges at each of them: that is,FM≅GS−B,aF\_\{M\}\\cong G\_\{S\-B,\\,a\}\. InGr,aG\_\{r,a\}witha≥2a\\geq 2therrclique vertices form a vertex cover, soν≤r\\nu\\leq r, while matching each clique vertex to a private leaf showsν≥r\\nu\\geq r\. Henceν⁡\(FM\)=S−B\\nu\(F\_\{M\}\)=S\-B\. ∎

###### Corollary C\.7\.7\.

With the notation of Lemma[C\.7\.6](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem6),

λ⁡\(Gn,a\)=𝔼⁡\[S\]a,σ2​\(Gn,a\)=\(a−22​a\)2​Var​\[S\]\+a−1a2​𝔼​\[S\]\.\\lambda\(G\_\{n,a\}\)=\\frac\{\\mathbb\{E\}\[S\]\}\{a\},\\qquad\\sigma^\{2\}\(G\_\{n,a\}\)=\\Bigl\(\\frac\{a\-2\}\{2a\}\\Bigr\)^\{2\}\\mathrm\{Var\}\[S\]\+\\frac\{a\-1\}\{a^\{2\}\}\\,\\mathbb\{E\}\[S\]\.\(20\)

###### Proof\.

By equation[19](https://arxiv.org/html/2608.16977#A3.E19)and𝔼⁡\[B∣S\]=q​S\\mathbb\{E\}\[B\\mid S\]=qSwe getλ⁡\(Gn,a\)=𝔼⁡\[ν⁡\(FM\)\]=𝔼⁡\[S−B\]=\(1−q\)​𝔼​\[S\]=𝔼⁡\[S\]/a\\lambda\(G\_\{n,a\}\)=\\mathbb\{E\}\[\\nu\(F\_\{M\}\)\]=\\mathbb\{E\}\[S\-B\]=\(1\-q\)\\mathbb\{E\}\[S\]=\\mathbb\{E\}\[S\]/a\. Also𝔼⁡\[ξ∣S\]=n2\+\(q−12\)​S\\mathbb\{E\}\[\\xi\\mid S\]=\\tfrac\{n\}\{2\}\+\(q\-\\tfrac\{1\}\{2\}\)SandVar⁡\[ξ∣S\]=q⁡\(1−q\)​S\\mathrm\{Var\}\[\\xi\\mid S\]=q\(1\-q\)S, so decomposing the variance overSS,

σ2=\(q−12\)2​Var​\[S\]\+q⁡\(1−q\)​𝔼​\[S\],\\sigma^\{2\}=\\bigl\(q\-\\tfrac\{1\}\{2\}\\bigr\)^\{2\}\\mathrm\{Var\}\[S\]\+q\(1\-q\)\\mathbb\{E\}\[S\],andq−12=\(a−2\)/\(2​a\)q\-\\tfrac\{1\}\{2\}=\(a\-2\)/\(2a\)whileq⁡\(1−q\)=\(a−1\)/a2q\(1\-q\)=\(a\-1\)/a^\{2\}\. ∎

Everything therefore reduces to the mean and the variance of the weight sequence equation[18](https://arxiv.org/html/2608.16977#A3.E18)\.

#### C\.7\.4The distribution of the free set

Throughout this subsection and the next we assume

4≤a≤n1/4,L:=a​n,4\\leq a\\leq n^\{1/4\},\\qquad L:=a\\sqrt\{n\},\(21\)and thatnnis larger than a suitable absolute constant; all implied constants below are absolute\. Sincea≤n1/4a\\leq n^\{1/4\}givesa3≤a​na^\{3\}\\leq a\\sqrt\{n\}, anda≥4a\\geq 4, we have

a2≤La≤L4,L≤n3/4,L≥4​n\.a^\{2\}\\leq\\frac\{L\}\{a\}\\leq\\frac\{L\}\{4\},\\qquad L\\leq n^\{3/4\},\\qquad L\\geq 4\\sqrt\{n\}\.\(22\)In particularL→∞L\\to\\inftywithnn, so every hypothesis of the form “LLlarger than an absolute constant” below is implied by “nnlarger than an absolute constant”; we use this without further comment\. We also record that2​L\+4≤n/22L\+4\\leq n/2fornnlarge, by the middle bound in equation[22](https://arxiv.org/html/2608.16977#A3.E22)\.

From equation[18](https://arxiv.org/html/2608.16977#A3.E18), for0≤s≤n−20\\leq s\\leq n\-2withs≡n\(mod2\)s\\equiv n\\pmod\{2\},

ρ⁡\(s\):=ws\+2ws=a2​\(n−s\)\(s\+1\)​\(s\+2\)\.\\rho\(s\):=\\frac\{w\_\{s\+2\}\}\{w\_\{s\}\}=\\frac\{a^\{2\}\(n\-s\)\}\{\(s\+1\)\(s\+2\)\}\.\(23\)The functionρ\\rhois strictly decreasing inss, so the sequence\(ws\)\(w\_\{s\}\)is strictly log\-concave along the arithmetic progressions≡n\(mod2\)s\\equiv n\\pmod\{2\}, and in particular unimodal\. Fix a modes0s\_\{0\}, that is, an admissibles0s\_\{0\}withws0=maxs⁡wsw\_\{s\_\{0\}\}=\\max\_\{s\}w\_\{s\}\.

###### Lemma C\.7\.8\.

Assume equation[21](https://arxiv.org/html/2608.16977#A3.E21)andnnlarge\. Then12​L≤s0≤2​L\+2\\tfrac\{1\}\{2\}L\\leq s\_\{0\}\\leq 2L\+2\.

###### Proof\.

Ifs≥2​Ls\\geq 2Lis admissible then\(s\+1\)​\(s\+2\)\>s2≥4​L2=4​a2​n\(s\+1\)\(s\+2\)\>s^\{2\}\\geq 4L^\{2\}=4a^\{2\}n, so equation[23](https://arxiv.org/html/2608.16977#A3.E23)gives

ρ⁡\(s\)≤a2​ns2≤14\.\\rho\(s\)\\ \\leq\\ \\frac\{a^\{2\}n\}\{s^\{2\}\}\\ \\leq\\ \\frac\{1\}\{4\}\.\(24\)Hencews\+2<wsw\_\{s\+2\}<w\_\{s\}for every admissibles≥2​Ls\\geq 2L\. If we hads0\>2​L\+2s\_\{0\}\>2L\+2thens0−2≥2​Ls\_\{0\}\-2\\geq 2Lwould be admissible andws0<ws0−2w\_\{s\_\{0\}\}<w\_\{s\_\{0\}\-2\}, contradicting maximality; therefores0≤2​L\+2s\_\{0\}\\leq 2L\+2\.

Sinces0\+2≤2​L\+4≤ns\_\{0\}\+2\\leq 2L\+4\\leq n, maximality also givesws0\+2≤ws0w\_\{s\_\{0\}\+2\}\\leq w\_\{s\_\{0\}\}, that isρ⁡\(s0\)≤1\\rho\(s\_\{0\}\)\\leq 1, that is

\(s0\+1\)​\(s0\+2\)≥a2​\(n−s0\)≥12​a2​n=12​L2,\(s\_\{0\}\+1\)\(s\_\{0\}\+2\)\\ \\geq\\ a^\{2\}\(n\-s\_\{0\}\)\\ \\geq\\ \\tfrac\{1\}\{2\}a^\{2\}n=\\tfrac\{1\}\{2\}L^\{2\},where we useds0≤2​L\+2≤n/2s\_\{0\}\\leq 2L\+2\\leq n/2\. Hences0\+2≥L/2s\_\{0\}\+2\\geq L/\\sqrt\{2\}, and sinceLLis large this yieldss0≥L/2−2≥L/2s\_\{0\}\\geq L/\\sqrt\{2\}\-2\\geq L/2\. ∎

###### Lemma C\.7\.9\.

Assume equation[21](https://arxiv.org/html/2608.16977#A3.E21)andnnlarge\. Then𝔼⁡\[S\]≤5​L\\mathbb\{E\}\[S\]\\leq 5L, and consequentlyλ⁡\(Gn,a\)≤5​n\\lambda\(G\_\{n,a\}\)\\leq 5\\sqrt\{n\}\.

###### Proof\.

LetTTbe the least integer withT≥2​LT\\geq 2LandT≡n\(mod2\)T\\equiv n\\pmod\{2\}, so thatT≤2​L\+2T\\leq 2L\+2\. By equation[24](https://arxiv.org/html/2608.16977#A3.E24)we haveρ⁡\(s\)≤14\\rho\(s\)\\leq\\frac\{1\}\{4\}for every admissibles≥Ts\\geq T, whencewT\+2​j≤4−j​wTw\_\{T\+2j\}\\leq 4^\{\-j\}w\_\{T\}for allj≥0j\\geq 0\. Splitting the expectation atTTand usingwT≤Ww\_\{T\}\\leq W,

𝔼⁡\[S\]≤T\+∑j≥0\(T\+2​j\)​wT\+2​jW≤T\+∑j≥0\(T\+2​j\)​4−j=73​T\+89\.\\mathbb\{E\}\[S\]\\ \\leq\\ T\+\\sum\_\{j\\geq 0\}\(T\+2j\)\\frac\{w\_\{T\+2j\}\}\{W\}\\ \\leq\\ T\+\\sum\_\{j\\geq 0\}\(T\+2j\)4^\{\-j\}=\\tfrac\{7\}\{3\}T\+\\tfrac\{8\}\{9\}\.SinceT≤2​L\+2T\\leq 2L\+2this is at most143​L\+509≤5​L\\tfrac\{14\}\{3\}L\+\\tfrac\{50\}\{9\}\\leq 5L, becauseLLis large\. Finallyλ⁡\(Gn,a\)=𝔼⁡\[S\]/a≤5​L/a=5​n\\lambda\(G\_\{n,a\}\)=\\mathbb\{E\}\[S\]/a\\leq 5L/a=5\\sqrt\{n\}by Corollary[C\.7\.7](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem7)\. ∎

The next lemma is the technical heart of the argument: the distribution ofSShas no heavy atom\. It is what forcesVar⁡\[S\]\\mathrm\{Var\}\[S\]to be large, and it also rules out the degenerate possibility thatλ\\lambdastays bounded\.

###### Lemma C\.7\.10\.

Assume equation[21](https://arxiv.org/html/2608.16977#A3.E21)andnnlarge\. Then

maxs⁡ℙ⁡\(S=s\)≤22L,L=a​n\.\\max\_\{s\}\\mathbb\{P\}\(S=s\)\\ \\leq\\ \\frac\{22\}\{\\sqrt\{L\}\},\\qquad L=a\\sqrt\{n\}\.

###### Proof\.

PutD=⌊L/4⌋D=\\lfloor\\sqrt\{L\}/4\\rfloor; sinceLLis large we haveD≥2D\\geq 2andD≥L/8D\\geq\\sqrt\{L\}/8\. By Lemma[C\.7\.8](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem8)and equation[22](https://arxiv.org/html/2608.16977#A3.E22),s0\+2​D≤2​L\+2\+L/2≤ns\_\{0\}\+2D\\leq 2L\+2\+\\sqrt\{L\}/2\\leq n, so the weightsws0,…,ws0\+2​Dw\_\{s\_\{0\}\},\\dots,w\_\{s\_\{0\}\+2D\}are all defined and positive\. Ass0s\_\{0\}is a mode and\(ws\)\(w\_\{s\}\)is unimodal,ws0\+2​j≥ws0\+2​Dw\_\{s\_\{0\}\+2j\}\\geq w\_\{s\_\{0\}\+2D\}for0≤j≤D0\\leq j\\leq D, whence

W≥∑j=0Dws0\+2​j≥\(D\+1\)​ws0\+2​DW\\ \\geq\\ \\sum\_\{j=0\}^\{D\}w\_\{s\_\{0\}\+2j\}\\ \\geq\\ \(D\+1\)\\,w\_\{s\_\{0\}\+2D\}and therefore

maxs⁡ℙ⁡\(S=s\)=ws0W≤1D\+1⋅ws0ws0\+2​D=1D\+1​∏j=0D−1ρ​\(s0\+2​j\)−1\.\\max\_\{s\}\\mathbb\{P\}\(S=s\)=\\frac\{w\_\{s\_\{0\}\}\}\{W\}\\ \\leq\\ \\frac\{1\}\{D\+1\}\\cdot\\frac\{w\_\{s\_\{0\}\}\}\{w\_\{s\_\{0\}\+2D\}\}=\\frac\{1\}\{D\+1\}\\prod\_\{j=0\}^\{D\-1\}\\rho\(s\_\{0\}\+2j\)^\{\-1\}\.\(25\)Sinceρ\\rhois decreasing,∏j=0D−1ρ⁡\(s0\+2​j\)≥ρ​\(s0\+2​D\)D\\prod\_\{j=0\}^\{D\-1\}\\rho\(s\_\{0\}\+2j\)\\geq\\rho\(s\_\{0\}\+2D\)^\{D\}, so it remains to boundρ⁡\(s0\+2​D\)\\rho\(s\_\{0\}\+2D\)from below\.

Ass0s\_\{0\}is a mode ands0≥12​L≥2s\_\{0\}\\geq\\tfrac\{1\}\{2\}L\\geq 2, we havews0−2≤ws0w\_\{s\_\{0\}\-2\}\\leq w\_\{s\_\{0\}\}, that isρ⁡\(s0−2\)≥1\\rho\(s\_\{0\}\-2\)\\geq 1, which readsa2​\(n−s0\+2\)≥\(s0−1\)​s0a^\{2\}\(n\-s\_\{0\}\+2\)\\geq\(s\_\{0\}\-1\)s\_\{0\}and hence

a2​\(n−s0\)≥s02−s0−2​a2\.a^\{2\}\(n\-s\_\{0\}\)\\ \\geq\\ s\_\{0\}^\{2\}\-s\_\{0\}\-2a^\{2\}\.Subtracting2​a2​D2a^\{2\}Dfrom both sides,

a2​\(n−s0−2​D\)≥s02−s0−2​a2​\(D\+1\)\.a^\{2\}\(n\-s\_\{0\}\-2D\)\\ \\geq\\ s\_\{0\}^\{2\}\-s\_\{0\}\-2a^\{2\}\(D\+1\)\.Nows0≥L/2s\_\{0\}\\geq L/2givess0≤2​s02/Ls\_\{0\}\\leq 2s\_\{0\}^\{2\}/L, whilea2≤L/4a^\{2\}\\leq L/4andD≤L/4D\\leq\\sqrt\{L\}/4give

2​a2​\(D\+1\)≤L2​\(L4\+1\)=L3/28\+L2≤s02​\(12​L\+2L\),2a^\{2\}\(D\+1\)\\ \\leq\\ \\frac\{L\}\{2\}\\Bigl\(\\frac\{\\sqrt\{L\}\}\{4\}\+1\\Bigr\)=\\frac\{L^\{3/2\}\}\{8\}\+\\frac\{L\}\{2\}\\ \\leq\\ s\_\{0\}^\{2\}\\Bigl\(\\frac\{1\}\{2\\sqrt\{L\}\}\+\\frac\{2\}\{L\}\\Bigr\),usings02≥L2/4s\_\{0\}^\{2\}\\geq L^\{2\}/4\. Therefore

a2​\(n−s0−2​D\)≥s02​\(1−12​L−4L\)\.a^\{2\}\(n\-s\_\{0\}\-2D\)\\ \\geq\\ s\_\{0\}^\{2\}\\Bigl\(1\-\\frac\{1\}\{2\\sqrt\{L\}\}\-\\frac\{4\}\{L\}\\Bigr\)\.On the other hand\(2​D\+2\)/s0≤\(12​L\+2\)/\(12​L\)=1/L\+4/L\(2D\+2\)/s\_\{0\}\\leq\(\\tfrac\{1\}\{2\}\\sqrt\{L\}\+2\)/\(\\tfrac\{1\}\{2\}L\)=1/\\sqrt\{L\}\+4/L, so

\(s0\+2​D\+1\)​\(s0\+2​D\+2\)≤\(s0\+2​D\+2\)2≤s02​\(1\+1L\+4L\)2\.\(s\_\{0\}\+2D\+1\)\(s\_\{0\}\+2D\+2\)\\ \\leq\\ \(s\_\{0\}\+2D\+2\)^\{2\}\\ \\leq\\ s\_\{0\}^\{2\}\\Bigl\(1\+\\frac\{1\}\{\\sqrt\{L\}\}\+\\frac\{4\}\{L\}\\Bigr\)^\{2\}\.Combining the last two displays,

ρ⁡\(s0\+2​D\)≥1−12​L−4L\(1\+1L\+4L\)2≥1−3L\\rho\(s\_\{0\}\+2D\)\\ \\geq\\ \\frac\{1\-\\tfrac\{1\}\{2\\sqrt\{L\}\}\-\\tfrac\{4\}\{L\}\}\{\\bigl\(1\+\\tfrac\{1\}\{\\sqrt\{L\}\}\+\\tfrac\{4\}\{L\}\\bigr\)^\{2\}\}\\ \\geq\\ 1\-\\frac\{3\}\{\\sqrt\{L\}\}forLLlarge\. Usinglog⁡\(1−x\)≥−43​x\\log\(1\-x\)\\geq\-\\tfrac\{4\}\{3\}xfor0≤x≤140\\leq x\\leq\\tfrac\{1\}\{4\}together withD≤L/4D\\leq\\sqrt\{L\}/4, we get

∏j=0D−1ρ⁡\(s0\+2​j\)≥\(1−3L\)D≥exp⁡\(−4​DL\)≥e−1\.\\prod\_\{j=0\}^\{D\-1\}\\rho\(s\_\{0\}\+2j\)\\ \\geq\\ \\Bigl\(1\-\\frac\{3\}\{\\sqrt\{L\}\}\\Bigr\)^\{D\}\\ \\geq\\ \\exp\\Bigl\(\-\\frac\{4D\}\{\\sqrt\{L\}\}\\Bigr\)\\ \\geq\\ e^\{\-1\}\.Substituting into equation[25](https://arxiv.org/html/2608.16977#A3.E25)and usingD\+1≥L/8D\+1\\geq\\sqrt\{L\}/8givesmaxs⁡ℙ⁡\(S=s\)≤8​e/L≤22/L\\max\_\{s\}\\mathbb\{P\}\(S=s\)\\leq 8e/\\sqrt\{L\}\\leq 22/\\sqrt\{L\}\. ∎

###### Lemma C\.7\.11\.

Assume equation[21](https://arxiv.org/html/2608.16977#A3.E21)andnnlarge\. Then𝔼⁡\[S\]≥L/4\\mathbb\{E\}\[S\]\\geq L/4, and consequentlyλ⁡\(Gn,a\)≥n/4\\lambda\(G\_\{n,a\}\)\\geq\\sqrt\{n\}/4\.

###### Proof\.

LetT′T^\{\\prime\}be the largest admissiblesswiths≤L/2s\\leq L/2, so thatT′≥L/2−2T^\{\\prime\}\\geq L/2\-2\. For admissibles≤T′s\\leq T^\{\\prime\}we haven−s≥n−L/2≥n/2n\-s\\geq n\-L/2\\geq n/2by equation[22](https://arxiv.org/html/2608.16977#A3.E22), hence

ρ⁡\(s\)=a2​\(n−s\)\(s\+1\)​\(s\+2\)≥a2​n/2\(L/2\+2\)2=L2/2\(L/2\+2\)2≥32\\rho\(s\)=\\frac\{a^\{2\}\(n\-s\)\}\{\(s\+1\)\(s\+2\)\}\\ \\geq\\ \\frac\{a^\{2\}n/2\}\{\(L/2\+2\)^\{2\}\}=\\frac\{L^\{2\}/2\}\{\(L/2\+2\)^\{2\}\}\\ \\geq\\ \\frac\{3\}\{2\}forLLlarge\. ThereforewT′−2​j≤\(2/3\)j​wT′w\_\{T^\{\\prime\}\-2j\}\\leq\(2/3\)^\{j\}w\_\{T^\{\\prime\}\}for allj≥0j\\geq 0, so that

ℙ⁡\(S≤T′\)=∑j≥0wT′−2​jW≤3​wT′W≤3​maxs⁡ℙ⁡\(S=s\)≤66L\\mathbb\{P\}\(S\\leq T^\{\\prime\}\)=\\sum\_\{j\\geq 0\}\\frac\{w\_\{T^\{\\prime\}\-2j\}\}\{W\}\\ \\leq\\ 3\\,\\frac\{w\_\{T^\{\\prime\}\}\}\{W\}\\ \\leq\\ 3\\max\_\{s\}\\mathbb\{P\}\(S=s\)\\ \\leq\\ \\frac\{66\}\{\\sqrt\{L\}\}by Lemma[C\.7\.10](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem10)\. Consequently𝔼⁡\[S\]≥T′​ℙ​\(S\>T′\)≥\(L/2−2\)​\(1−66/L\)≥L/4\\mathbb\{E\}\[S\]\\geq T^\{\\prime\}\\mathbb\{P\}\(S\>T^\{\\prime\}\)\\geq\(L/2\-2\)\(1\-66/\\sqrt\{L\}\)\\geq L/4forLLlarge\. Dividing byaagivesλ⁡\(Gn,a\)=𝔼⁡\[S\]/a≥n/4\\lambda\(G\_\{n,a\}\)=\\mathbb\{E\}\[S\]/a\\geq\\sqrt\{n\}/4\. ∎

#### C\.7\.5A variance lower bound

We use the following elementary fact, which converts a bound on the largest atom of an integer\-valued random variable into a lower bound on its variance\.

###### Lemma C\.7\.12\.

LetXXbe an integer\-valued random variable and putθ:=maxk∈ℤ⁡ℙ⁡\(X=k\)\\theta:=\\max\_\{k\\in\\mathbb\{Z\}\}\\mathbb\{P\}\(X=k\)\. Then

Var⁡\[X\]≥\(1−θ\)312​θ2\.\\mathrm\{Var\}\[X\]\\ \\geq\\ \\frac\{\(1\-\\theta\)^\{3\}\}\{12\\,\\theta^\{2\}\}\.

###### Proof\.

Fixc∈ℝc\\in\\mathbb\{R\}\. Foru≥0u\\geq 0the interval\[c−u,c\+u\]\[c\-u,c\+u\]contains at most2​u\+12u\+1integers, soℙ⁡\(\|X−c\|≤u\)≤\(2​u\+1\)​θ\\mathbb\{P\}\(\|X\-c\|\\leq u\)\\leq\(2u\+1\)\\thetaandℙ⁡\(\|X−c\|\>u\)≥1−\(2​u\+1\)​θ\\mathbb\{P\}\(\|X\-c\|\>u\)\\geq 1\-\(2u\+1\)\\theta\. PutU=\(1−θ\)/\(2​θ\)U=\(1\-\\theta\)/\(2\\theta\), the point at which the last expression vanishes, so that the integrand below is nonnegative on\[0,U\]\[0,U\]\. Then

𝔼⁡\[\(X−c\)2\]=∫0∞2​u​ℙ​\(\|X−c\|\>u\)​𝑑u≥∫0U2​u​\(1−\(2​u\+1\)​θ\)​𝑑u=U2​\(1−θ\)−43​θ​U3\.\\mathbb\{E\}\\bigl\[\(X\-c\)^\{2\}\\bigr\]=\\int\_\{0\}^\{\\infty\}2u\\,\\mathbb\{P\}\(\|X\-c\|\>u\)\\,du\\ \\geq\\ \\int\_\{0\}^\{U\}2u\\bigl\(1\-\(2u\+1\)\\theta\\bigr\)\\,du=U^\{2\}\(1\-\\theta\)\-\\tfrac\{4\}\{3\}\\theta U^\{3\}\.SubstitutingU=\(1−θ\)/\(2​θ\)U=\(1\-\\theta\)/\(2\\theta\)gives

\(1−θ\)34​θ2−\(1−θ\)36​θ2=\(1−θ\)312​θ2,\\frac\{\(1\-\\theta\)^\{3\}\}\{4\\theta^\{2\}\}\-\\frac\{\(1\-\\theta\)^\{3\}\}\{6\\theta^\{2\}\}=\\frac\{\(1\-\\theta\)^\{3\}\}\{12\\theta^\{2\}\},and takingc=𝔼⁡\[X\]c=\\mathbb\{E\}\[X\]proves the claim\. ∎

###### Corollary C\.7\.13\.

Assume equation[21](https://arxiv.org/html/2608.16977#A3.E21)andnnlarge\. ThenVar⁡\[S\]≥L/8000\\mathrm\{Var\}\[S\]\\geq L/8000\.

###### Proof\.

Writeθ=maxs⁡ℙ⁡\(S=s\)\\theta=\\max\_\{s\}\\mathbb\{P\}\(S=s\), soθ≤22/L\\theta\\leq 22/\\sqrt\{L\}by Lemma[C\.7\.10](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem10), andθ≤110\\theta\\leq\\tfrac\{1\}\{10\}sinceLLis large\. The functionx↦\(1−x\)3/\(12​x2\)x\\mapsto\(1\-x\)^\{3\}/\(12x^\{2\}\)is decreasing on\(0,1\)\(0,1\), so Lemma[C\.7\.12](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem12)gives

Var⁡\[S\]≥\(1−θ\)312​θ2≥\(9/10\)312⋅L484≥L8000\.∎\\mathrm\{Var\}\[S\]\\ \\geq\\ \\frac\{\(1\-\\theta\)^\{3\}\}\{12\\theta^\{2\}\}\\ \\geq\\ \\frac\{\(9/10\)^\{3\}\}\{12\}\\cdot\\frac\{L\}\{484\}\\ \\geq\\ \\frac\{L\}\{8000\}\.\\qed

#### C\.7\.6Proof of the main theorem

###### Proof of Theorem[C\.7\.2](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem2)\.

Letn0n\_\{0\}be an absolute constant large enough for the finitely many hypotheses “nnlarge” invoked above, and assumen≥n0n\\geq n\_\{0\}and4≤a≤n1/44\\leq a\\leq n^\{1/4\}\. Lemmas[C\.7\.9](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem9)and[C\.7\.11](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem11)give14​n≤λ⁡\(Gn,a\)≤5​n\\tfrac\{1\}\{4\}\\sqrt\{n\}\\leq\\lambda\(G\_\{n,a\}\)\\leq 5\\sqrt\{n\}\. For the variance,a≥4a\\geq 4gives\(\(a−2\)/\(2​a\)\)2≥\(1/4\)2=1/16\\bigl\(\(a\-2\)/\(2a\)\\bigr\)^\{2\}\\geq\(1/4\)^\{2\}=1/16, so equation[20](https://arxiv.org/html/2608.16977#A3.E20)and Corollary[C\.7\.13](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem13)give

σ2​\(Gn,a\)≥116​Var​\[S\]≥L128000≥10−6​a​n,\\sigma^\{2\}\(G\_\{n,a\}\)\\ \\geq\\ \\frac\{1\}\{16\}\\mathrm\{Var\}\[S\]\\ \\geq\\ \\frac\{L\}\{128000\}\\ \\geq\\ 10^\{\-6\}a\\sqrt\{n\},which is the bound onσ2​\(Gn,a\)\\sigma^\{2\}\(G\_\{n,a\}\)withc=10−6c=10^\{\-6\}\. Dividing the two bounds givesσ2​\(Gn,a\)/λ⁡\(Gn,a\)≥2⋅10−7​a\\sigma^\{2\}\(G\_\{n,a\}\)/\\lambda\(G\_\{n,a\}\)\\geq 2\\cdot 10^\{\-7\}a\. ∎

###### Proof of Corollary[C\.7\.3](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem3)\.

Forn≥max⁡\{n0,256\}n\\geq\\max\\\{n\_\{0\},256\\\}the choicea=an=⌊n1/4⌋a=a\_\{n\}=\\lfloor n^\{1/4\}\\rfloorsatisfies4≤an≤n1/44\\leq a\_\{n\}\\leq n^\{1/4\}, so Theorem[C\.7\.2](https://arxiv.org/html/2608.16977#A3.SS7.Thmtheorem2)applies toHn=Gn,anH\_\{n\}=G\_\{n,a\_\{n\}\}\. It givesλ⁡\(Hn\)≥n/4→∞\\lambda\(H\_\{n\}\)\\geq\\sqrt\{n\}/4\\to\\infty, thenσ2​\(Hn\)≥c​an​n→∞\\sigma^\{2\}\(H\_\{n\}\)\\geq c\\,a\_\{n\}\\sqrt\{n\}\\to\\infty, and finallyσ2​\(Hn\)/λ⁡\(Hn\)≥c​an/5=Ω⁡\(n1/4\)\\sigma^\{2\}\(H\_\{n\}\)/\\lambda\(H\_\{n\}\)\\geq c\\,a\_\{n\}/5=\\Omega\(n^\{1/4\}\)\. A boundσ2​\(G\)≤C​λ​\(G\)\\sigma^\{2\}\(G\)\\leq C\\lambda\(G\)valid for all finite simple graphsGGwould forcean≤5​C/ca\_\{n\}\\leq 5C/cfor everynn, which is absurd\. ∎

## References\.

- Godsil \(1981\)Chris D Godsil\.Matching behaviour is asymptotically normal\.*Combinatorica*, 1\(4\):369–376, 1981\.
- Heilmann & Lieb \(1972\)Ole J Heilmann and Elliott H Lieb\.Theory of monomer\-dimer systems\.*Communications in mathematical Physics*, 25\(3\):190–232, 1972\.
- Kahn \(2000\)Jeff Kahn\.A normal law for matchings\.*Combinatorica*, 20\(3\):339–392, 2000\.
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### C\.8An ordered hypergraph extremal function of ordern​log⁡nn\\log n

For a fixed patternFF, Klazar’s extremal functionexe⁡\(F,n\)\\operatorname\{ex\}\_\{e\}\(F,n\)is the largest number of edges of a simple hypergraph on at mostnnlinearly ordered vertices that contains no order\-preserving Berge copy ofFF\. For the five\-vertex ordered graphG1G\_\{1\}below, the corresponding ordered*graph*extremal function isΘ⁡\(n​log⁡n\)\\Theta\(n\\log n\), while Klazar’s bounds for the hypergraph function differ by a factor oflog⁡n​\(log⁡log⁡n\)3\\log n\(\\log\\log n\)^\{3\}\. We show that the two functions have the same order of magnitude, so thatexe⁡\(G1,n\)=Θ⁡\(n​log⁡n\)\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)=\\Theta\(n\\log n\)\. The proof compresses aG1G\_\{1\}\-free hypergraph into a singleG1G\_\{1\}\-free ordered graph, at the cost of a factor of five and an additivenn\.

#### C\.8\.1Introduction

We follow the setup of[Klazar 2004a](https://arxiv.org/html/2608.16977#bibh.bib5)\. A*hypergraph*is a finite listH=\(Ej:j∈I\)H=\(E\_\{j\}:j\\in I\)of finite nonempty subsets ofℕ\\mathbb\{N\}, called*edges*\. Its vertex set isV⁡\(H\)=⋃j∈IEjV\(H\)=\\bigcup\_\{j\\in I\}E\_\{j\}, so thatHHhas no isolated vertices, and we writee⁡\(H\)=\|I\|e\(H\)=\|I\|for the number of edges andi⁡\(H\)=∑j∈I\|Ej\|i\(H\)=\\sum\_\{j\\in I\}\|E\_\{j\}\|for the number of vertex–edge incidences\. The hypergraphHHis*simple*if its edges are pairwise distinct, and it is a*graph*if every edge has exactly two elements\. Vertices always carry the linear order inherited fromℕ\\mathbb\{N\}\.

Given hypergraphsH=\(Ej:j∈I\)H=\(E\_\{j\}:j\\in I\)andF=\(Fk:k∈K\)F=\(F\_\{k\}:k\\in K\), we writeH≻FH\\succ F, and say thatHH*contains*FF, if there are an increasing injectionϕ:V⁡\(F\)→V⁡\(H\)\\phi\\colon V\(F\)\\to V\(H\)and an injectionψ:K→I\\psi\\colon K\\to Isuch that

ϕ⁡\(Fk\)⊆Eψ⁡\(k\)for every​k∈K\.\\phi\(F\_\{k\}\)\\subseteq E\_\{\\psi\(k\)\}\\qquad\\text\{for every \}k\\in K\.OtherwiseHHis*FF\-free*, writtenH⊁FH\\not\\succ F\. Thus a copy ofFFinHHconsists of an order\-preserving image of the vertices ofFFtogether with a choice of pairwise distinct hyperedges ofHH, one for each edge ofFF, each containing the image of the edge assigned to it; the chosen hyperedges are allowed to contain further vertices\. This is a Berge\-type containment that respects the vertex order, and it restricts to ordinary ordered subgraph containment whenHHandFFare both graphs, since thenϕ⁡\(Fk\)\\phi\(F\_\{k\}\)andEψ⁡\(k\)E\_\{\\psi\(k\)\}both have two elements and the inclusion forces equality\. The associated extremal functions are

exe⁡\(F,n\)\\displaystyle\\operatorname\{ex\}\_\{e\}\(F,n\)=max\{e\(H\):Hsimple,\|V\(H\)\|≤n,H⊁F\},\\displaystyle=\\max\\\{e\(H\):\\ H\\text\{ simple\},\\ \|V\(H\)\|\\leq n,\\ H\\not\\succ F\\\},exi⁡\(F,n\)\\displaystyle\\operatorname\{ex\}\_\{i\}\(F,n\)=max\{i\(H\):Hsimple,\|V\(H\)\|≤n,H⊁F\},\\displaystyle=\\max\\\{i\(H\):\\ H\\text\{ simple\},\\ \|V\(H\)\|\\leq n,\\ H\\not\\succ F\\\},andgex⁡\(F,n\)\\operatorname\{gex\}\(F,n\)denotes the same maximum ofe⁡\(G\)e\(G\)taken over simple ordered*graphs*GGalone\. Of coursegex⁡\(F,n\)≤exe⁡\(F,n\)≤exi⁡\(F,n\)\\operatorname\{gex\}\(F,n\)\\leq\\operatorname\{ex\}\_\{e\}\(F,n\)\\leq\\operatorname\{ex\}\_\{i\}\(F,n\)for everyFFandnn\.

Throughout,G1G\_\{1\}denotes the ordered graph

G1=\(\{1,3\},\{1,5\},\{2,3\},\{2,4\}\)G\_\{1\}=\\bigl\(\\\{1,3\\\},\\\{1,5\\\},\\\{2,3\\\},\\\{2,4\\\}\\bigr\)on the vertices1<2<3<4<51<2<3<4<5\. Reading\{1,2\}\\\{1,2\\\}as rows and\{3,4,5\}\\\{3,4,5\\\}as columns,G1G\_\{1\}is the ordered bipartite graph of the2×32\\times 3zero\-one matrix\(101110\)\\left\(\\begin\{smallmatrix\}1&0&1\\\\ 1&1&0\\end\{smallmatrix\}\\right\), and[Füredi 1990](https://arxiv.org/html/2608.16977#bibh.bib3)determined the extremal function of that matrix in the course of bounding the number of unit distances among the vertices of a convexnn\-gon\. The upper bound was obtained independently by[Bienstock & Györi 1991](https://arxiv.org/html/2608.16977#bibh.bib2)\.[Füredi & Hajnal 1992](https://arxiv.org/html/2608.16977#bibh.bib4)began the systematic study of the resulting matrix extremal problems, and[Tardos 2019](https://arxiv.org/html/2608.16977#bibh.bib9)surveys the ordered graph theory that grew out of them\. As[Klazar 2004a](https://arxiv.org/html/2608.16977#bibh.bib5)records, in this particular case the bounds pass from ordered bipartite graphs to all ordered graphs, so that

gex⁡\(G1,n\)=Θ⁡\(n​log⁡n\)\.\\operatorname\{gex\}\(G\_\{1\},n\)=\\Theta\(n\\log n\)\.\(26\)
[Klazar 2004a](https://arxiv.org/html/2608.16977#bibh.bib5)introduced the containment≻\\succand the functionsexe\\operatorname\{ex\}\_\{e\}andexi\\operatorname\{ex\}\_\{i\}, and asked how much of equation[26](https://arxiv.org/html/2608.16977#A3.E26)survives the passage from ordered graphs to ordered hypergraphs\. His Theorem 3\.3 gives, in its two parts,

n​log⁡n≪exe⁡\(G1,n\)≤exi⁡\(G1,n\)≪n​\(log⁡n\)2​\(log⁡log⁡n\)3,n\\log n\\ll\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)\\leq\\operatorname\{ex\}\_\{i\}\(G\_\{1\},n\)\\ll n\(\\log n\)^\{2\}\(\\log\\log n\)^\{3\},the upper bound being obtained by iterating a recursion that controlsexe⁡\(G1,n\)\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)in terms of the ordered graph extremal function of blow\-ups ofG1G\_\{1\}\. He then asked, as Problem 3\.4:

*What is the exact asymptotics ofexe⁡\(G1,n\)\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)?*

We answer this up to the implied constants: the hypergraph function has the same order of magnitude as the graph function\.

The closest previous work we are aware of on the hypergraph function is Klazar’s own\. In a companion paper,[Klazar 2004b](https://arxiv.org/html/2608.16977#bibh.bib6)determinesexe⁡\(F,n\)\\operatorname\{ex\}\_\{e\}\(F,n\)andexi⁡\(F,n\)\\operatorname\{ex\}\_\{i\}\(F,n\)exactly for the5555patternsFFwith at most four incidences, whereasG1G\_\{1\}has eight\. That paper normalizes by\|V⁡\(H\)\|=n\|V\(H\)\|=nin place of\|V⁡\(H\)\|≤n\|V\(H\)\|\\leq n; by its Proposition 2\.4 the two conventions give the sameexe⁡\(F,n\)\\operatorname\{ex\}\_\{e\}\(F,n\)unlessFFconsists of distinct singleton edges\.[Klazar & Marcus 2007](https://arxiv.org/html/2608.16977#bibh.bib7)and, independently,[Balogh et al\. 2006](https://arxiv.org/html/2608.16977#bibh.bib1)carry the linear bound of[Marcus & Tardos 2004](https://arxiv.org/html/2608.16977#bibh.bib8)for excluded permutation matrices from matrices over to ordered hypergraphs, but that theorem applies to permutation patterns, andG1G\_\{1\}is not one: by equation[26](https://arxiv.org/html/2608.16977#A3.E26)its extremal function is already superlinear at the graph level\.

###### Theorem C\.8\.1\.

For everyn≥1n\\geq 1,

gex⁡\(G1,n\)≤exe⁡\(G1,n\)≤n\+5​gex⁡\(G1,n\)\.\\operatorname\{gex\}\(G\_\{1\},n\)\\ \\leq\\ \\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)\\ \\leq\\ n\+5\\operatorname\{gex\}\(G\_\{1\},n\)\.In particularexe⁡\(G1,n\)=Θ⁡\(n​log⁡n\)\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)=\\Theta\(n\\log n\)\.

The key point is that aG1G\_\{1\}\-free simple ordered hypergraph is controlled by a singleG1G\_\{1\}\-free ordered graph\. Process the hyperedges one at a time and record, for each, one pair of its vertices that has not been recorded before\. The recorded pairs form an ordered graphBB, and a copy ofG1G\_\{1\}inBBpulls back to four distinct hyperedges forming a copy ofG1G\_\{1\}in the host, soBBisG1G\_\{1\}\-free\. The hyperedges from which nothing new was recorded are cliques ofBB, and the crucial observation is that aG1G\_\{1\}\-free ordered graphBBhas at most4​e​\(B\)4e\(B\)cliques of size at least two, because in such a graph any edge\{a,b\}\\\{a,b\\\}witha<ba<bhas at most two common neighbors to the right ofbb\.

#### C\.8\.2Cliques in aG1G\_\{1\}\-free ordered graph

We use “clique” to mean any set of pairwise adjacent vertices, not necessarily a maximal one\.

###### Lemma C\.8\.2\.

LetBBbe aG1G\_\{1\}\-free simple ordered graph\. Then for every edge\{a,b\}\\\{a,b\\\}ofBBwitha<ba<b, the verticesaaandbbhave at most two common neighbors greater thanbb\. ConsequentlyBBhas at most4​e​\(B\)4e\(B\)cliques of size at least two\.

###### Proof\.

Fix an edge\{a,b\}\\\{a,b\\\}ofBBwitha<ba<b, and suppose thataaandbbhave three common neighbors greater thanbb\. Choose three such verticesc<d<fc<d<f, so that the five vertices

ofBBcarry in particular the four edges

\{a,c\},\{a,f\},\{b,c\},\{b,d\}\.\\\{a,c\\\},\\quad\\\{a,f\\\},\\quad\\\{b,c\\\},\\quad\\\{b,d\\\}\.The increasing injection1↦a1\\mapsto a,2↦b2\\mapsto b,3↦c3\\mapsto c,4↦d4\\mapsto d,5↦f5\\mapsto fcarries the four edges ofG1G\_\{1\}to these four edges ofBB, soB≻G1B\\succ G\_\{1\}, a contradiction\.

Now letKKbe a clique ofBBwith\|K\|≥2\|K\|\\geq 2and leta<ba<bbe its two smallest vertices\. Then\{a,b\}\\\{a,b\\\}is an edge ofBB, and every remaining vertex ofKKis a common neighbor ofaaandbbgreater thanbb\. HenceKKis determined by the edge\{a,b\}\\\{a,b\\\}together with a subset of the set of common neighbors ofaaandbbto the right ofbb, a set of size at most two\. So at most22=42^\{2\}=4cliques of size at least two haveaaandbbas their two smallest vertices, and summing over the edges ofBBgives the bound\. ∎

#### C\.8\.3Proof of the main theorem

###### Proof of Theorem[C\.8\.1](https://arxiv.org/html/2608.16977#A3.SS8.Thmtheorem1)\.

The lower bound is immediate: a simple ordered graph is a simple ordered hypergraph, and the containment relation≻\\succused to definegex\\operatorname\{gex\}is the one used to defineexe\\operatorname\{ex\}\_\{e\}, so the maximum definingexe⁡\(G1,n\)\\operatorname\{ex\}\_\{e\}\(G\_\{1\},n\)is taken over a larger family\.

For the upper bound, letHHbe a simple ordered hypergraph with\|V⁡\(H\)\|≤n\|V\(H\)\|\\leq nandH⊁G1H\\not\\succ G\_\{1\}\. SinceHHis simple, its singleton edges are pairwise distinct one\-element subsets ofV⁡\(H\)V\(H\), so there are at mostnnof them\.

We build an ordered graphBBfrom the remaining edges\. List the edges ofHHof size at least two in an arbitrary order and start withBBempty\. Processing them one at a time, if the current edgeEEcontains a pair\{a,b\}\\\{a,b\\\}that is not yet an edge ofBB, then add one such pair toBBand callEE*assigned*to that pair; otherwise callEE*unassigned*and change nothing\. Every assigned edge creates exactly one new edge ofBB, so assignment is a bijection between the assigned edges ofHHand the edges ofBB, and in particular the number of assigned edges ise⁡\(B\)e\(B\)\. AlsoV⁡\(B\)⊆V⁡\(H\)V\(B\)\\subseteq V\(H\), so\|V⁡\(B\)\|≤n\|V\(B\)\|\\leq n\.

We claim thatBBisG1G\_\{1\}\-free\. Suppose instead that there are verticesx1<x2<x3<x4<x5x\_\{1\}<x\_\{2\}<x\_\{3\}<x\_\{4\}<x\_\{5\}ofBBsuch that

\{x1,x3\},\{x1,x5\},\{x2,x3\},\{x2,x4\}\\\{x\_\{1\},x\_\{3\}\\\},\\quad\\\{x\_\{1\},x\_\{5\}\\\},\\quad\\\{x\_\{2\},x\_\{3\}\\\},\\quad\\\{x\_\{2\},x\_\{4\}\\\}are all edges ofBB\. These are four distinct edges ofBB, so the hyperedges assigned to them are four distinct edgesD1,D2,D3,D4D\_\{1\},D\_\{2\},D\_\{3\},D\_\{4\}ofHHwith

\{x1,x3\}⊆D1,\{x1,x5\}⊆D2,\{x2,x3\}⊆D3,\{x2,x4\}⊆D4\.\\\{x\_\{1\},x\_\{3\}\\\}\\subseteq D\_\{1\},\\quad\\\{x\_\{1\},x\_\{5\}\\\}\\subseteq D\_\{2\},\\quad\\\{x\_\{2\},x\_\{3\}\\\}\\subseteq D\_\{3\},\\quad\\\{x\_\{2\},x\_\{4\}\\\}\\subseteq D\_\{4\}\.Takingϕ⁡\(i\)=xi\\phi\(i\)=x\_\{i\}for1≤i≤51\\leq i\\leq 5and sending the edges\{1,3\},\{1,5\},\{2,3\},\{2,4\}\\\{1,3\\\},\\\{1,5\\\},\\\{2,3\\\},\\\{2,4\\\}ofG1G\_\{1\}toD1,D2,D3,D4D\_\{1\},D\_\{2\},D\_\{3\},D\_\{4\}respectively givesH≻G1H\\succ G\_\{1\}, contrary to hypothesis\. HenceBBisG1G\_\{1\}\-free, and therefore

e⁡\(B\)≤gex⁡\(G1,n\)\.e\(B\)\\leq\\operatorname\{gex\}\(G\_\{1\},n\)\.
It remains to count the unassigned edges\. If an edgeEEofHHwith\|E\|≥2\|E\|\\geq 2is unassigned, then at the momentEEwas processed every pair contained inEEwas already an edge ofBB; since edges are only ever added toBB, the same holds at the end of the process, soEEis a clique ofBBof size at least two\. Distinct edges ofHHare distinct sets becauseHHis simple, so distinct unassigned edges give distinct cliques ofBB\. By Lemma[C\.8\.2](https://arxiv.org/html/2608.16977#A3.SS8.Thmtheorem2)the number of unassigned edges is therefore at most4​e​\(B\)4e\(B\)\.

Collecting the three types of edge,

e⁡\(H\)≤n\+e⁡\(B\)\+4​e​\(B\)≤n\+5​gex⁡\(G1,n\),e\(H\)\\ \\leq\\ n\+e\(B\)\+4e\(B\)\\ \\leq\\ n\+5\\operatorname\{gex\}\(G\_\{1\},n\),which is the upper bound\. The final assertion follows by combining the two bounds with equation[26](https://arxiv.org/html/2608.16977#A3.E26)\. ∎

## References\.

- Balogh et al\. \(2006\)József Balogh, Béla Bollobás, and Robert Morris\.Hereditary properties of partitions, ordered graphs and ordered hypergraphs\.*European Journal of combinatorics*, 27\(8\):1263–1281, 2006\.
- Bienstock & Györi \(1991\)Dan Bienstock and Ervin Györi\.An extremal problem on sparse 0\-1 matrices\.*SIAM Journal on Discrete Mathematics*, 4\(1\):17–27, 1991\.
- Füredi \(1990\)Zoltán Füredi\.The maximum number of unit distances in a convex n\-gon\.*Journal of Combinatorial Theory, Series A*, 55\(2\):316–320, 1990\.
- Füredi & Hajnal \(1992\)Zoltán Füredi and Péter Hajnal\.Davenport\-schinzel theory of matrices\.*Discrete Mathematics*, 103\(3\):233–251, 1992\.
- Klazar \(2004a\)Martin Klazar\.Extremal problems for ordered \(hyper\) graphs: applications of davenport–schinzel sequences\.*European Journal of Combinatorics*, 25\(1\):125–140, 2004a\.
- Klazar \(2004b\)Martin Klazar\.Extremal problems for ordered hypergraphs: small patterns and some enumeration\.*Discrete applied mathematics*, 143\(1\-3\):144–154, 2004b\.
- Klazar & Marcus \(2007\)Martin Klazar and Adam Marcus\.Extensions of the linear bound in the füredi–hajnal conjecture\.*Advances in Applied Mathematics*, 38\(2\):258–266, 2007\.
- Marcus & Tardos \(2004\)Adam Marcus and Gábor Tardos\.Excluded permutation matrices and the stanley–wilf conjecture\.*Journal of Combinatorial Theory, Series A*, 107\(1\):153–160, 2004\.
- Tardos \(2019\)Gábor Tardos\.Extremal theory of vertex or edge ordered graphs1\.*Surveys in combinatorics 2019*, 456:221, 2019\.

### C\.9Eigenvalues below−2\-2under repeated subdivision

This problem concerns the eigenvalues that a graph retains below−2\-2when a fixed set of its edges is subdivided over and over\. We show that their number is eventually constant, and that the constant is the number of negative eigenvalues of a single fixed matrix on the original vertex set, namely2​In\+AR−DS2I\_\{n\}\+A\_\{R\}\-D\_\{S\}, whereARA\_\{R\}records the edges that are never subdivided andDSD\_\{S\}the degrees in the edges that are\. The proof eliminates the subdivision vertices by a Schur complement, which is legitimate because the block they contribute is positive definite, and then compares the resultingn×nn\\times nmatrices in the Loewner order\.

#### C\.9\.1Introduction

LetGGbe a finite simple graph onn=\|V⁡\(G\)\|n=\|V\(G\)\|vertices, letS⊆E⁡\(G\)S\\subseteq E\(G\)be a fixed set of edges, and writeR=E⁡\(G\)∖SR=E\(G\)\\setminus Sfor the rest\. Fort≥1t\\geq 1letGt=Gt​\(S\)G\_\{t\}=G\_\{t\}\(S\)be the graph obtained fromGGby replacing every edgeu​v∈Suv\\in Swith a path of lengthttfromuutovv, the*tt\-stretch*ofu​vuv, whoset−1t\-1internal vertices are new and lie on no other stretch\. The edges ofRRare left untouched, andG1=GG\_\{1\}=G\. We writeλ1​\(X\)≥⋯≥λ\|V⁡\(X\)\|​\(X\)\\lambda\_\{1\}\(X\)\\geq\\cdots\\geq\\lambda\_\{\|V\(X\)\|\}\(X\)for the adjacency eigenvalues of a graphXXand

mX​\(a,b\):=\#⁡\{i:λi​\(X\)∈\(a,b\)\}m\_\{X\}\(a,b\):=\\\#\\\{i:\\lambda\_\{i\}\(X\)\\in\(a,b\)\\\}for the number of them in an interval, counted with multiplicity\. For a real symmetric matrixMMwe writen−​\(M\)n\_\{\-\}\(M\)for its number of negative eigenvalues, counted with multiplicity, that is, for its*negative index of inertia*\. FinallyARA\_\{R\}andASA\_\{S\}denote the adjacency matrices of the spanning subgraphs\(V⁡\(G\),R\)\(V\(G\),R\)and\(V⁡\(G\),S\)\(V\(G\),S\), andDSD\_\{S\}the diagonal matrix of the degreesdegS⁡\(v\)\\deg\_\{S\}\(v\)in\(V⁡\(G\),S\)\(V\(G\),S\), all indexed byV⁡\(G\)V\(G\); the degree ofvvin\(V⁡\(G\),R\)\(V\(G\),R\)is writtendegR⁡\(v\)\\deg\_\{R\}\(v\)\.

The window\[−2,2\]\[\-2,2\]is the one that matters here\. DeletingQ=\{v∈V⁡\(G\):degG⁡\(v\)≥3\}Q=\\\{v\\in V\(G\):\\deg\_\{G\}\(v\)\\geq 3\\\}fromGtG\_\{t\}leaves a disjoint union of paths and cycles, whose eigenvalues all lie in\[−2,2\]\[\-2,2\], so interlacing bounds the number of eigenvalues ofGtG\_\{t\}above22, and likewise the number below−2\-2, by\|Q\|\|Q\|, uniformly intt\([Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9)\)\. The value−2\-2is also the classical threshold on the negative side: a connected graph withmG​\(−∞,−2\)=0m\_\{G\}\(\-\\infty,\-2\)=0is a generalized line graph or one of finitely many exceptional graphs representable in the root systemE8E\_\{8\}, by the classification of[Cameron et al\. 1991](https://arxiv.org/html/2608.16977#bibi.bib2), and the structure theory of graphs with least eigenvalue−2\-2is built around that dichotomy\([Cvetkovic et al\. 2004](https://arxiv.org/html/2608.16977#bibi.bib3)\)\. SomGt​\(−∞,−2\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\)measures how far the subdivided graph is from being a generalized line graph, and the question is whether repeated subdivision settles this quantity down\.

Subdivision has been studied spectrally since the theorem of[Hoffman & Smith 1974](https://arxiv.org/html/2608.16977#bibi.bib6)on the spectral radii of topologically equivalent graphs, and it is the basic operation in Hoffman’s program on limit points of spectral radii, surveyed by[Wang et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib10)\. Subdividing a*subset*of the edges is exactly the operation used by[Haiman et al\. 2022](https://arxiv.org/html/2608.16977#bibi.bib4)to construct graphs with high approximate second eigenvalue multiplicity, showing that the multiplicity bound of[Jiang et al\. 2021](https://arxiv.org/html/2608.16977#bibi.bib8)behind the resolution of the equiangular lines problem is sharp in that relaxed sense\. Motivated by this,[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9)analyze the whole spectrum ofGtG\_\{t\}ast→∞t\\to\\infty\. They study the four sequencesmGt​\(2,∞\)m\_\{G\_\{t\}\}\(2,\\infty\),mGt​\(−∞,−2\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\),mHt​\(2,∞\)m\_\{H\_\{t\}\}\(2,\\infty\), andmHt​\(−∞,−2\)m\_\{H\_\{t\}\}\(\-\\infty,\-2\), whereHt=Ht​\(S\)H\_\{t\}=H\_\{t\}\(S\)is obtained fromG2​t\+1​\(S\)G\_\{2t\+1\}\(S\)by deleting the middle edge of every stretch\. SinceHtH\_\{t\}is an induced subgraph ofHt\+1H\_\{t\+1\}, the twoHH\-sequences are nondecreasing, and subdividing a single edge cannot decrease the number of eigenvalues above22, somGt​\(2,∞\)m\_\{G\_\{t\}\}\(2,\\infty\)is nondecreasing as well\. Being nondecreasing and bounded by\|Q\|\|Q\|, those three sequences are eventually constant\. The fourth sequence is not monotone, and is left open as[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9):

*There existst0∈ℕt\_\{0\}\\in\\mathbb\{N\}such thatmGt​\(−∞,−2\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\)is constant for allt≥t0t\\geq t\_\{0\}\.*

The failure of monotonicity is genuine and is caused by parity: subdividing changes the length of every cycle through a stretch, and Figure 1 of[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9)exhibits a graph whose successive subdivisions have11,00, and11eigenvalues below−2\-2\. Example[C\.9\.4](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem4)below returns to that graph and follows the sequence to the end\.

In the special caseS=E⁡\(G\)S=E\(G\)the conjecture already follows from the same paper\. There[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9)show that for each fixedk≤nk\\leq nthe eigenvalueλ\|V⁡\(Gt\)\|−k\+1​\(Gt\)\\lambda\_\{\|V\(G\_\{t\}\)\|\-k\+1\}\(G\_\{t\}\)tends to−dk/dk−1\-d\_\{k\}/\\sqrt\{d\_\{k\}\-1\}whendk≥3d\_\{k\}\\geq 3and to−2\-2otherwise, whered1≥⋯≥dnd\_\{1\}\\geq\\cdots\\geq d\_\{n\}is the degree sequence ofGG, so at least\|Q\|\|Q\|eigenvalues ofGtG\_\{t\}lie below−2\-2oncettis large, while their interlacing bound caps the count at\|Q\|\|Q\|for everytt\. For a generalSSthe upper half of this argument survives, since that cap is proved for an arbitrarySS; the lower half does not\. They do prove that each individual eigenvalueλ\|V⁡\(Gt\)\|−k\+1​\(Gt\)\\lambda\_\{\|V\(G\_\{t\}\)\|\-k\+1\}\(G\_\{t\}\)converges, but for a generalSSthe limit is not known in closed form and may be−2\-2itself, and an eigenvalue converging to−2\-2is free to cross the threshold again and again\. That boundary case is the whole difficulty\. Since the count never exceeds\|Q\|\|Q\|, it equals the number ofk≤\|Q\|k\\leq\|Q\|withλ\|V⁡\(Gt\)\|−k\+1​\(Gt\)<−2\\lambda\_\{\|V\(G\_\{t\}\)\|\-k\+1\}\(G\_\{t\}\)<\-2; the indices whose limit differs from−2\-2contribute a constant from some point on, and Conjecture 20 asserts that the remaining ones do too\. We are aware of no work on the conjecture itself; the papers citing[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9)pursue other lines, among them the Hoffman program\([Wang et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib10)\)and the effect of subdivision on other spectral parameters, such as the Perron eigenvalue of the Ricci matrix of a tree\([Bai et al\. 2026](https://arxiv.org/html/2608.16977#bibi.bib1)\), which is also handled by eliminating the internal structure with a Schur complement\.

We prove the conjecture for everyGGand everySS, and identify the constant\.

###### Theorem C\.9\.1\.

LetGGbe a finite simple graph onnnvertices, letS⊆E⁡\(G\)S\\subseteq E\(G\), letR=E⁡\(G\)∖SR=E\(G\)\\setminus S, and put

K∞:=2​In\+AR−DS\.K\_\{\\infty\}:=2I\_\{n\}\+A\_\{R\}\-D\_\{S\}\.\(27\)Then

mGt​\(−∞,−2\)≤n−​\(K∞\)for every​t≥2,m\_\{G\_\{t\}\}\(\-\\infty,\-2\)\\leq n\_\{\-\}\(K\_\{\\infty\}\)\\qquad\\text\{for every \}t\\geq 2,with equality for all sufficiently largett\. In particularmGt​\(−∞,−2\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\)is eventually constant, and its eventual value isn−​\(K∞\)n\_\{\-\}\(K\_\{\\infty\}\)\.

When every edge is subdivided the matrixK∞K\_\{\\infty\}is diagonal and the eventual value can be read off the degree sequence\.

###### Corollary C\.9\.2\.

IfS=E⁡\(G\)S=E\(G\), thenmGt​\(−∞,−2\)=\|Q\|m\_\{G\_\{t\}\}\(\-\\infty,\-2\)=\|Q\|for all sufficiently largett, whereQ=\{v∈V⁡\(G\):degG⁡\(v\)≥3\}Q=\\\{v\\in V\(G\):\\deg\_\{G\}\(v\)\\geq 3\\\}\.

This is the value predicted by[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9), and it makes explicit the negative\-side case of the tightness they assert for their interlacing bound\.

###### Example C\.9\.4\.

LetGGbe the cyclev1​v2​v3​v4v\_\{1\}v\_\{2\}v\_\{3\}v\_\{4\}together with a pendant edgev1​v5v\_\{1\}v\_\{5\}, and letS=\{v2​v3\}S=\\\{v\_\{2\}v\_\{3\}\\\}, so thatGtG\_\{t\}is the cycleCt\+3C\_\{t\+3\}with a pendant edge attached\. This is the family in Figure 1 of[Kumar et al\. 2025](https://arxiv.org/html/2608.16977#bibi.bib9), whose first three members have11,00, and11eigenvalues below−2\-2\. HereDS=diag⁡\(0,1,1,0,0\)D\_\{S\}=\\operatorname\{diag\}\(0,1,1,0,0\)and

K∞=\(2101111000001101012010002\),K\_\{\\infty\}=\\begin\{pmatrix\}2&1&0&1&1\\\\ 1&1&0&0&0\\\\ 0&0&1&1&0\\\\ 1&0&1&2&0\\\\ 1&0&0&0&2\\end\{pmatrix\},whose eigenvalues are approximately−0\.1388\-0\.1388,0\.54010\.5401,1\.51021\.5102,2\.38352\.3835and3\.70503\.7050, son−​\(K∞\)=1n\_\{\-\}\(K\_\{\\infty\}\)=1\. Theorem[C\.9\.1](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem1)therefore predicts the value11, and indeed the sequencemGt​\(−∞,−2\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\)fort=1,2,3,…t=1,2,3,\\dotsis

1,0,1,0,1,1,1,1,…,1,\\;0,\\;1,\\;0,\\;1,\\;1,\\;1,\\;1,\\;\\dots,oscillating with the parity of the cycle length untilt=5t=5and constant thereafter\. The effective threshold of Remark[C\.9\.3](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem3)gives onlyt\>2⋅3/γt\>2\\cdot 3/\\gammawithγ≈0\.1388\\gamma\\approx 0\.1388, that ist≥44t\\geq 44, so the bound is far from sharp on this example\.

The proof has two steps\. The first is exact and holds for everyt≥2t\\geq 2: eliminating the subdivision vertices replacesA⁡\(Gt\)\+2​IA\(G\_\{t\}\)\+2Iwith ann×nn\\times nmatrix

Kt=K∞\+1t​\(DS−\(−1\)t​AS\)K\_\{t\}=K\_\{\\infty\}\+\\frac\{1\}\{t\}\\bigl\(D\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}\\bigr\)\(28\)of the same negative inertia\. The second is a comparison: the perturbation in equation[28](https://arxiv.org/html/2608.16977#A3.E28)is positive semidefinite for every parity oftt, being the Laplacian of\(V⁡\(G\),S\)\(V\(G\),S\)for eventtand its signless Laplacian for oddtt, and it tends to00\. SoKtK\_\{t\}approachesK∞K\_\{\\infty\}from above in the Loewner order, which pins the negative inertia from both sides\.

#### C\.9\.2Eliminating the subdivision vertices

WritePℓP\_\{\\ell\}for the path onℓ\\ellvertices and putTℓ:=2​Iℓ\+A⁡\(Pℓ\)T\_\{\\ell\}:=2I\_\{\\ell\}\+A\(P\_\{\\ell\}\)\.

###### Lemma C\.9\.5\.

For everyℓ≥1\\ell\\geq 1the matrixTℓT\_\{\\ell\}is positive definite,detTℓ=ℓ\+1\\det T\_\{\\ell\}=\\ell\+1, and

\(Tℓ−1\)1,1=\(Tℓ−1\)ℓ,ℓ=ℓℓ\+1,\(Tℓ−1\)1,ℓ=\(Tℓ−1\)ℓ,1=\(−1\)ℓ\+1ℓ\+1\.\(T\_\{\\ell\}^\{\-1\}\)\_\{1,1\}=\(T\_\{\\ell\}^\{\-1\}\)\_\{\\ell,\\ell\}=\\frac\{\\ell\}\{\\ell\+1\},\\qquad\(T\_\{\\ell\}^\{\-1\}\)\_\{1,\\ell\}=\(T\_\{\\ell\}^\{\-1\}\)\_\{\\ell,1\}=\\frac\{\(\-1\)^\{\\ell\+1\}\}\{\\ell\+1\}\.

###### Proof\.

Fory=\(y1,…,yℓ\)𝖳∈ℝℓy=\(y\_\{1\},\\dots,y\_\{\\ell\}\)^\{\\mathsf\{T\}\}\\in\\mathbb\{R\}^\{\\ell\},

y𝖳​Tℓ​y=2​∑i=1ℓyi2\+2​∑i=1ℓ−1yi​yi\+1=y12\+yℓ2\+∑i=1ℓ−1\(yi\+yi\+1\)2\.y^\{\\mathsf\{T\}\}T\_\{\\ell\}y=2\\sum\_\{i=1\}^\{\\ell\}y\_\{i\}^\{2\}\+2\\sum\_\{i=1\}^\{\\ell\-1\}y\_\{i\}y\_\{i\+1\}=y\_\{1\}^\{2\}\+y\_\{\\ell\}^\{2\}\+\\sum\_\{i=1\}^\{\\ell\-1\}\(y\_\{i\}\+y\_\{i\+1\}\)^\{2\}\.If this vanishes theny1=0y\_\{1\}=0andyi\+1=−yiy\_\{i\+1\}=\-y\_\{i\}for allii, soy=0y=0; henceTℓT\_\{\\ell\}is positive definite\. ExpandingdetTℓ\\det T\_\{\\ell\}along the first row givesdetTℓ=2​detTℓ−1−detTℓ−2\\det T\_\{\\ell\}=2\\det T\_\{\\ell\-1\}\-\\det T\_\{\\ell\-2\}withdetT0=1\\det T\_\{0\}=1anddetT1=2\\det T\_\{1\}=2, whencedetTℓ=ℓ\+1\\det T\_\{\\ell\}=\\ell\+1\. By Cramer’s rule,\(Tℓ−1\)1,1\(T\_\{\\ell\}^\{\-1\}\)\_\{1,1\}is the determinant of the matrix obtained by deleting the first row and column, namelydetTℓ−1=ℓ\\det T\_\{\\ell\-1\}=\\ell, divided bydetTℓ=ℓ\+1\\det T\_\{\\ell\}=\\ell\+1; the entry\(Tℓ−1\)ℓ,ℓ\(T\_\{\\ell\}^\{\-1\}\)\_\{\\ell,\\ell\}is the same by symmetry\. For the corner entry,\(Tℓ−1\)1,ℓ=\(−1\)ℓ\+1​det\(N′\)/detTℓ\(T\_\{\\ell\}^\{\-1\}\)\_\{1,\\ell\}=\(\-1\)^\{\\ell\+1\}\\det\(N^\{\\prime\}\)/\\det T\_\{\\ell\}, whereN′N^\{\\prime\}isTℓT\_\{\\ell\}with its last row and first column deleted\. The rows ofN′N^\{\\prime\}are indexed by1≤i≤ℓ−11\\leq i\\leq\\ell\-1and its columns by2≤j≤ℓ2\\leq j\\leq\\ell, and the entry in position\(i,j\)\(i,j\)vanishes unless\|i−j\|≤1\|i\-j\|\\leq 1; reindexing the columns byk=j−1k=j\-1makesN′N^\{\\prime\}lower triangular with every diagonal entry equal to\(Tℓ\)i,i\+1=1\(T\_\{\\ell\}\)\_\{i,i\+1\}=1\. HencedetN′=1\\det N^\{\\prime\}=1and\(Tℓ−1\)1,ℓ=\(−1\)ℓ\+1/\(ℓ\+1\)\(T\_\{\\ell\}^\{\-1\}\)\_\{1,\\ell\}=\(\-1\)^\{\\ell\+1\}/\(\\ell\+1\)\. ∎

uuw1w\_\{1\}w2w\_\{2\}⋯\\cdotswt−1w\_\{t\-1\}vvTt−1T\_\{t\-1\}uuvv−\(−1\)t/t\-\(\-1\)^\{t\}/t−\(t−1\)/t\-\(t\-1\)/t−\(t−1\)/t\-\(t\-1\)/tFigure 10:Thett\-stretch of an edgeu​v∈Suv\\in SinsideGtG\_\{t\}, on the left, and its contribution to the Schur complement, on the right\. Thet−1t\-1internal vertices, drawn hollow, carry the positive definite blockTt−1T\_\{t\-1\}ofA⁡\(Gt\)\+2​IA\(G\_\{t\}\)\+2I; eliminating them subtracts\(t−1\)/t\(t\-1\)/tfrom each of the diagonal entries atuuand atvvand subtracts\(−1\)t/t\(\-1\)^\{t\}/tfrom the entries atu​vuvand atv​uvu, which is the content of Proposition[C\.9\.6](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem6)\.###### Proposition C\.9\.6\.

For everyt≥2t\\geq 2,

mGt​\(−∞,−2\)=n−​\(Kt\),Kt:=2​In\+AR−t−1t​DS−\(−1\)tt​AS,m\_\{G\_\{t\}\}\(\-\\infty,\-2\)=n\_\{\-\}\(K\_\{t\}\),\\qquad K\_\{t\}:=2I\_\{n\}\+A\_\{R\}\-\\frac\{t\-1\}\{t\}D\_\{S\}\-\\frac\{\(\-1\)^\{t\}\}\{t\}A\_\{S\},andKtK\_\{t\}satisfies equation[28](https://arxiv.org/html/2608.16977#A3.E28)\.

###### Proof\.

PutBt:=A⁡\(Gt\)\+2​IB\_\{t\}:=A\(G\_\{t\}\)\+2I\. An eigenvalueλ\\lambdaofA⁡\(Gt\)A\(G\_\{t\}\)satisfiesλ<−2\\lambda<\-2exactly whenλ\+2<0\\lambda\+2<0, so

mGt​\(−∞,−2\)=n−​\(Bt\)\.m\_\{G\_\{t\}\}\(\-\\infty,\-2\)=n\_\{\-\}\(B\_\{t\}\)\.PartitionV⁡\(Gt\)V\(G\_\{t\}\)into the original verticesV=V⁡\(G\)V=V\(G\)and the setWtW\_\{t\}of internal vertices of the stretches\. Two internal vertices are adjacent inGtG\_\{t\}only if they lie on the same stretch and are consecutive on it, soBt​\[Wt,Wt\]B\_\{t\}\[W\_\{t\},W\_\{t\}\]is block diagonal with one blockTt−1T\_\{t\-1\}for each edge ofSS, as in Figure[10](https://arxiv.org/html/2608.16977#A3.F10)\. By Lemma[C\.9\.5](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem5)this block is positive definite, so Haynsworth’s inertia additivity formula\([Haynsworth 1968](https://arxiv.org/html/2608.16977#bibi.bib5)\)applies and gives

n−​\(Bt\)=n−​\(Bt​\[V,V\]−Bt​\[V,Wt\]​Bt​\[Wt,Wt\]−1​Bt​\[Wt,V\]\)\.n\_\{\-\}\(B\_\{t\}\)=n\_\{\-\}\\bigl\(B\_\{t\}\[V,V\]\-B\_\{t\}\[V,W\_\{t\}\]\\,B\_\{t\}\[W\_\{t\},W\_\{t\}\]^\{\-1\}\\,B\_\{t\}\[W\_\{t\},V\]\\bigr\)\.It remains to identify the Schur complement on the right\.

No edge ofSSsurvives inGtG\_\{t\}fort≥2t\\geq 2, soBt​\[V,V\]=2​In\+ARB\_\{t\}\[V,V\]=2I\_\{n\}\+A\_\{R\}\. Label the stretch ofu​v∈Suv\\in Sasu,w1,…,wℓ,vu,w\_\{1\},\\dots,w\_\{\\ell\},vwithℓ=t−1\\ell=t\-1, so that the column ofBt​\[V,Wt\]B\_\{t\}\[V,W\_\{t\}\]atw1w\_\{1\}iseue\_\{u\}, the column atwℓw\_\{\\ell\}iseve\_\{v\}, and all other columns of that stretch vanish; hereeu,ev∈ℝVe\_\{u\},e\_\{v\}\\in\\mathbb\{R\}^\{V\}are standard basis vectors, and forℓ=1\\ell=1the single column iseu\+eve\_\{u\}\+e\_\{v\}\. WritingC=Tℓ−1C=T\_\{\\ell\}^\{\-1\}, the stretch ofu​vuvtherefore contributes

C1,1​eu​eu𝖳\+Cℓ,ℓ​ev​ev𝖳\+C1,ℓ​\(eu​ev𝖳\+ev​eu𝖳\)C\_\{1,1\}e\_\{u\}e\_\{u\}^\{\\mathsf\{T\}\}\+C\_\{\\ell,\\ell\}e\_\{v\}e\_\{v\}^\{\\mathsf\{T\}\}\+C\_\{1,\\ell\}\\bigl\(e\_\{u\}e\_\{v\}^\{\\mathsf\{T\}\}\+e\_\{v\}e\_\{u\}^\{\\mathsf\{T\}\}\\bigr\)toBt​\[V,Wt\]​Bt​\[Wt,Wt\]−1​Bt​\[Wt,V\]B\_\{t\}\[V,W\_\{t\}\]B\_\{t\}\[W\_\{t\},W\_\{t\}\]^\{\-1\}B\_\{t\}\[W\_\{t\},V\], and this formula is correct forℓ=1\\ell=1as well, since thenC1,1=Cℓ,ℓ=C1,ℓ=12C\_\{1,1\}=C\_\{\\ell,\\ell\}=C\_\{1,\\ell\}=\\tfrac\{1\}\{2\}\. By Lemma[C\.9\.5](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem5)withℓ=t−1\\ell=t\-1we haveC1,1=Cℓ,ℓ=\(t−1\)/tC\_\{1,1\}=C\_\{\\ell,\\ell\}=\(t\-1\)/tandC1,ℓ=\(−1\)t/tC\_\{1,\\ell\}=\(\-1\)^\{t\}/t\. Summing overSSand using∑u​v∈S\(eu​eu𝖳\+ev​ev𝖳\)=DS\\sum\_\{uv\\in S\}\(e\_\{u\}e\_\{u\}^\{\\mathsf\{T\}\}\+e\_\{v\}e\_\{v\}^\{\\mathsf\{T\}\}\)=D\_\{S\}and∑u​v∈S\(eu​ev𝖳\+ev​eu𝖳\)=AS\\sum\_\{uv\\in S\}\(e\_\{u\}e\_\{v\}^\{\\mathsf\{T\}\}\+e\_\{v\}e\_\{u\}^\{\\mathsf\{T\}\}\)=A\_\{S\}, the Schur complement equalsKtK\_\{t\}as displayed\. Finally

Kt=2​In\+AR−DS\+1t​DS−\(−1\)tt​AS=K∞\+1t​\(DS−\(−1\)t​AS\),K\_\{t\}=2I\_\{n\}\+A\_\{R\}\-D\_\{S\}\+\\frac\{1\}\{t\}D\_\{S\}\-\\frac\{\(\-1\)^\{t\}\}\{t\}A\_\{S\}=K\_\{\\infty\}\+\\frac\{1\}\{t\}\\bigl\(D\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}\\bigr\),which is equation[28](https://arxiv.org/html/2608.16977#A3.E28)\. ∎

#### C\.9\.3Comparison in the Loewner order

###### Lemma C\.9\.8\.

For everyt≥2t\\geq 2we haveKt⪰K∞K\_\{t\}\\succeq K\_\{\\infty\}in the Loewner order and‖Kt−K∞‖≤2​Δ​\(G\)/t\\\|K\_\{t\}\-K\_\{\\infty\}\\\|\\leq 2\\Delta\(G\)/t, where∥⋅∥\\\|\\cdot\\\|is the spectral norm\.

###### Proof\.

Fix an arbitrary orientation of each edge ofSS\. Forx∈ℝV⁡\(G\)x\\in\\mathbb\{R\}^\{V\(G\)\},

x𝖳​\(DS−\(−1\)t​AS\)​x=∑u​v∈S\(xu−\(−1\)t​xv\)2≥0,x^\{\\mathsf\{T\}\}\\bigl\(D\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}\\bigr\)x=\\sum\_\{uv\\in S\}\\bigl\(x\_\{u\}\-\(\-1\)^\{t\}x\_\{v\}\\bigr\)^\{2\}\\geq 0,since expanding the squares returnsx𝖳​DS​x−\(−1\)t​x𝖳​AS​xx^\{\\mathsf\{T\}\}D\_\{S\}x\-\(\-1\)^\{t\}x^\{\\mathsf\{T\}\}A\_\{S\}xand the summands do not depend on the chosen orientation\. ThusDS−\(−1\)t​ASD\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}is positive semidefinite, being the Laplacian of\(V⁡\(G\),S\)\(V\(G\),S\)for eventtand its signless Laplacian for oddtt, and equation[28](https://arxiv.org/html/2608.16977#A3.E28)givesKt−K∞=1t​\(DS−\(−1\)t​AS\)⪰0K\_\{t\}\-K\_\{\\infty\}=\\frac\{1\}\{t\}\\bigl\(D\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}\\bigr\)\\succeq 0\. For the norm bound, the spectral norm of a real symmetric matrix is at most its largest absolute row sum, and the row ofDS−\(−1\)t​ASD\_\{S\}\-\(\-1\)^\{t\}A\_\{S\}indexed byvvhas diagonal entrydegS⁡\(v\)\\deg\_\{S\}\(v\)and a furtherdegS⁡\(v\)\\deg\_\{S\}\(v\)entries of absolute value11, one for eachSS\-neighbor ofvv, so that row sum is2​degS⁡\(v\)≤2​Δ​\(G\)2\\deg\_\{S\}\(v\)\\leq 2\\Delta\(G\), and the factor1/t1/tyields the stated bound\. ∎

#### C\.9\.4Proof of the main theorem

###### Proof of Theorem[C\.9\.1](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem1)\.

Putr=n−​\(K∞\)r=n\_\{\-\}\(K\_\{\\infty\}\)and fixt≥2t\\geq 2\. By Proposition[C\.9\.6](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem6)it suffices to prove thatn−​\(Kt\)≤rn\_\{\-\}\(K\_\{t\}\)\\leq ralways, and thatn−​\(Kt\)≥rn\_\{\-\}\(K\_\{t\}\)\\geq roncettis large\.

For the upper bound we may assumer<nr<n, since otherwise there is nothing to prove\. By definition ofrrwe haveλn−r​\(K∞\)≥0\\lambda\_\{n\-r\}\(K\_\{\\infty\}\)\\geq 0, andKt⪰K∞K\_\{t\}\\succeq K\_\{\\infty\}givesλj​\(Kt\)≥λj​\(K∞\)\\lambda\_\{j\}\(K\_\{t\}\)\\geq\\lambda\_\{j\}\(K\_\{\\infty\}\)for everyjjby Weyl monotonicity\([Horn & Johnson 2012](https://arxiv.org/html/2608.16977#bibi.bib7)\)\. Henceλn−r​\(Kt\)≥0\\lambda\_\{n\-r\}\(K\_\{t\}\)\\geq 0, so at mostrreigenvalues ofKtK\_\{t\}are negative, that is,n−​\(Kt\)≤rn\_\{\-\}\(K\_\{t\}\)\\leq r\. This holds for everyt≥2t\\geq 2and is the asserted inequality\.

For the lower bound we may assumer≥1r\\geq 1, since forr=0r=0the upper bound already givesn−​\(Kt\)=0n\_\{\-\}\(K\_\{t\}\)=0for everyt≥2t\\geq 2\. Letγ:=−λn−r\+1​\(K∞\)\>0\\gamma:=\-\\lambda\_\{n\-r\+1\}\(K\_\{\\infty\}\)\>0be the gap of Remark[C\.9\.3](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem3), so thatλj​\(K∞\)≤−γ\\lambda\_\{j\}\(K\_\{\\infty\}\)\\leq\-\\gammafor everyj≥n−r\+1j\\geq n\-r\+1\. Weyl’s perturbation inequality together with Lemma[C\.9\.8](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem8)gives

λj​\(Kt\)≤λj​\(K∞\)\+‖Kt−K∞‖≤−γ\+2​Δ​\(G\)t\\lambda\_\{j\}\(K\_\{t\}\)\\leq\\lambda\_\{j\}\(K\_\{\\infty\}\)\+\\\|K\_\{t\}\-K\_\{\\infty\}\\\|\\leq\-\\gamma\+\\frac\{2\\Delta\(G\)\}\{t\}for thosejj, which is negative as soon ast\>2​Δ​\(G\)/γt\>2\\Delta\(G\)/\\gamma\. For suchttat leastrreigenvalues ofKtK\_\{t\}are negative, son−​\(Kt\)≥rn\_\{\-\}\(K\_\{t\}\)\\geq r\.

Combining the two bounds,n−​\(Kt\)=rn\_\{\-\}\(K\_\{t\}\)=rfor everyt\>max⁡\{2,2​Δ​\(G\)/γ\}t\>\\max\\\{2,2\\Delta\(G\)/\\gamma\\\}, and Proposition[C\.9\.6](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem6)turns this intomGt​\(−∞,−2\)=r=n−​\(K∞\)m\_\{G\_\{t\}\}\(\-\\infty,\-2\)=r=n\_\{\-\}\(K\_\{\\infty\}\), as desired\. ∎

###### Proof of Remark[C\.9\.3](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem3)\.

The first assertion is the thresholdt\>2​Δ​\(G\)/γt\>2\\Delta\(G\)/\\gammadisplayed in the proof above\. For the second, the spectral norm ofK∞K\_\{\\infty\}is at most its largest absolute row sum, which ismaxv⁡\(\|2−degS⁡\(v\)\|\+degR⁡\(v\)\)≤2\+Δ⁡\(G\)\\max\_\{v\}\\bigl\(\|2\-\\deg\_\{S\}\(v\)\|\+\\deg\_\{R\}\(v\)\\bigr\)\\leq 2\+\\Delta\(G\), so every eigenvalue ofK∞K\_\{\\infty\}has absolute value at most2\+Δ⁡\(G\)2\+\\Delta\(G\)\. On the other handK∞K\_\{\\infty\}has integer entries, so the product of its nonzero eigenvalues is, up to sign, the last nonvanishing coefficient of its characteristic polynomial and hence a nonzero integer\. The absolute values of those at mostnneigenvalues therefore multiply to at least11, and each factor is at most2\+Δ⁡\(G\)2\+\\Delta\(G\), so the smallest of them, and in particularγ\\gamma, is at least\(2\+Δ⁡\(G\)\)−\(n−1\)\(2\+\\Delta\(G\)\)^\{\-\(n\-1\)\}\. ∎

###### Proof of Corollary[C\.9\.2](https://arxiv.org/html/2608.16977#A3.SS9.Thmtheorem2)\.

WhenS=E⁡\(G\)S=E\(G\)we haveAR=0A\_\{R\}=0andDSD\_\{S\}equal to the degree matrixDDofGG, so equation[27](https://arxiv.org/html/2608.16977#A3.E27)makesK∞=2​In−DK\_\{\\infty\}=2I\_\{n\}\-Ddiagonal with entries2−degG⁡\(v\)2\-\\deg\_\{G\}\(v\)\. Such an entry is negative exactly whendegG⁡\(v\)≥3\\deg\_\{G\}\(v\)\\geq 3, son−​\(K∞\)=\|Q\|n\_\{\-\}\(K\_\{\\infty\}\)=\|Q\|\. ∎

## References\.

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### C\.10Small unions of lines closing a route to the Nikodym bound

[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)conjectured that a family ofΩ⁡\(q3\)\\Omega\(q^\{3\}\)lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}, no plane of which contains a superlinear number of them, must cover all but a vanishing proportion of𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}\. We show that the union can have density1/2\+o⁡\(1\)1/2\+o\(1\)\. For every odd prime powerqqthere is a familyLLofq2​\(q\+1\)/2q^\{2\}\(q\+1\)/2lines, at mostq\+1q\+1of which lie in any single plane, whose union has exactlyq2​\(q\+1\)/2q^\{2\}\(q\+1\)/2points\. The family consists of one half of the tangent lines to the paraboloidz=x2−ν​y2z=x^\{2\}\-\\nu y^\{2\}, whereν\\nuis a fixed nonsquare, the half being selected by the quadratic character of the direction\. The same family also refutes the sharper quantitative conjecture that they state alongside the first one\.

#### C\.10\.1Introduction

For a setLLof affine lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}, write

P⁡\(L\)=⋃ℓ∈LℓP\(L\)=\\bigcup\_\{\\ell\\in L\}\\ellfor the set of points lying on some line ofLL\. The problem is to bound\|P⁡\(L\)\|\|P\(L\)\|from below whenLLis large and no plane of𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}carries too much of it\. In Section 1\.2\.1 of their paper on Kakeya and Nikodym sets in three dimensions,[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)propose the following, their Conjecture 1\.4\.

*LetC\>0C\>0be a constant independent ofqq, and letα⁡\(q\)∈ω⁡\(q\)\\alpha\(q\)\\in\\omega\(q\)\. IfLLis a set of at leastC​q3Cq^\{3\}lines and no plane containsα⁡\(q\)\\alpha\(q\)lines ofLL, then\|P⁡\(L\)\|≥\(1−o⁡\(1\)\)​q3\|P\(L\)\|\\geq\(1\-o\(1\)\)q^\{3\}\. Theo⁡\(1\)o\(1\)is a function ofqqthat depends onCCandα\\alpha\.*

In the same place they propose the following sharper form, their Conjecture 1\.5\.

*Letε\>0\\varepsilon\>0be any constant and letqqbe a sufficiently large prime power\. LetLLbe a set of at leastq5/2\+εq^\{5/2\+\\varepsilon\}lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}such that no plane contains more than\(1/2\)​q3/2\(1/2\)q^\{3/2\}lines ofLL\. Then,\|P⁡\(L\)\|≥q3−O⁡\(q5/2\)\|P\(L\)\|\\geq q^\{3\}\-O\(q^\{5/2\}\)\.*

These two conjectures are the geometric heart of[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)\. Forqqsufficiently large, their Theorem 1\.1 gives\|K\|≥0\.2107​q3\|K\|\\geq 0\.2107q^\{3\}for every Kakeya setK⊆𝔽q3K\\subseteq\\mathbb\{F\}\_\{q\}^\{3\}and their Theorem 1\.3 gives\|𝒩\|≥0\.38​q3\|\\mathcal\{N\}\|\\geq 0\.38q^\{3\}for every Nikodym set𝒩⊆𝔽q3\\mathcal\{N\}\\subseteq\\mathbb\{F\}\_\{q\}^\{3\}, while their Theorem 3\.8 shows that Conjecture 1\.4 would upgrade the second of these to the conjectured optimal bound\(1−o⁡\(1\)\)​q3\(1\-o\(1\)\)q^\{3\}\. Conjecture 1\.5 is carefully calibrated against two constructions of[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5): from a nondegenerate Hermitian variety, for squareq=p2q=p^\{2\}and each0<θ<10<\\theta<1, they produce a setLLof\(θ\+o⁡\(1\)\)​q7/2\(\\theta\+o\(1\)\)q^\{7/2\}lines with no plane containing more than\(θ\+o⁡\(1\)\)​q3/2\(\\theta\+o\(1\)\)q^\{3/2\}of them and with\|P⁡\(L\)\|≤q3−\(1−θ\+o⁡\(1\)\)​q5/2\|P\(L\)\|\\leq q^\{3\}\-\(1\-\\theta\+o\(1\)\)q^\{5/2\}, so the error termO⁡\(q5/2\)O\(q^\{5/2\}\)in the conclusion of Conjecture 1\.5 cannot be improved; a second construction of theirs, the union ofO⁡\(q1/2\)O\(q^\{1/2\}\)Hermitian varieties, shows that the hypothesis\|L\|≥q5/2\+ε\|L\|\\geq q^\{5/2\+\\varepsilon\}cannot be substantially relaxed\. They also prove that any0\.62​q30\.62q^\{3\}lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}already satisfy\|P⁡\(L\)\|≥\(0\.38−o⁡\(1\)\)​q3\|P\(L\)\|\\geq\(0\.38\-o\(1\)\)q^\{3\}with no hypothesis on planes at all, and that without such a hypothesis this is close to sharp: for largeqq, they take all lines lying in a union of0\.62​q0\.62qplanes through a common point and obtain\(1−o⁡\(1\)\)​0\.62​q3\(1\-o\(1\)\)0\.62q^\{3\}lines whose union has fewer than0\.43​q30\.43q^\{3\}points\. It is this flat example that the plane hypothesis of Conjecture 1\.4 is designed to exclude, and our family shows that excluding it does not help: the half\-tangent family has at mostq\+1q\+1lines in any plane, against orderq2q^\{2\}for that example\.

The same question for the much smaller count\|L\|=q2\|L\|=q^\{2\}has a longer history, coming from work on Kakeya sets\.[Wolff 1999](https://arxiv.org/html/2608.16977#bibj.bib8)showed that a setLLofq2q^\{2\}lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}, at mostO⁡\(q\)O\(q\)of which lie in any plane, has\|P⁡\(L\)\|=Ω⁡\(q5/2\)\|P\(L\)\|=\\Omega\(q^\{5/2\}\), and[Mockenhaupt & Tao 2004](https://arxiv.org/html/2608.16977#bibj.bib6)showed that this is sharp whenqqis a square\. Over prime fields the exponent improves:[Ellenberg & Hablicsek 2016](https://arxiv.org/html/2608.16977#bibj.bib3)prove that such anLLhas\|P⁡\(L\)\|≥c​q3\|P\(L\)\|\\geq cq^\{3\}for an absolute constantc\>0c\>0, once no plane contains more thanqqof its lines\. Conjecture 1\.4 assumes many more lines,C​q3Cq^\{3\}rather thanq2q^\{2\}, allowsω⁡\(q\)\\omega\(q\)rather thanO⁡\(q\)O\(q\)of them in a plane, and asks for the constant1−o⁡\(1\)1\-o\(1\)in place ofcc\. We are aware of no counterexample in the literature respecting its plane hypothesis, and[Tao 2025](https://arxiv.org/html/2608.16977#bibj.bib7), who constructs a Nikodym set in𝔽qd\\mathbb\{F\}\_\{q\}^\{d\}of sizeqd−\(\(d−2\)/log⁡2\+1\+o⁡\(1\)\)​qd−1​log⁡qq^\{d\}\-\\bigl\(\(d\-2\)/\\log 2\+1\+o\(1\)\\bigr\)q^\{d\-1\}\\log qfor each fixedd≥3d\\geq 3and each odd prime powerqq, still records Conjecture 1\.4 as the route to the three\-dimensional case of the Nikodym conjecture\.

The construction below is not new as a piece of finite geometry\. The projective closure of the paraboloid is the quadricX2−ν​Y2−Z​W=0X^\{2\}\-\\nu Y^\{2\}\-ZW=0ofPG⁡\(3,q\)\\mathrm\{PG\}\(3,q\), an elliptic quadric withq2\+1q^\{2\}\+1points, and the splitting of theq\+1q\+1tangent lines at each of its points into two halves of size\(q\+1\)/2\(q\+1\)/2, according to whether the quadratic form takes square or nonsquare values on the tangent line away from the quadric, is precisely the splitting used by[Bruen & Drudge 1999](https://arxiv.org/html/2608.16977#bibj.bib1)to build the first infinite family of Cameron–Liebler line classes ofPG⁡\(3,q\)\\mathrm\{PG\}\(3,q\),qqodd, with parameter\(q2\+1\)/2\(q^\{2\}\+1\)/2\.[Gavrilyuk et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib4)write the splitting out explicitly as a partition of the tangent lines into two classes, either of which yields a Cameron–Liebler line class of that parameter when adjoined to the secant lines or to the external lines, and they derive a further such family from it\.[Cossidente & Pavese 2017](https://arxiv.org/html/2608.16977#bibj.bib2)construct yet more families with the same parameter for oddq≥7q\\geq 7\. What is new here is only the observation that this classical half\-tangent family, read in affine coordinates, is a counterexample to the line\-union conjecture\.

Our counterexample is at the opposite extreme from the Hermitian family of[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5): its union misses a positive proportion of𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}rather than aq−1/2q^\{\-1/2\}proportion, and it exists for every odd prime power, including the prime fields for which no Hermitian variety is available\.

###### Theorem C\.10\.1\.

Letqqbe an odd prime power\. There is a setLLof affine lines in𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}such that

\|L\|=q2​\(q\+1\)2,maxΠ⁡\|\{ℓ∈L:ℓ⊆Π\}\|≤q\+1,\|P⁡\(L\)\|=q2​\(q\+1\)2,\|L\|=\\frac\{q^\{2\}\(q\+1\)\}\{2\},\\qquad\\max\_\{\\Pi\}\\bigl\|\\\{\\ell\\in L:\\ell\\subseteq\\Pi\\\}\\bigr\|\\leq q\+1,\\qquad\|P\(L\)\|=\\frac\{q^\{2\}\(q\+1\)\}\{2\},where the maximum is taken over all affine planesΠ⊆𝔽q3\\Pi\\subseteq\\mathbb\{F\}\_\{q\}^\{3\}\.

###### Corollary C\.10\.2\.

Conjecture 1\.4 of[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)is false, and so is Conjecture 1\.5 of[Lund et al\. 2018](https://arxiv.org/html/2608.16977#bibj.bib5)for every fixed0<ε<1/20<\\varepsilon<1/2\.

We now briefly summarize the construction\. Fix a nonsquareν∈𝔽q∗\\nu\\in\\mathbb\{F\}\_\{q\}^\{\*\}and letQ⁡\(x,y\)=x2−ν​y2Q\(x,y\)=x^\{2\}\-\\nu y^\{2\}, an anisotropic binary form, and slice𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}into theqqlevel sets ofQ⁡\(x,y\)−zQ\(x,y\)\-z\. The key point is that the tangent line to the paraboloidz=Q⁡\(x,y\)z=Q\(x,y\)in a direction\[u:v\]\[u:v\]meets only the level sets\{Q\(x,y\)−z=s\}\\\{Q\(x,y\)\-z=s\\\}withs=0s=0orssin the square class ofQ⁡\(u,v\)Q\(u,v\), so keeping only the\(q\+1\)/2\(q\+1\)/2directions withQ⁡\(u,v\)Q\(u,v\)a square confines the whole family to the level sets indexed by00and by the nonzero squares\. The plane bound comes from the fact that a plane meets the paraboloid in at mostq\+1q\+1points and, unless it is a tangent plane, determines the direction of a tangent line at each of them\.

#### C\.10\.2The half\-tangent family

Fix an odd prime powerqqand a nonsquareν∈𝔽q∗\\nu\\in\\mathbb\{F\}\_\{q\}^\{\*\}, and set

Q⁡\(x,y\)=x2−ν​y2\.Q\(x,y\)=x^\{2\}\-\\nu y^\{2\}\.The binary formQQis*anisotropic*: ifQ⁡\(x,y\)=0Q\(x,y\)=0with\(x,y\)≠\(0,0\)\(x,y\)\\neq\(0,0\), theny≠0y\\neq 0andν=\(x/y\)2\\nu=\(x/y\)^\{2\}would be a square\. Let

S=\{\(a,b,Q\(a,b\)\):a,b∈𝔽q\}⊆𝔽q3S=\\\{\(a,b,Q\(a,b\)\):a,b\\in\\mathbb\{F\}\_\{q\}\\\}\\subseteq\\mathbb\{F\}\_\{q\}^\{3\}be the associated affine paraboloid, a set ofq2q^\{2\}points\.

Call a projective direction\[u:v\]∈PG\(1,q\)\[u:v\]\\in\\mathrm\{PG\}\(1,q\)*square*ifQ⁡\(u,v\)Q\(u,v\)is a nonzero square in𝔽q\\mathbb\{F\}\_\{q\}, and letD⊆PG⁡\(1,q\)D\\subseteq\\mathrm\{PG\}\(1,q\)be the set of square directions\. This is well defined:Q⁡\(u,v\)≠0Q\(u,v\)\\neq 0for\(u,v\)≠\(0,0\)\(u,v\)\\neq\(0,0\)by anisotropy, and replacing\(u,v\)\(u,v\)byλ⁡\(u,v\)\\lambda\(u,v\)multipliesQ⁡\(u,v\)Q\(u,v\)by the squareλ2\\lambda^\{2\}\.

###### Lemma C\.10\.5\.

We have\|D\|=\(q\+1\)/2\|D\|=\(q\+1\)/2\.

###### Proof\.

Sinceν\\nuis a nonsquare we have𝔽q​\(ν\)=𝔽q2\\mathbb\{F\}\_\{q\}\(\\sqrt\{\\nu\}\)=\\mathbb\{F\}\_\{q^\{2\}\}, andQQis the norm form of this extension:

N⁡\(u\+v​ν\)=\(u\+v​ν\)​\(u−v​ν\)=u2−ν​v2=Q⁡\(u,v\)\.N\(u\+v\\sqrt\{\\nu\}\)=\(u\+v\\sqrt\{\\nu\}\)\(u\-v\\sqrt\{\\nu\}\)=u^\{2\}\-\\nu v^\{2\}=Q\(u,v\)\.The norm mapN:𝔽q2∗→𝔽q∗N:\\mathbb\{F\}\_\{q^\{2\}\}^\{\*\}\\to\\mathbb\{F\}\_\{q\}^\{\*\}is surjective with kernel of sizeq\+1q\+1, so everyc∈𝔽q∗c\\in\\mathbb\{F\}\_\{q\}^\{\*\}has exactlyq\+1q\+1preimages\. Hence exactlyq−12​\(q\+1\)\\tfrac\{q\-1\}\{2\}\(q\+1\)pairs\(u,v\)≠\(0,0\)\(u,v\)\\neq\(0,0\)haveQ⁡\(u,v\)Q\(u,v\)a nonzero square\. Each projective direction accounts for exactlyq−1q\-1such pairs, so\|D\|=\(q\+1\)/2\|D\|=\(q\+1\)/2\. ∎

For\(a,b\)∈𝔽q2\(a,b\)\\in\\mathbb\{F\}\_\{q\}^\{2\}and\[u:v\]∈PG\(1,q\)\[u:v\]\\in\\mathrm\{PG\}\(1,q\)set

ℓa,b,\[u:v\]=\{\(a\+tu,b\+tv,Q\(a,b\)\+2t\(au−νbv\)\):t∈𝔽q\}\.\\ell\_\{a,b,\[u:v\]\}=\\bigl\\\{\\bigl\(a\+tu,\\ b\+tv,\\ Q\(a,b\)\+2t\(au\-\\nu bv\)\\bigr\):t\\in\\mathbb\{F\}\_\{q\}\\bigr\\\}\.Replacing\(u,v\)\(u,v\)byλ⁡\(u,v\)\\lambda\(u,v\)only reparametrizes this set, so it depends on\[u:v\]\[u:v\]alone, and it is a line because\(u,v\)≠\(0,0\)\(u,v\)\\neq\(0,0\)\. ExpandingQQgives the identity

Q⁡\(a\+t​u,b\+t​v\)−\(Q⁡\(a,b\)\+2​t​\(a​u−ν​b​v\)\)=t2​Q​\(u,v\),Q\(a\+tu,\\,b\+tv\)\-\\bigl\(Q\(a,b\)\+2t\(au\-\\nu bv\)\\bigr\)=t^\{2\}Q\(u,v\),\(29\)so alongℓa,b,\[u:v\]\\ell\_\{a,b,\[u:v\]\}the functionQ⁡\(x,y\)−zQ\(x,y\)\-zequalst2​Q​\(u,v\)t^\{2\}Q\(u,v\)and vanishes to order two att=0t=0\. In other words,ℓa,b,\[u:v\]\\ell\_\{a,b,\[u:v\]\}is the tangent line toSSat\(a,b,Q⁡\(a,b\)\)\(a,b,Q\(a,b\)\)in the direction\[u:v\]\[u:v\]\. Define the*half\-tangent family*

L=\{ℓa,b,\[u:v\]:\(a,b\)∈𝔽q2,\[u:v\]∈D\}\.L=\\bigl\\\{\\ell\_\{a,b,\[u:v\]\}:\(a,b\)\\in\\mathbb\{F\}\_\{q\}^\{2\},\\ \[u:v\]\\in D\\bigr\\\}\.
###### Lemma C\.10\.6\.

Everyℓ∈L\\ell\\in LmeetsSSexactly in its point of tangency, and\|L\|=q2​\(q\+1\)/2\|L\|=q^\{2\}\(q\+1\)/2\.

###### Proof\.

By equation[29](https://arxiv.org/html/2608.16977#A3.E29), the point ofℓa,b,\[u:v\]\\ell\_\{a,b,\[u:v\]\}with parameterttlies onSSif and only ift2​Q​\(u,v\)=0t^\{2\}Q\(u,v\)=0\. SinceQ⁡\(u,v\)≠0Q\(u,v\)\\neq 0, this forcest=0t=0\. Soℓ∩S\\ell\\cap Sis the single point\(a,b,Q⁡\(a,b\)\)\(a,b,Q\(a,b\)\), which is therefore determined byℓ\\ell, as is the direction\[u:v\]\[u:v\]\. Hence\(a,b,\[u:v\]\)↦ℓa,b,\[u:v\]\(a,b,\[u:v\]\)\\mapsto\\ell\_\{a,b,\[u:v\]\}is injective on𝔽q2×D\\mathbb\{F\}\_\{q\}^\{2\}\\times D, and Lemma[C\.10\.5](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem5)gives\|L\|=q2⋅q\+12\|L\|=q^\{2\}\\cdot\\tfrac\{q\+1\}\{2\}\. ∎

Figure[11](https://arxiv.org/html/2608.16977#A3.F11)shows the selected directions at a point ofSS, and the resulting union, in the caseq=7q=7\.

pp\[1:0\]\[1\{:\}0\]\[1:1\]\[1\{:\}1\]\[1:2\]\[1\{:\}2\]\[1:3\]\[1\{:\}3\]\[1:4\]\[1\{:\}4\]\[1:5\]\[1\{:\}5\]\[1:6\]\[1\{:\}6\]\[0:1\]\[0\{:\}1\]s=0s=0s=1s=1s=2s=2s=3s=3s=4s=4s=5s=5s=6s=6Figure 11:The two counts behind Theorem[C\.10\.1](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem1), drawn forq=7q=7andν=3\\nu=3\. On the left are the eight tangent lines toSSat a pointpp, all lying in the tangent plane atpp; the four solid ones are those whose direction\[u:v\]\[u:v\]hasQ⁡\(u,v\)=u2−3​v2Q\(u,v\)=u^\{2\}\-3v^\{2\}a nonzero square, and these are exactly the lines ofLLthroughpp\. On the right are the seven points of𝔽q3\\mathbb\{F\}\_\{q\}^\{3\}lying over a fixed\(x,y\)\(x,y\), labeled by the values=Q⁡\(x,y\)−zs=Q\(x,y\)\-z; the four solid ones are those belonging toP⁡\(L\)P\(L\), namely those withs∈\{0,1,2,4\}s\\in\\\{0,1,2,4\\\}\.
#### C\.10\.3Lines contained in a plane

###### Lemma C\.10\.7\.

Every affine planeΠ⊆𝔽q3\\Pi\\subseteq\\mathbb\{F\}\_\{q\}^\{3\}contains at mostq\+1q\+1lines ofLL\.

###### Proof\.

Write

Π=\{\(x,y,z\)∈𝔽q3:c1​x\+c2​y\+c3​z=c0\},\(c1,c2,c3\)≠\(0,0,0\)\.\\Pi=\\\{\(x,y,z\)\\in\\mathbb\{F\}\_\{q\}^\{3\}:c\_\{1\}x\+c\_\{2\}y\+c\_\{3\}z=c\_\{0\}\\\},\\qquad\(c\_\{1\},c\_\{2\},c\_\{3\}\)\\neq\(0,0,0\)\.Ifℓa,b,\[u:v\]⊆Π\\ell\_\{a,b,\[u:v\]\}\\subseteq\\Pi, then its point of tangency lies inΠ∩S\\Pi\\cap Sand its direction vector\(u,v,2​\(a​u−ν​b​v\)\)\(u,v,2\(au\-\\nu bv\)\)lies in the direction plane ofΠ\\Pi, the latter condition reading

\(c1\+2​c3​a\)​u\+\(c2−2​ν​c3​b\)​v=0\.\(c\_\{1\}\+2c\_\{3\}a\)u\+\(c\_\{2\}\-2\\nu c\_\{3\}b\)v=0\.\(30\)
The size ofΠ∩S\\Pi\\cap S\.Ifc3=0c\_\{3\}=0, thenΠ∩S\\Pi\\cap Sis parametrized by theqqsolutions\(x,y\)\(x,y\)ofc1​x\+c2​y=c0c\_\{1\}x\+c\_\{2\}y=c\_\{0\}, so\|Π∩S\|=q\|\\Pi\\cap S\|=q\. Ifc3≠0c\_\{3\}\\neq 0, then substitutingz=Q⁡\(x,y\)z=Q\(x,y\)into the equation ofΠ\\Piand completing the square gives

Q⁡\(x\+c12​c3,y−c22​ν​c3\)=c0c3\+c124​c32−c224​ν​c32\.Q\\Bigl\(x\+\\frac\{c\_\{1\}\}\{2c\_\{3\}\},\\ y\-\\frac\{c\_\{2\}\}\{2\\nu c\_\{3\}\}\\Bigr\)=\\frac\{c\_\{0\}\}\{c\_\{3\}\}\+\\frac\{c\_\{1\}^\{2\}\}\{4c\_\{3\}^\{2\}\}\-\\frac\{c\_\{2\}^\{2\}\}\{4\\nu c\_\{3\}^\{2\}\}\.\(31\)If the right\-hand side of equation[31](https://arxiv.org/html/2608.16977#A3.E31)is zero, then anisotropy ofQQgives exactly one solution, and otherwise the norm count in the proof of Lemma[C\.10\.5](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem5)gives exactlyq\+1q\+1solutions\. Hence\|Π∩S\|∈\{1,q,q\+1\}\|\\Pi\\cap S\|\\in\\\{1,q,q\+1\\\}, and in particular\|Π∩S\|≤q\+1\|\\Pi\\cap S\|\\leq q\+1\.

Tangent planes\.SinceQ⁡\(x,y\)−zQ\(x,y\)\-zhas gradient\(2​a,−2​ν​b,−1\)\(2a,\-2\\nu b,\-1\)at\(a,b,Q⁡\(a,b\)\)\(a,b,Q\(a,b\)\), the tangent plane toSSat that point is

Ta,b:2​a​x−2​ν​b​y−z=Q⁡\(a,b\)\.T\_\{a,b\}:\\qquad 2ax\-2\\nu by\-z=Q\(a,b\)\.We claim that, for\(a,b,Q⁡\(a,b\)\)∈Π∩S\(a,b,Q\(a,b\)\)\\in\\Pi\\cap S, the two coefficients in equation[30](https://arxiv.org/html/2608.16977#A3.E30)both vanish if and only ifΠ=Ta,b\\Pi=T\_\{a,b\}\. Indeed, they vanish exactly whenc1=−2​c3​ac\_\{1\}=\-2c\_\{3\}aandc2=2​ν​c3​bc\_\{2\}=2\\nu c\_\{3\}b, which forcesc3≠0c\_\{3\}\\neq 0; dividing the equation ofΠ\\Piby−c3\-c\_\{3\}then turns it into2ax−2νby−z=−c0/c32ax\-2\\nu by\-z=\-c\_\{0\}/c\_\{3\}, and evaluating at\(a,b,Q⁡\(a,b\)\)∈Π\(a,b,Q\(a,b\)\)\\in\\Pigives−c0/c3=Q\(a,b\)\-c\_\{0\}/c\_\{3\}=Q\(a,b\), soΠ=Ta,b\\Pi=T\_\{a,b\}\. Conversely, ifΠ=Ta,b\\Pi=T\_\{a,b\}then\(c1,c2,c3\)=λ⁡\(2​a,−2​ν​b,−1\)\(c\_\{1\},c\_\{2\},c\_\{3\}\)=\\lambda\(2a,\-2\\nu b,\-1\)for someλ∈𝔽q∗\\lambda\\in\\mathbb\{F\}\_\{q\}^\{\*\}, whencec1=−2​c3​ac\_\{1\}=\-2c\_\{3\}aandc2=2​ν​c3​bc\_\{2\}=2\\nu c\_\{3\}b\. Comparing with equation[31](https://arxiv.org/html/2608.16977#A3.E31), whose right\-hand side vanishes exactly when the unique point ofΠ∩S\\Pi\\cap Shas\(a,b\)=\(−c1/\(2c3\),c2/\(2νc3\)\)\(a,b\)=\(\-c\_\{1\}/\(2c\_\{3\}\),\\,c\_\{2\}/\(2\\nu c\_\{3\}\)\), we conclude thatΠ\\Piis a tangent plane ofSSif and only if\|Π∩S\|=1\|\\Pi\\cap S\|=1\.

The two cases\.Suppose first thatΠ\\Piis not a tangent plane ofSS\. Then for every\(a,b,Q⁡\(a,b\)\)∈Π∩S\(a,b,Q\(a,b\)\)\\in\\Pi\\cap Sthe two coefficients in equation[30](https://arxiv.org/html/2608.16977#A3.E30)are not both zero, so at most one\[u:v\]∈PG\(1,q\)\[u:v\]\\in\\mathrm\{PG\}\(1,q\)satisfies equation[30](https://arxiv.org/html/2608.16977#A3.E30), and therefore at most one line ofLLtangent at that point is contained inΠ\\Pi\. Since every line ofLLcontained inΠ\\Piis tangent at some point ofΠ∩S\\Pi\\cap S, the planeΠ\\Picontains at most\|Π∩S\|≤q\+1\|\\Pi\\cap S\|\\leq q\+1lines ofLL\. Suppose instead thatΠ=Tp\\Pi=T\_\{p\}is the tangent plane at a pointp∈Sp\\in S\. ThenΠ∩S=\{p\}\\Pi\\cap S=\\\{p\\\}, so every line ofLLinsideΠ\\Piis tangent atpp, and there are exactly\|D\|=\(q\+1\)/2\|D\|=\(q\+1\)/2of these\. This exhausts all possibilities, and in every caseΠ\\Picontains at mostq\+1q\+1lines ofLL\. ∎

#### C\.10\.4The union

###### Lemma C\.10\.8\.

We have

P⁡\(L\)=\{\(x,y,z\)∈𝔽q3:Q⁡\(x,y\)−z∈\{0\}∪\(𝔽q∗\)2\},P\(L\)=\\bigl\\\{\(x,y,z\)\\in\\mathbb\{F\}\_\{q\}^\{3\}:\\ Q\(x,y\)\-z\\in\\\{0\\\}\\cup\(\\mathbb\{F\}\_\{q\}^\{\*\}\)^\{2\}\\bigr\\\},and consequently\|P⁡\(L\)\|=q2​\(q\+1\)/2\|P\(L\)\|=q^\{2\}\(q\+1\)/2\.

###### Proof\.

Writes=Q⁡\(x,y\)−zs=Q\(x,y\)\-zfor the value of the defining function at a point\(x,y,z\)\(x,y,z\)\. By equation[29](https://arxiv.org/html/2608.16977#A3.E29), at the point ofℓa,b,\[u:v\]\\ell\_\{a,b,\[u:v\]\}with parameterttwe haves=t2​Q​\(u,v\)s=t^\{2\}Q\(u,v\)\. For\[u:v\]∈D\[u:v\]\\in Dthe valueQ⁡\(u,v\)Q\(u,v\)is a nonzero square, sos∈\{0\}∪\(𝔽q∗\)2s\\in\\\{0\\\}\\cup\(\\mathbb\{F\}\_\{q\}^\{\*\}\)^\{2\}, which gives one inclusion\.

Conversely, supposes=Q⁡\(x,y\)−z∈\{0\}∪\(𝔽q∗\)2s=Q\(x,y\)\-z\\in\\\{0\\\}\\cup\(\\mathbb\{F\}\_\{q\}^\{\*\}\)^\{2\}, and fix any\[u:v\]∈D\[u:v\]\\in D, which exists by Lemma[C\.10\.5](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem5)\. Ifs=0s=0, then\(x,y,z\)∈S\(x,y,z\)\\in Sand\(x,y,z\)\(x,y,z\)is the point ofℓx,y,\[u:v\]\\ell\_\{x,y,\[u:v\]\}witht=0t=0\. Ifs≠0s\\neq 0, thens/Q⁡\(u,v\)s/Q\(u,v\)is a nonzero square, says/Q⁡\(u,v\)=λ2s/Q\(u,v\)=\\lambda^\{2\}withλ∈𝔽q∗\\lambda\\in\\mathbb\{F\}\_\{q\}^\{\*\}, and we set

a=x−λ​u,b=y−λ​v\.a=x\-\\lambda u,\\qquad b=y\-\\lambda v\.At parametert=λt=\\lambdathe lineℓa,b,\[u:v\]\\ell\_\{a,b,\[u:v\]\}has first two coordinates\(a\+λ​u,b\+λ​v\)=\(x,y\)\(a\+\\lambda u,b\+\\lambda v\)=\(x,y\), and by equation[29](https://arxiv.org/html/2608.16977#A3.E29)its third coordinate is

Q⁡\(a,b\)\+2​λ​\(a​u−ν​b​v\)=Q⁡\(x,y\)−λ2​Q​\(u,v\)=Q⁡\(x,y\)−s=z\.Q\(a,b\)\+2\\lambda\(au\-\\nu bv\)=Q\(x,y\)\-\\lambda^\{2\}Q\(u,v\)=Q\(x,y\)\-s=z\.Hence\(x,y,z\)∈ℓa,b,\[u:v\]⊆P\(L\)\(x,y,z\)\\in\\ell\_\{a,b,\[u:v\]\}\\subseteq P\(L\), which gives the other inclusion\.

Finally, for each fixed\(x,y\)∈𝔽q2\(x,y\)\\in\\mathbb\{F\}\_\{q\}^\{2\}the mapz↦Q⁡\(x,y\)−zz\\mapsto Q\(x,y\)\-zis a bijection of𝔽q\\mathbb\{F\}\_\{q\}, and

\|\{0\}∪\(𝔽q∗\)2\|=1\+q−12=q\+12\.\\bigl\|\\\{0\\\}\\cup\(\\mathbb\{F\}\_\{q\}^\{\*\}\)^\{2\}\\bigr\|=1\+\\frac\{q\-1\}\{2\}=\\frac\{q\+1\}\{2\}\.Therefore\|P⁡\(L\)\|=q2⋅q\+12\|P\(L\)\|=q^\{2\}\\cdot\\tfrac\{q\+1\}\{2\}, as desired\. ∎

#### C\.10\.5Proof of the main theorem

###### Proof of Theorem[C\.10\.1](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem1)\.

LetLLbe the half\-tangent family\. Lemma[C\.10\.6](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem6)gives\|L\|=q2​\(q\+1\)/2\|L\|=q^\{2\}\(q\+1\)/2, Lemma[C\.10\.7](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem7)givesmaxΠ⁡\|\{ℓ∈L:ℓ⊆Π\}\|≤q\+1\\max\_\{\\Pi\}\|\\\{\\ell\\in L:\\ell\\subseteq\\Pi\\\}\|\\leq q\+1, and Lemma[C\.10\.8](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem8)gives\|P⁡\(L\)\|=q2​\(q\+1\)/2\|P\(L\)\|=q^\{2\}\(q\+1\)/2\. ∎

###### Proof of Corollary[C\.10\.2](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem2)\.

Letqqrange over the odd prime powers and letL=LqL=L\_\{q\}be the family of Theorem[C\.10\.1](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem1), so that

\|Lq\|=q2​\(q\+1\)2≥q32,\|P⁡\(Lq\)\|=q2​\(q\+1\)2=\(12\+o⁡\(1\)\)​q3\.\|L\_\{q\}\|=\\frac\{q^\{2\}\(q\+1\)\}\{2\}\\geq\\frac\{q^\{3\}\}\{2\},\\qquad\|P\(L\_\{q\}\)\|=\\frac\{q^\{2\}\(q\+1\)\}\{2\}=\\Bigl\(\\frac\{1\}\{2\}\+o\(1\)\\Bigr\)q^\{3\}\.For Conjecture 1\.4, takeC=1/2C=1/2\. Given any functionα\\alphawithα⁡\(q\)/q→∞\\alpha\(q\)/q\\to\\inftywe haveα⁡\(q\)\>q\+1\\alpha\(q\)\>q\+1for all largeqq, so by Theorem[C\.10\.1](https://arxiv.org/html/2608.16977#A3.SS10.Thmtheorem1)no plane containsα⁡\(q\)\\alpha\(q\)lines ofLqL\_\{q\}\. The hypotheses therefore hold along the odd prime powers, while the conclusion\|P⁡\(Lq\)\|≥\(1−o⁡\(1\)\)​q3\|P\(L\_\{q\}\)\|\\geq\(1\-o\(1\)\)q^\{3\}fails\.

For Conjecture 1\.5, fix0<ε<1/20<\\varepsilon<1/2\. Then\|Lq\|≥q3/2≥q5/2\+ε\|L\_\{q\}\|\\geq q^\{3\}/2\\geq q^\{5/2\+\\varepsilon\}as soon asq1/2−ε≥2q^\{1/2\-\\varepsilon\}\\geq 2, andq\+1≤12​q3/2q\+1\\leq\\tfrac\{1\}\{2\}q^\{3/2\}for everyq≥7q\\geq 7, so the hypotheses hold for all sufficiently large oddqq\. But

q3−\|P⁡\(Lq\)\|=q32−q22,q^\{3\}\-\|P\(L\_\{q\}\)\|=\\frac\{q^\{3\}\}\{2\}\-\\frac\{q^\{2\}\}\{2\},which is notO⁡\(q5/2\)O\(q^\{5/2\}\), so the conclusion fails\. ∎

## References\.

- Bruen & Drudge \(1999\)Aiden A Bruen and Keldon Drudge\.The construction of Cameron–Liebler line classes inPG⁡\(3,q\)\\mathrm\{PG\}\(3,q\)\.*Finite Fields and Their Applications*, 5\(1\):35–45, 1999\.
- Cossidente & Pavese \(2017\)Antonio Cossidente and Francesco Pavese\.New Cameron–Liebler line classes with parameterq2\+12\\frac\{q^\{2\}\+1\}\{2\}\.*arXiv preprint arXiv:1707\.01878*, 2017\.
- Ellenberg & Hablicsek \(2016\)Jordan S Ellenberg and Marton Hablicsek\.An incidence conjecture of bourgain over fields of positive characteristic\.In*Forum of Mathematics, Sigma*, volume 4, pp\. e23\. Cambridge University Press, 2016\.
- Gavrilyuk et al\. \(2018\)Alexander L Gavrilyuk, Ilia Matkin, and Tim Penttila\.Derivation of cameron–liebler line classes\.*Designs, Codes and Cryptography*, 86\(1\):231–236, 2018\.
- Lund et al\. \(2018\)Ben Lund, Shubhangi Saraf, and Charles Wolf\.Finite field kakeya and nikodym sets in three dimensions\.*SIAM Journal on Discrete Mathematics*, 32\(4\):2836–2849, 2018\.
- Mockenhaupt & Tao \(2004\)Gerd Mockenhaupt and Terence Tao\.Restriction and Kakeya phenomena for finite fields\.*Duke Mathematical Journal*, 121\(1\):35–74, 2004\.
- Tao \(2025\)Terence Tao\.New nikodym set constructions over finite fields\.*arXiv preprint arXiv:2511\.07721*, 2025\.
- Wolff \(1999\)Thomas Wolff\.Recent work connected with the kakeya problem\.*Prospects in mathematics \(Princeton, NJ, 1996\)*, 2:129–162, 1999\.

### C\.11High\-girth graphs attainingχcs​\(G\)=2​χ​\(G\)\\chi^\{s\}\_\{c\}\(G\)=2\\chi\(G\)

For every graphGGthe signed circular chromatic number satisfiesχc​\(G\)≤χcs​\(G\)≤2​χc​\(G\)\\chi\_\{c\}\(G\)\\leq\\chi^\{s\}\_\{c\}\(G\)\\leq 2\\chi\_\{c\}\(G\), and[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)showed that the upper bound is approached bykk\-chromatic graphs of arbitrarily large girth\. Whether it is reached by a finite such graph was left open\. We show that it is: for all integersk,g≥2k,g\\geq 2there is a finite simple graph of chromatic numberkkand girth at leastggwhose signed circular chromatic number is exactly2​k2k\. The witness is a sparse randomkk\-partite signed graph with its short cycles deleted, and the point of the proof is a union bound that rules out every admissible ratiop/q<2​kp/q<2kat once\.

#### C\.11\.1Introduction

A*signed graph*\(G,σ\)\(G,\\sigma\)is a graphGGtogether with a*signature*σ:E⁡\(G\)→\{\+,−\}\\sigma:E\(G\)\\to\\\{\+,\-\\\}\. Following[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8), for an even integerppand an integerqqwith1≤q≤p/21\\leq q\\leq p/2, a*\(p,q\)\(p,q\)\-coloring*of\(G,σ\)\(G,\\sigma\)is a mapf:V⁡\(G\)→ℤpf:V\(G\)\\to\\mathbb\{Z\}\_\{p\}such that

dp​\(f⁡\(u\),f⁡\(v\)\)≥qfor every positive edge​u​v,d\_\{p\}\\bigl\(f\(u\),f\(v\)\\bigr\)\\geq q\\quad\\text\{for every positive edge \}uv,dp​\(f⁡\(u\),f⁡\(v\)\+p2\)≥qfor every negative edge​u​v,d\_\{p\}\\bigl\(f\(u\),f\(v\)\+\\tfrac\{p\}\{2\}\\bigr\)\\geq q\\quad\\text\{for every negative edge \}uv,wheredp​\(a,b\)=min⁡\{\|a−b\|,p−\|a−b\|\}d\_\{p\}\(a,b\)=\\min\\\{\|a\-b\|,\\,p\-\|a\-b\|\\\}is the distance in the cycleℤp\\mathbb\{Z\}\_\{p\}\. The*circular chromatic number*of\(G,σ\)\(G,\\sigma\)isχc​\(G,σ\)=inf\{p/q:\(G,σ\)​has a​\(p,q\)​\-coloring\}\\chi\_\{c\}\(G,\\sigma\)=\\inf\\\{p/q:\(G,\\sigma\)\\text\{ has a \}\(p,q\)\\text\{\-coloring\}\\\}, and the*signed circular chromatic number*of a graphGGis

χcs​\(G\)=max⁡\{χc​\(G,σ\):σ​a signature of​G\}\.\\chi^\{s\}\_\{c\}\(G\)=\\max\\bigl\\\{\\chi\_\{c\}\(G,\\sigma\):\\sigma\\text\{ a signature of \}G\\bigr\\\}\.For the all\-positive signature one recovers the ordinary circular chromatic number surveyed by[Zhu 2001](https://arxiv.org/html/2608.16977#bibk.bib14), soχc​\(G\)≤χcs​\(G\)\\chi\_\{c\}\(G\)\\leq\\chi^\{s\}\_\{c\}\(G\), and[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)give the matching upper boundχcs​\(G\)≤2​χc​\(G\)\\chi^\{s\}\_\{c\}\(G\)\\leq 2\\chi\_\{c\}\(G\)\. Switching at a vertex setAA, that is, reversing the signs of the edges of the cut\(A,V⁡\(G\)∖A\)\(A,V\(G\)\\setminus A\), in the sense of[Zaslavsky 1982b](https://arxiv.org/html/2608.16977#bibk.bib12), does not changeχc​\(G,σ\)\\chi\_\{c\}\(G,\\sigma\), since it amounts to replacingf⁡\(v\)f\(v\)byf⁡\(v\)\+p/2f\(v\)\+p/2onAA; quantifying over*all*mapsf:V⁡\(G\)→ℤpf:V\(G\)\\to\\mathbb\{Z\}\_\{p\}therefore already accounts for switching\. Throughout, the girth of a signed graph is the girth of its underlying graph, and all graphs are simple, so girth at least33is automatic and digons do not arise\.

Naserasr, Wang and Zhu proved that the boundχcs​\(G\)≤2​χc​\(G\)\\chi^\{s\}\_\{c\}\(G\)\\leq 2\\chi\_\{c\}\(G\)remains tight when the girth is prescribed: for all integersk,g≥2k,g\\geq 2and everyε\>0\\varepsilon\>0there is a graphGGof girth at leastggwithχ⁡\(G\)=k\\chi\(G\)=kandχcs​\(G\)\>2​k−ε\\chi^\{s\}\_\{c\}\(G\)\>2k\-\\varepsilon\([Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8), Theorem 30\)\. For each integerppthey produce a graph on the vertex set of a bipartite augmented tree of[Alon et al\. 2016](https://arxiv.org/html/2608.16977#bibk.bib1), with one new edge for each leaf joining two of that leaf’s ancestors, carrying a signature that admits no\(2​k​p,p\+1\)\(2kp,p\+1\)\-coloring\. The resulting lower bounds2​k​p/\(p\+1\)2kp/\(p\+1\)increase to2​k2kbut never equal it\. In the remark immediately following that proof they ask whether the supremum is attained:

*It is not known whether there is a finitekk\-chromatic graph of girth at leastggand withχcs​\(G\)=2​k\\chi^\{s\}\_\{c\}\(G\)=2k\.*

We answer this affirmatively, for every pairk,g≥2k,g\\geq 2, by a random construction\.

Attainment questions of this shape go both ways in this subject\. On the one hand,[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)show that the supremum ofχc​\(G,σ\)\\chi\_\{c\}\(G,\\sigma\)over signeddd\-degenerate simple graphs equals2​⌊d/2⌋\+22\\lfloor d/2\\rfloor\+2, and that it is attained by\(Kd\+1,\+\)\(K\_\{d\+1\},\+\)for oddddand by an explicit signed graphΩd\\Omega\_\{d\}for evend≥4d\\geq 4; ford=2d=2they produce only a sequence of signed graphs whose circular chromatic numbers tend to44\.[Kardoš et al\. 2023](https://arxiv.org/html/2608.16977#bibk.bib6)then showed that in the cased=2d=2the supremum is genuinely never attained, by proving that every signed22\-degenerate simple graph onnnvertices has circular chromatic number at most4−2/⌊\(n\+1\)/2⌋4\-2/\\lfloor\(n\+1\)/2\\rfloor, and that this bound is tight for everyn≥2n\\geq 2\. On the other hand[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)attain the supremum10/310/3for signed series\-parallel simple graphs with an explicit signed outerplanar graph, and[Pan & Zhu 2022](https://arxiv.org/html/2608.16977#bibk.bib9)went on to show that every rational in\[2,10/3\]\[2,10/3\]occurs;[Zhu & Zhu 2023](https://arxiv.org/html/2608.16977#bibk.bib13)carry the same analysis out for signed series\-parallel graphs in which every cycle with an odd number of positive edges is long, a signed analogue of odd girth rather than the girth of the underlying graph\. The girth family of[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)sat on neither side of this divide\.

Coloring of signed graphs goes back to[Zaslavsky 1982a](https://arxiv.org/html/2608.16977#bibk.bib11), whose00\-free2​k2k\-colorings are exactly the circular2​k2k\-colorings\([Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)\), and to[Máčajová et al\. 2014](https://arxiv.org/html/2608.16977#bibk.bib7), who proposed a chromatic number for signed graphs and conjectured that every signed planar simple graph is44\-colorable, equivalently thatχcs​\(G\)≤4\\chi^\{s\}\_\{c\}\(G\)\\leq 4for every planarGG\([Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)\);[Kardoš & Narboni 2021](https://arxiv.org/html/2608.16977#bibk.bib5)refuted this, and[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)exhibit a signed planar simple graph withχc=4\+23\\chi\_\{c\}=4\+\\tfrac\{2\}\{3\}\. A different circular refinement of signed graph coloring was introduced earlier by[Kang & Steffen 2018](https://arxiv.org/html/2608.16977#bibk.bib4); the two notions are compared by[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8), and the area as a whole is surveyed in[Wang 2022](https://arxiv.org/html/2608.16977#bibk.bib10)\. We are not aware of any work that settles the attainment question above for every pairk,gk,g\. Our construction adapts the deletion argument of[Erdös 1959](https://arxiv.org/html/2608.16977#bibk.bib2)to a randomkk\-partite signed graph\.

#### C\.11\.2Statement and small cases

One half of the problem is immediate\.

###### Lemma C\.11\.1\.

Ifχ⁡\(G\)≤k\\chi\(G\)\\leq kthen every signatureσ\\sigmaofGGadmits a\(2​k,1\)\(2k,1\)\-coloring, and henceχcs​\(G\)≤2​k\\chi^\{s\}\_\{c\}\(G\)\\leq 2k\.

###### Proof\.

LetV1,…,VkV\_\{1\},\\dots,V\_\{k\}be the color classes of a properkk\-coloring ofGGand putf⁡\(v\)=i−1∈ℤ2​kf\(v\)=i\-1\\in\\mathbb\{Z\}\_\{2k\}forv∈Viv\\in V\_\{i\}\. Letu​v∈E⁡\(G\)uv\\in E\(G\)\. Thenf⁡\(u\)≠f⁡\(v\)f\(u\)\\neq f\(v\), sod2​k​\(f⁡\(u\),f⁡\(v\)\)≥1d\_\{2k\}\(f\(u\),f\(v\)\)\\geq 1; andf⁡\(v\)\+k∈\{k,…,2​k−1\}f\(v\)\+k\\in\\\{k,\\dots,2k\-1\\\}is distinct fromf⁡\(u\)∈\{0,…,k−1\}f\(u\)\\in\\\{0,\\dots,k\-1\\\}, sod2​k​\(f⁡\(u\),f⁡\(v\)\+k\)≥1d\_\{2k\}\(f\(u\),f\(v\)\+k\)\\geq 1\. Both edge conditions hold regardless ofσ\\sigma, soχc​\(G,σ\)≤2​k\\chi\_\{c\}\(G,\\sigma\)\\leq 2kfor everyσ\\sigma\. ∎

For smallkkandggthe remaining half can already be settled by a finite search\.

###### Example C\.11\.2\.

Takek=2k=2andg=4g=4, and letK3,4K\_\{3,4\}have parts\{u1,u2,u3\}\\\{u\_\{1\},u\_\{2\},u\_\{3\}\\\}and\{w1,w2,w3,w4\}\\\{w\_\{1\},w\_\{2\},w\_\{3\},w\_\{4\}\\\}\. The signature whose negative edges are exactlyu2​w2u\_\{2\}w\_\{2\},u2​w4u\_\{2\}w\_\{4\},u3​w2u\_\{3\}w\_\{2\}andu3​w3u\_\{3\}w\_\{3\}has circular chromatic number44, soχcs​\(K3,4\)=4=2​χ​\(K3,4\)\\chi^\{s\}\_\{c\}\(K\_\{3,4\}\)=4=2\\chi\(K\_\{3,4\}\)\. Such a value is certified by an exhaustive check over the finitely many pairs\(p,q\)\(p,q\)left admissible by[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8), which states thatχc​\(G,σ\)=p/q\\chi\_\{c\}\(G,\\sigma\)=p/qfor some evenp≤2​\|V⁡\(G\)\|p\\leq 2\|V\(G\)\|and that the infimum in its definition is a minimum\. No graph on fewer than seven vertices hasχ=2\\chi=2andχcs=4\\chi^\{s\}\_\{c\}=4\. Indeed, ifH⊆GH\\subseteq Gthen every signature ofHHextends toGGand every\(p,q\)\(p,q\)\-coloring of the extension restricts toHH, soχcs​\(H\)≤χcs​\(G\)\\chi^\{s\}\_\{c\}\(H\)\\leq\\chi^\{s\}\_\{c\}\(G\); every bipartite graph on at most six vertices is a subgraph ofK3,3K\_\{3,3\}, ofK2,4K\_\{2,4\}or ofK1,5K\_\{1,5\}; andχcs​\(K3,3\)=3\\chi^\{s\}\_\{c\}\(K\_\{3,3\}\)=3,χcs​\(K2,4\)=8/3\\chi^\{s\}\_\{c\}\(K\_\{2,4\}\)=8/3andχcs​\(K1,5\)=2\\chi^\{s\}\_\{c\}\(K\_\{1,5\}\)=2\. The valueχcs​\(K3,4\)=4\\chi^\{s\}\_\{c\}\(K\_\{3,4\}\)=4is also due to[Gujgiczer et al\. 2023](https://arxiv.org/html/2608.16977#bibk.bib3), whose signed graphB​Q^​\(2,3\)\\widehat\{BQ\}\(2,3\)isK3,4K\_\{3,4\}with a maximum matching positive and the remaining nine edges negative; naming that matchingu1​w1u\_\{1\}w\_\{1\},u2​w3u\_\{2\}w\_\{3\},u3​w4u\_\{3\}w\_\{4\}and switching at\{u1,w1\}\\\{u\_\{1\},w\_\{1\}\\\}turns it into the signature above\. Atk=3k=3andg=3g=3the same happens: ifK3,3,4K\_\{3,3,4\}has parts\{x1,x2,x3\}\\\{x\_\{1\},x\_\{2\},x\_\{3\}\\\},\{y1,y2,y3\}\\\{y\_\{1\},y\_\{2\},y\_\{3\}\\\}and\{z1,z2,z3,z4\}\\\{z\_\{1\},z\_\{2\},z\_\{3\},z\_\{4\}\\\}, then the signature whose negative edges are exactly

x1​y1,x1​y3,x3​y3,x1​z4,x2​z1,x2​z4,y1​z2,y1​z4,y2​z2x\_\{1\}y\_\{1\},\\ x\_\{1\}y\_\{3\},\\ x\_\{3\}y\_\{3\},\\ x\_\{1\}z\_\{4\},\\ x\_\{2\}z\_\{1\},\\ x\_\{2\}z\_\{4\},\\ y\_\{1\}z\_\{2\},\\ y\_\{1\}z\_\{4\},\\ y\_\{2\}z\_\{2\}has circular chromatic number66, soχcs​\(K3,3,4\)=6=2​χ​\(K3,3,4\)\\chi^\{s\}\_\{c\}\(K\_\{3,3,4\}\)=6=2\\chi\(K\_\{3,3,4\}\)\.

###### Theorem C\.11\.3\.

For all integersk,g≥2k,g\\geq 2there is a finite simple graphGGwithχ⁡\(G\)=k\\chi\(G\)=k, girth at leastgg, andχcs​\(G\)=2​k\\chi^\{s\}\_\{c\}\(G\)=2k\.

By Lemma[C\.11\.1](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem1)the whole content of Theorem[C\.11\.3](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem3)is the lower bound: one must exhibit a*single*signatureσ\\sigmaon akk\-chromatic graph of girth at leastggwithχc​\(G,σ\)≥2​k\\chi\_\{c\}\(G,\\sigma\)\\geq 2k, that is, with no\(p,q\)\(p,q\)\-coloring for any admissible\(p,q\)\(p,q\)withp/q<2​kp/q<2k\. Two features of the problem make this delicate\. First, akk\-critical graph satisfiesχcs​\(G\)≤2​k−2\\chi^\{s\}\_\{c\}\(G\)\\leq 2k\-2\([Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)\), so a witness must be far from critical; the construction below in fact produces graphs in which every single edge can be deleted without lowering the chromatic number\. Second, by[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)the valueχc​\(G,σ\)\\chi\_\{c\}\(G,\\sigma\)is a ratiop/qp/qwithppeven andp≤2​\|V⁡\(G\)\|p\\leq 2\|V\(G\)\|, so the number of ratios below2​k2kthat have to be excluded grows with the graph\. Hence an argument that excludes one fixed target ratio per construction cannot suffice, and the union bound has to range over all admissible\(p,q\)\(p,q\)at once\.

The proof splits into two independent pieces\. The first is a deterministic statement about colors: ifp/q<2​kp/q<2kthen among anykkcolors inℤp\\mathbb\{Z\}\_\{p\}there are two that one of the two signs forbids, so everykk\-tuple of colors carries a blocking pair\. The second is the usual Erdős deletion argument, applied to akk\-partite random signed graph sparse enough that short cycles are rare but dense enough that the failure probability for a fixed coloring beats the number of colorings\.

#### C\.11\.3Neutral color pairs

Fix an even integerppand an integerqqwith1≤q≤p/21\\leq q\\leq p/2\. Call an unordered pair\{a,b\}⊆ℤp\\\{a,b\\\}\\subseteq\\mathbb\{Z\}\_\{p\}*neutral*if it obstructs neither sign, that is, ifdp​\(a,b\)≥qd\_\{p\}\(a,b\)\\geq qanddp​\(a,b\+p2\)≥qd\_\{p\}\(a,b\+\\frac\{p\}\{2\}\)\\geq q\. Sincedp​\(a,b\+p2\)=p2−dp​\(a,b\)d\_\{p\}\(a,b\+\\frac\{p\}\{2\}\)=\\frac\{p\}\{2\}\-d\_\{p\}\(a,b\), neutrality says exactly that

q≤dp​\(a,b\)≤p2−q\.q\\ \\leq\\ d\_\{p\}\(a,b\)\\ \\leq\\ \\tfrac\{p\}\{2\}\-q\.\(32\)A pair that is not neutral is*blocking*: at least one of the two signs placed on it violates the\(p,q\)\(p,q\)\-condition\. Note that\{a,a\}\\\{a,a\\\}is blocking, becausedp​\(a,a\)=0<qd\_\{p\}\(a,a\)=0<q\.

aabbAaA\_\{a\}AaA\_\{a\}AbA\_\{b\}AbA\_\{b\}Figure 12:The two arcs making upAaA\_\{a\}and the two arcs making upAbA\_\{b\}, drawn on the cycleℤp\\mathbb\{Z\}\_\{p\}\. Each arc has lengthqq, so\|Aa\|=\|Ab\|=2​q\|A\_\{a\}\|=\|A\_\{b\}\|=2q; as in the proof of Lemma[C\.11\.5](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem5), the pair\{a,b\}\\\{a,b\\\}is neutral precisely when the four arcs are pairwise disjoint, which is the situation drawn\.###### Lemma C\.11\.5\.

The graph on vertex setℤp\\mathbb\{Z\}\_\{p\}whose edges are the neutral pairs has clique number at most⌊p/\(2​q\)⌋\\lfloor p/\(2q\)\\rfloor\. In particular, ifp/q<2​kp/q<2kthen it contains no clique of sizekk\.

###### Proof\.

Fora∈ℤpa\\in\\mathbb\{Z\}\_\{p\}letAa=\[a,a\+q\)∪\[a\+p2,a\+p2\+q\)A\_\{a\}=\[a,a\+q\)\\cup\[a\+\\frac\{p\}\{2\},a\+\\frac\{p\}\{2\}\+q\), a union of two arcs ofℤp\\mathbb\{Z\}\_\{p\}of lengthqqeach, as in Figure[12](https://arxiv.org/html/2608.16977#A3.F12)\. The two arcs are disjoint becauseq≤p/2q\\leq p/2, so\|Aa\|=2​q\|A\_\{a\}\|=2q\. Suppose\{a,b\}\\\{a,b\\\}is neutral\. The arcs\[a,a\+q\)\[a,a\+q\)and\[b,b\+q\)\[b,b\+q\)meet only ifdp​\(a,b\)<qd\_\{p\}\(a,b\)<q, which equation[32](https://arxiv.org/html/2608.16977#A3.E32)forbids\. The arcs\[a,a\+q\)\[a,a\+q\)and\[b\+p2,b\+p2\+q\)\[b\+\\frac\{p\}\{2\},b\+\\frac\{p\}\{2\}\+q\)meet only ifdp​\(a,b\+p2\)<qd\_\{p\}\(a,b\+\\frac\{p\}\{2\}\)<q, likewise forbidden, and the two remaining pairs of arcs give back these same two conditions\. HenceAa∩Ab=∅A\_\{a\}\\cap A\_\{b\}=\\varnothing\. Ifa1,…,ata\_\{1\},\\dots,a\_\{t\}is a neutral clique, the setsAa1,…,AatA\_\{a\_\{1\}\},\\dots,A\_\{a\_\{t\}\}are therefore pairwise disjoint subsets ofℤp\\mathbb\{Z\}\_\{p\}, so2​t​q≤p2tq\\leq pandt≤⌊p/\(2​q\)⌋t\\leq\\lfloor p/\(2q\)\\rfloor\. Finallyp/q<2​kp/q<2kgives⌊p/\(2​q\)⌋≤k−1\\lfloor p/\(2q\)\\rfloor\\leq k\-1\. ∎

Lemma[C\.11\.5](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem5)is the only place where the hypothesisp/q<2​kp/q<2kis used, and it is what forces everykk\-tuple of colors to contain a blocking pair\.

#### C\.11\.4The random construction

Fixk≥2k\\geq 2andg≥2g\\geq 2, and fix a real numberα\\alphawith

0<α<1,α⁡\(g−2\)<1\.0<\\alpha<1,\\qquad\\alpha\(g\-2\)<1\.For a large integermmsetn=k​mn=kmandρ=m−1\+α∈\(0,1\]\\rho=m^\{\-1\+\\alpha\}\\in\(0,1\]\. LetV=V1∪⋯∪VkV=V\_\{1\}\\cup\\dots\\cup V\_\{k\}with\|Vi\|=m\|V\_\{i\}\|=m, so\|V\|=n\|V\|=n, and call a pair of vertices lying in distinct parts a*cross pair*\. Independently for each cross pair\{u,v\}\\\{u,v\\\}, place

a positive edge with probability​ρ2,a negative edge with probability​ρ2,\\displaystyle\\text\{a positive edge with probability \}\\tfrac\{\\rho\}\{2\},\\qquad\\text\{a negative edge with probability \}\\tfrac\{\\rho\}\{2\},no edge with probability​1−ρ\.\\displaystyle\\text\{no edge with probability \}1\-\\rho\.No edge is placed inside a part\. This yields a random signed simple graph\(G0,σ0\)\(G\_\{0\},\\sigma\_\{0\}\)onVVwhose underlying graph iskk\-partite\.

###### Lemma C\.11\.6\.

Letppbe even, let1≤q≤p/21\\leq q\\leq p/2withp/q<2​kp/q<2k, and letf:V→ℤpf:V\\to\\mathbb\{Z\}\_\{p\}be arbitrary\. Then at leastm2m^\{2\}cross pairs\{u,v\}\\\{u,v\\\}have\{f⁡\(u\),f⁡\(v\)\}\\\{f\(u\),f\(v\)\\\}blocking\. Likewise, for every mapc:V→\[k−1\]c:V\\to\[k\-1\]at leastm2m^\{2\}cross pairs are monochromatic undercc\.

###### Proof\.

Call a tuple\(v1,…,vk\)∈V1×⋯×Vk\(v\_\{1\},\\dots,v\_\{k\}\)\\in V\_\{1\}\\times\\dots\\times V\_\{k\}a transversal; there aremkm^\{k\}of them\. If all\(k2\)\\binom\{k\}\{2\}pairs of a transversal were neutral, then the colorsf⁡\(v1\),…,f⁡\(vk\)f\(v\_\{1\}\),\\dots,f\(v\_\{k\}\)would be pairwise distinct, since a repeated color gives a blocking pair, and they would form a neutral clique of sizekk, contradicting Lemma[C\.11\.5](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem5)\. So every transversal contains a blocking cross pair\. Each cross pair lies in exactlymk−2m^\{k\-2\}transversals, obtained by choosing one vertex from each of the remainingk−2k\-2parts\. Hence the number of blocking cross pairs is at leastmk/mk−2=m2m^\{k\}/m^\{k\-2\}=m^\{2\}\. The second statement is the same count, with the pigeonhole principle in place of Lemma[C\.11\.5](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem5)\. ∎

#### C\.11\.5Proof of the main theorem

###### Proof of Theorem[C\.11\.3](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem3)\.

We consider three events for\(G0,σ0\)\(G\_\{0\},\\sigma\_\{0\}\)\.

No cheap circular coloring\.Fix an evenppand an integer1≤q≤p/21\\leq q\\leq p/2withp/q<2​kp/q<2k, and fixf:V→ℤpf:V\\to\\mathbb\{Z\}\_\{p\}\. LetXfX\_\{f\}be the number of edges of\(G0,σ0\)\(G\_\{0\},\\sigma\_\{0\}\)violatingff\. By Lemma[C\.11\.6](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem6)we may fix a set of exactlym2m^\{2\}blocking cross pairs; for each of them one of the two signs violatesff, and that particular signed edge is present with probabilityρ/2\\rho/2, independently over pairs\. HenceXfX\_\{f\}stochastically dominates aBin⁡\(m2,ρ/2\)\\mathrm\{Bin\}\(m^\{2\},\\rho/2\)variable, whose mean is12​m1\+α\\tfrac\{1\}\{2\}m^\{1\+\\alpha\}, and the Chernoff bound gives

ℙ\[Xf<14m1\+α\]≤exp\(−116m1\+α\)\.\\mathbb\{P\}\\Bigl\[X\_\{f\}<\\tfrac\{1\}\{4\}m^\{1\+\\alpha\}\\Bigr\]\\ \\leq\\ \\exp\\bigl\(\-\\tfrac\{1\}\{16\}m^\{1\+\\alpha\}\\bigr\)\.By[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)the circular chromatic number of any signed graph onnnvertices equalsp/qp/qfor some evenp≤2​np\\leq 2n, and this applies to every spanning subgraph ofG0G\_\{0\}, so it suffices to range over evenp≤2​np\\leq 2n\. The number of triples\(p,q,f\)\(p,q,f\)withppeven,p≤2​np\\leq 2n,1≤q≤p/21\\leq q\\leq p/2andf:V→ℤpf:V\\to\\mathbb\{Z\}\_\{p\}is at mostn⋅n⋅\(2​n\)n=exp⁡\(O⁡\(m​log⁡m\)\)n\\cdot n\\cdot\(2n\)^\{n\}=\\exp\(O\(m\\log m\)\), andm1\+α≫m​log⁡mm^\{1\+\\alpha\}\\gg m\\log m, so with probability1−o⁡\(1\)1\-o\(1\)every such triple withp/q<2​kp/q<2ksatisfiesXf≥14​m1\+αX\_\{f\}\\geq\\frac\{1\}\{4\}m^\{1\+\\alpha\}\.

No cheap\(k−1\)\(k\-1\)\-coloring\.For a mapc:V→\[k−1\]c:V\\to\[k\-1\]letYcY\_\{c\}be the number ofcc\-monochromatic edges ofG0G\_\{0\}\. By Lemma[C\.11\.6](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem6)and the same argument,YcY\_\{c\}stochastically dominatesBin⁡\(m2,ρ\)\\mathrm\{Bin\}\(m^\{2\},\\rho\), whose mean ism1\+αm^\{1\+\\alpha\}, soℙ\[Yc<14m1\+α\]≤exp\(−18m1\+α\)\\mathbb\{P\}\[Y\_\{c\}<\\frac\{1\}\{4\}m^\{1\+\\alpha\}\]\\leq\\exp\(\-\\frac\{1\}\{8\}m^\{1\+\\alpha\}\)\. There are\(k−1\)n=exp⁡\(O⁡\(m\)\)\(k\-1\)^\{n\}=\\exp\(O\(m\)\)such maps, so with probability1−o⁡\(1\)1\-o\(1\)everycchasYc≥14​m1\+αY\_\{c\}\\geq\\frac\{1\}\{4\}m^\{1\+\\alpha\}\.

Few short cycles\.LetZZbe the number of cycles ofG0G\_\{0\}of length less thangg\. Since a cycle of lengthℓ\\ellis present only if allℓ\\ellof its pairs receive edges,

𝔼⁡\[Z\]≤∑ℓ=3g−1nℓ​ρℓ2​ℓ≤∑ℓ=3g−1\(k​mα\)ℓ=O⁡\(mα⁡\(g−1\)\),\\mathbb\{E\}\[Z\]\\ \\leq\\ \\sum\_\{\\ell=3\}^\{g\-1\}\\frac\{n^\{\\ell\}\\rho^\{\\ell\}\}\{2\\ell\}\\ \\leq\\ \\sum\_\{\\ell=3\}^\{g\-1\}\(km^\{\\alpha\}\)^\{\\ell\}\\ =\\ O\\bigl\(m^\{\\alpha\(g\-1\)\}\\bigr\),where the implicit constant depends onkkandggonly\. Nowα⁡\(g−1\)<1\+α\\alpha\(g\-1\)<1\+\\alphaprecisely becauseα⁡\(g−2\)<1\\alpha\(g\-2\)<1, so𝔼⁡\[Z\]=o⁡\(m1\+α\)\\mathbb\{E\}\[Z\]=o\(m^\{1\+\\alpha\}\), and Markov’s inequality givesZ<18​m1\+αZ<\\frac\{1\}\{8\}m^\{1\+\\alpha\}with probability1−o⁡\(1\)1\-o\(1\)\.

Formmlarge all three events hold simultaneously, and we fix such an outcome\. Delete one edge from each cycle ofG0G\_\{0\}of length less thangg, obtaining a signed graph\(G,σ\)\(G,\\sigma\)on the same vertex setVVwith fewer than18​m1\+α\\frac\{1\}\{8\}m^\{1\+\\alpha\}edges removed\. A cycle ofGGof length less thanggwould be a cycle ofG0G\_\{0\}of length less thanggall of whose edges survived the deletion, and there is none, soGGhas girth at leastgg; moreoverGGis simple andkk\-partite\.

Letppbe even withp≤2​np\\leq 2n, let1≤q≤p/21\\leq q\\leq p/2satisfyp/q<2​kp/q<2k, and letf:V→ℤpf:V\\to\\mathbb\{Z\}\_\{p\}be arbitrary\. Thenffstill violates at least14​m1\+α−18​m1\+α\>0\\frac\{1\}\{4\}m^\{1\+\\alpha\}\-\\frac\{1\}\{8\}m^\{1\+\\alpha\}\>0edges of\(G,σ\)\(G,\\sigma\), so\(G,σ\)\(G,\\sigma\)has no\(p,q\)\(p,q\)\-coloring for any such pair\. Since\|V⁡\(G\)\|=n\|V\(G\)\|=n,[Naserasr et al\. 2020](https://arxiv.org/html/2608.16977#bibk.bib8)givesχc​\(G,σ\)=p0/q0\\chi\_\{c\}\(G,\\sigma\)=p\_\{0\}/q\_\{0\}for a pair\(p0,q0\)\(p\_\{0\},q\_\{0\}\)withp0p\_\{0\}even,p0≤2​np\_\{0\}\\leq 2nand1≤q0≤p0/21\\leq q\_\{0\}\\leq p\_\{0\}/2for which\(G,σ\)\(G,\\sigma\)has a\(p0,q0\)\(p\_\{0\},q\_\{0\}\)\-coloring, and the previous sentence forcesp0/q0≥2​kp\_\{0\}/q\_\{0\}\\geq 2k, soχc​\(G,σ\)≥2​k\\chi\_\{c\}\(G,\\sigma\)\\geq 2k\. Similarly everyc:V→\[k−1\]c:V\\to\[k\-1\]leaves a monochromatic edge, soχ⁡\(G\)≥k\\chi\(G\)\\geq k, whileχ⁡\(G\)≤k\\chi\(G\)\\leq kbecauseGGiskk\-partite\. Finally Lemma[C\.11\.1](https://arxiv.org/html/2608.16977#A3.SS11.Thmtheorem1)givesχcs​\(G\)≤2​k\\chi^\{s\}\_\{c\}\(G\)\\leq 2k, whence

2​k≤χc​\(G,σ\)≤χcs​\(G\)≤2​k\.∎2k\\ \\leq\\ \\chi\_\{c\}\(G,\\sigma\)\\ \\leq\\ \\chi^\{s\}\_\{c\}\(G\)\\ \\leq\\ 2k\.\\qed

## References\.

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### C\.12Counting 4\-critical linear triple systems

Rödl and Siggers conjectured that the number ofkk\-critical\(r,l\)\(r,l\)\-systems onnnvertices is exponential innln^\{l\}\. We disprove this for\(k,r,l\)=\(4,3,2\)\(k,r,l\)=\(4,3,2\): there is a constantc\>0c\>0such that for infinitely manynnthere are at leastexp⁡\(c​n2​log⁡n\)\\exp\(cn^\{2\}\\log n\)pairwise nonisomorphic44\-critical linear triple systems onnnvertices\. The construction lifts a dense44\-critical graph to a triple system along a proper edge coloring of that graph, in such a way that the edge coloring can be read off from any minimal non\-33\-colorable subsystem of the lift\. The extralog⁡n\\log nin the exponent is exactly the entropy of the edge coloring, and it survives the passage from labeled systems to isomorphism classes\.

#### C\.12\.1Introduction

A hypergraphHHis*kk\-colorable*if its vertices can be colored withkkcolors so that no edge is monochromatic, and*kk\-chromatic*ifkkis the least such number\. Following[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4),HHis*kk\-critical*if it iskk\-chromatic, has no isolated vertices, andH−eH\-eis\(k−1\)\(k\-1\)\-colorable for every edgeeeofHH\. An*\(r,l\)\(r,l\)\-system*is anrr\-uniform hypergraph in which noll\-set of vertices lies in more than one edge; a\(3,2\)\(3,2\)\-system is a*linear triple system*\. WriteT⁡\(k,r,l,n\)T\(k,r,l,n\)for the number of nonisomorphickk\-critical\(r,l\)\(r,l\)\-systems onnnvertices\.

The first bound on this count is due to[Abbott et al\. 1980](https://arxiv.org/html/2608.16977#bibl.bib2), who showed that for allk,r≥3k,r\\geq 3there is a constantb=b⁡\(k,r\)\>1b=b\(k,r\)\>1withT⁡\(k,r,2,n\)\>bnT\(k,r,2,n\)\>b^\{\\,n\}for all largenn\.[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)proved that for allk≥3k\\geq 3andr\>l≥2r\>l\\geq 2and all largennthere is akk\-critical\(r,l\)\(r,l\)\-system onnnvertices with at leastc0​nlc\_\{0\}\\,n^\{l\}edges, which is optimal up to the constant since an\(r,l\)\(r,l\)\-system onnnvertices has at most\(nl\)/\(rl\)\\binom\{n\}\{l\}\\big/\\binom\{r\}\{l\}edges\. Feeding that density into a construction indexed by the bipartitions of the edge set, they improved the Abbott–Liu–Toft bound to

T⁡\(k,r,l,n\)\>αnlT\(k,r,l,n\)\>\\alpha^\{\\,n^\{l\}\}for allk≥3k\\geq 3,r\>l≥2r\>l\\geq 2and all largenn, with a constantα=α⁡\(k,r,l\)\>1\\alpha=\\alpha\(k,r,l\)\>1\([Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4), Theorem 6\.1\)\. Every system they produce has at leastc′​nlc^\{\\prime\}n^\{l\}edges for a constantc′\>0c^\{\\prime\}\>0, and Section 6 of[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)closes with a trivial upper bound and a conjecture:

*A trivial upper bound for the number of\(r,l\)\(r,l\)\-systems withc′​nlc^\{\\prime\}n^\{l\}edges is\(\(nr\)c′​nl\)<\(nrc′​nl\)≈\(nr−l\)nl=O⁡\(dnl​log⁡nr−l\)\\binom\{\\binom\{n\}\{r\}\}\{c^\{\\prime\}n^\{l\}\}<\\binom\{n^\{r\}\}\{c^\{\\prime\}n^\{l\}\}\\approx\(n^\{r\-l\}\)^\{n^\{l\}\}=O\(d^\{\\,n^\{l\}\\log n^\{r\-l\}\}\)\. We conjecture that the actual number is in fact exponential innln^\{l\}\.*

We read “the actual number” asT⁡\(k,r,l,n\)T\(k,r,l,n\), which the preceding theorem bounds from below byαnl\\alpha^\{\\,n^\{l\}\}; the content of the conjecture is then the matching upper boundT⁡\(k,r,l,n\)≤CnlT\(k,r,l,n\)\\leq C^\{n^\{l\}\}for some constantC=C⁡\(k,r,l\)C=C\(k,r,l\), that is, that the factorlog⁡nr−l\\log n^\{r\-l\}in the trivial bound is an artifact of the counting\. For graphs the corresponding assertion is immediate, since there are only2\(n2\)2^\{\\binom\{n\}\{2\}\}graphs onnnlabeled vertices\. Oncer\>lr\>l, the trivial count of\(r,l\)\(r,l\)\-systems withΘ⁡\(nl\)\\Theta\(n^\{l\}\)edges carries a logarithm in the exponent, and the conjecture asserts that criticality removes it\. We show that criticality does not remove it, already for linear triple systems andk=4k=4\.

On the competing reading, in which “the actual number” counts all\(r,l\)\(r,l\)\-systems onnnvertices withΩ⁡\(nl\)\\Omega\(n^\{l\}\)edges, criticality plays no role and the logarithm is present for a direct counting reason, so it is the reading above that is at issue\.

Beyond the lower boundαnl\\alpha^\{\\,n^\{l\}\}of[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)we are aware of no further work on the growth ofT⁡\(k,r,l,n\)T\(k,r,l,n\), and the problem is not recorded in the survey of color\-critical hypergraphs of[Kostochka 2006](https://arxiv.org/html/2608.16977#bibl.bib3), which concerns critical systems with few edges rather than their number\.

###### Theorem C\.12\.1\.

There is a constantc\>0c\>0and there are infinitely many integersnnsuch that

T⁡\(4,3,2,n\)≥exp⁡\(c​n2​log⁡n\)\.T\(4,3,2,n\)\\geq\\exp\\bigl\(c\\,n^\{2\}\\log n\\bigr\)\.In particular there is no constantCCfor whichT⁡\(4,3,2,n\)≤Cn2T\(4,3,2,n\)\\leq C^\{n^\{2\}\}for allnn\.

We now summarize the construction\. Fix a44\-critical graphGGonmmvertices withΩ⁡\(m2\)\\Omega\(m^\{2\}\)edges, which exists by a theorem of Toft, and fix a palette\[t\]\[t\]witht=4​mt=4m\. Every proper edge coloringβ:E⁡\(G\)→\[t\]\\beta\\colon E\(G\)\\to\[t\]is turned into a linear triple systemJβJ\_\{\\beta\}on a vertex set that does not depend onβ\\beta: take three copiesv1,v2,v3v^\{1\},v^\{2\},v^\{3\}of each vertexvvofGGtogether with one*palette vertex*wjw\_\{j\}for eachj∈\[t\]j\\in\[t\], and place over each edgeu​vuvofGGthe three triples\{ui,vi,wβ⁡\(u​v\)\}\\\{u^\{i\},v^\{i\},w\_\{\\beta\(uv\)\}\\\}\. Two forcing gadgets of[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)are then attached: one chains the palette vertices so that they all receive a common color, and one tiesv1,v2,v3v^\{1\},v^\{2\},v^\{3\}together so that they receive three distinct colors\. A proper33\-coloring ofJβJ\_\{\\beta\}would therefore select, for eachvv, the unique copyviv^\{i\}carrying the palette color, andv↦iv\\mapsto iwould be a proper33\-coloring ofGG\. The key point is that a minimal non\-33\-colorable subsystemKβK\_\{\\beta\}ofJβJ\_\{\\beta\}must retain a triple over every edge ofGG, sinceGGis44\-critical; that triple exhibits the valueβ⁡\(u​v\)\\beta\(uv\), soβ↦Kβ\\beta\\mapsto K\_\{\\beta\}is injective on labeled systems\. There areexp⁡\(Ω⁡\(m2​log⁡m\)\)\\exp\(\\Omega\(m^\{2\}\\log m\)\)choices ofβ\\beta, whileJβJ\_\{\\beta\}has onlyO⁡\(m\)O\(m\)vertices, so passing to isomorphism classes costs a factorexp⁡\(O⁡\(m​log⁡m\)\)\\exp\(O\(m\\log m\)\)and the bound survives\.

#### C\.12\.2The construction

We use two known ingredients\. The first is a theorem of[Toft 1970](https://arxiv.org/html/2608.16977#bibl.bib5): there is a constanta\>0a\>0such that for every sufficiently largemmthere is a44\-critical graph onmmvertices with at leasta​m2a\\,m^\{2\}edges\. As recorded by[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4), one may takea=1/16a=1/16here\.

The second is the pair of forcing gadgets of[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)\. We recall only the properties used below\.

###### Lemma C\.12\.2\(Forcing gadgets\)\.

There are fixed finite linear triple systemsSSandDDwith the following properties\.

1. 1\.SShas two designated verticesx,yx,y, and no edge ofSScontains both of them\. In every proper33\-coloring ofSSthe verticesx,yx,yreceive the same color, and conversely every assignment of a common color toxxandyyextends to a proper33\-coloring ofSS\.
2. 2\.DDhas three designated verticesu1,u2,u3u\_\{1\},u\_\{2\},u\_\{3\}\. In every proper33\-coloring ofDDthese three vertices receive three distinct colors, and conversely every assignment of three distinct colors tou1,u2,u3u\_\{1\},u\_\{2\},u\_\{3\}extends to a proper33\-coloring ofDD\.

###### Proof\.

TakeS=S⁡\(3,3\)S=S\(3,3\)andD=D⁡\(3,3\)D=D\(3,3\)from[Rödl & Siggers 2006](https://arxiv.org/html/2608.16977#bibl.bib4)\. Both are33\-chromatic\(3,2\)\(3,2\)\-systems, so each has at least one proper33\-coloring, and the two forward implications are exactly the forcing properties for which those gadgets are built\. That no edge ofSScontains both designated vertices is noted in the proof of that construction:S⁡\(3,3\)S\(3,3\)is obtained by deleting edges from the system left when one edge throughxxandyyis removed from a44\-critical linear triple system, and such a system exists by[Abbott & Liu 1978](https://arxiv.org/html/2608.16977#bibl.bib1)\.

For the converse statements, fix a proper33\-coloringψ0\\psi\_\{0\}ofSS\. Thenψ0​\(x\)=ψ0​\(y\)\\psi\_\{0\}\(x\)=\\psi\_\{0\}\(y\), and given any target colorγ\\gammawe may composeψ0\\psi\_\{0\}with a permutation of\[3\]\[3\]carryingψ0​\(x\)\\psi\_\{0\}\(x\)toγ\\gamma\. Likewise fix a proper33\-coloringψ0\\psi\_\{0\}ofDD; thenψ0​\(u1\),ψ0​\(u2\),ψ0​\(u3\)\\psi\_\{0\}\(u\_\{1\}\),\\psi\_\{0\}\(u\_\{2\}\),\\psi\_\{0\}\(u\_\{3\}\)are distinct, and given distinct target colorsγ1,γ2,γ3\\gamma\_\{1\},\\gamma\_\{2\},\\gamma\_\{3\}the assignmentψ0​\(ui\)↦γi\\psi\_\{0\}\(u\_\{i\}\)\\mapsto\\gamma\_\{i\}is a well\-defined permutation of\[3\]\[3\], with which we composeψ0\\psi\_\{0\}\. ∎

Fix from now on an integermmlarge enough for Toft’s theorem, and letGGbe a44\-critical graph onmmvertices with

e⁡\(G\)≥a​m2\.e\(G\)\\geq a\\,m^\{2\}\.Sett=4​mt=4m, and let𝒜\\mathcal\{A\}be the set of proper edge coloringsβ:E⁡\(G\)→\[t\]\\beta\\colon E\(G\)\\to\[t\], that is, of maps assigning distinct colors to any two edges sharing an endpoint\.

###### Lemma C\.12\.3\.

\|𝒜\|≥\(2​m\)e⁡\(G\)\|\\mathcal\{A\}\|\\geq\(2m\)^\{e\(G\)\}\.

###### Proof\.

OrderE⁡\(G\)E\(G\)arbitrarily and color greedily\. When the edgeu​vuvis reached, the colors already used on edges meetingu​vuvnumber at most

\(deg⁡u−1\)\+\(deg⁡v−1\)≤2​m−4,\(\\deg u\-1\)\+\(\\deg v\-1\)\\leq 2m\-4,since every degree is at mostm−1m\-1\. Ast=4​mt=4m, at least2​m\+42m\+4colors are available at every step\. Distinct sequences of choices produce distinct colorings, since the edge order is fixed, so\|𝒜\|≥\(2​m\)e⁡\(G\)\|\\mathcal\{A\}\|\\geq\(2m\)^\{e\(G\)\}\. ∎

We now define the ambient vertex set\. Let

V∗=\{v1,v2,v3:v∈V\(G\)\}∪\{w1,…,wt\}∪W,V^\{\\ast\}=\\\{v^\{1\},v^\{2\},v^\{3\}:v\\in V\(G\)\\\}\\cup\\\{w\_\{1\},\\dots,w\_\{t\}\\\}\\cup W,whereWWconsists oft−1t\-1disjoint sets of fresh internal vertices, one for a copy ofSSattached to each consecutive pairwj,wj\+1w\_\{j\},w\_\{j\+1\}, together withmmfurther disjoint sets of fresh internal vertices, one for a copy ofDDattached to each triplev1,v2,v3v^\{1\},v^\{2\},v^\{3\}\. The setV∗V^\{\\ast\}does not depend onβ\\beta\.

Forβ∈𝒜\\beta\\in\\mathcal\{A\}letJβJ\_\{\\beta\}be the hypergraph onV∗V^\{\\ast\}whose edges are the following\. For everyu​v∈E⁡\(G\)uv\\in E\(G\)and everyi∈\[3\]i\\in\[3\]there is the*lifted triple*

\{ui,vi,wβ⁡\(u​v\)\}\.\\\{u^\{i\},v^\{i\},w\_\{\\beta\(uv\)\}\\\}\.\(33\)For everyj∈\[t−1\]j\\in\[t\-1\]there are the edges of a copy ofSSwhose designated vertices are identified withwjw\_\{j\}andwj\+1w\_\{j\+1\}and whose remaining vertices are the corresponding fresh set inWW\. For everyv∈V⁡\(G\)v\\in V\(G\)there are the edges of a copy ofDDwhose designated vertices are identified withv1,v2,v3v^\{1\},v^\{2\},v^\{3\}and whose remaining vertices are the corresponding fresh set inWW\. Figure[13](https://arxiv.org/html/2608.16977#A3.F13)shows the three pieces\.

u1u^\{1\}u2u^\{2\}u3u^\{3\}v1v^\{1\}v2v^\{2\}v3v^\{3\}wβ⁡\(u​v\)w\_\{\\beta\(uv\)\}the three triples overu​vuvSSSSSSw1w\_\{1\}w2w\_\{2\}w3w\_\{3\}wtw\_\{t\}one common color on the paletteDDv1v^\{1\}v2v^\{2\}v3v^\{3\}three distinct colors on the copies ofvvFigure 13:The three ingredients ofJβJ\_\{\\beta\}\. On the left, the three lifted triples equation[33](https://arxiv.org/html/2608.16977#A3.E33)sitting over one edgeu​vuvofGG, drawn as triangles sharing the palette vertexwβ⁡\(u​v\)w\_\{\\beta\(uv\)\}\. On the upper right, the chain of copies ofSSthat forcesw1,…,wtw\_\{1\},\\dots,w\_\{t\}to receive a single common color\. On the lower right, the copy ofDDthat forces the three copies of a vertexvvto receive three distinct colors\.###### Lemma C\.12\.4\.

For everyβ∈𝒜\\beta\\in\\mathcal\{A\}the hypergraphJβJ\_\{\\beta\}is a linear triple system, and\|V∗\|≤C0​m\|V^\{\\ast\}\|\\leq C\_\{0\}mfor a constantC0C\_\{0\}depending only onSSandDD\.

###### Proof\.

Writes=\|V⁡\(S\)\|s=\|V\(S\)\|andd0=\|V⁡\(D\)\|d\_\{0\}=\|V\(D\)\|\. Counting the vertices ofV∗V^\{\\ast\}layer by layer,

\|V∗\|=3​m\+t\+\(t−1\)​\(s−2\)\+m⁡\(d0−3\)≤\(4​s\+d0−4\)​m,\|V^\{\\ast\}\|=3m\+t\+\(t\-1\)\(s\-2\)\+m\(d\_\{0\}\-3\)\\leq\(4s\+d\_\{0\}\-4\)\\,m,sincet=4​mt=4m; takeC0=4​s\+d0−4C\_\{0\}=4s\+d\_\{0\}\-4\.

Every edge ofJβJ\_\{\\beta\}has three vertices, so it remains to check that no pair of vertices lies in two edges\. Consider first a pair of the form\{ui,vi\}\\\{u^\{i\},v^\{i\}\\\}withu≠vu\\neq v\. No edge of a gadget copy contains copies of two distinct vertices ofGG, and a lifted triple containing bothuiu^\{i\}andviv^\{i\}lies over the edgeu​vuvin layerii, so at most one edge contains the pair\. Next consider a pair\{vi,wj\}\\\{v^\{i\},w\_\{j\}\\\}\. Gadget copies contribute no such edge, and a lifted triple containing this pair lies over an edge ofGGincident tovvand coloredjj; sinceβ\\betais proper there is at most one such edge, and the layeriiis determined as well\. A pair\{wj,wj′\}\\\{w\_\{j\},w\_\{j^\{\\prime\}\}\\\}lies in no lifted triple, and it lies in no gadget edge either, because the copies ofSSmeet each other only in single palette vertices and no edge ofSScontains both designated vertices\. A pair\{vi,vi′\}\\\{v^\{i\},v^\{i^\{\\prime\}\}\\\}withi≠i′i\\neq i^\{\\prime\}lies in no lifted triple, and the only gadget copy containing it is the copy ofDDattached tovv\. A pair\{ui,vi′\}\\\{u^\{i\},v^\{i^\{\\prime\}\}\\\}withu≠vu\\neq vandi≠i′i\\neq i^\{\\prime\}lies in no edge at all: every lifted triple uses a single layer, and no gadget edge contains copies of two distinct vertices ofGG\. Every remaining pair contains a fresh internal vertex, which lies in a single gadget copy, so every edge containing the pair is an edge of that copy; each copy is itself a linear triple system, so there is at most one such edge\. Hence no pair of vertices lies in two edges ofJβJ\_\{\\beta\}\. ∎

###### Lemma C\.12\.5\.

For everyβ∈𝒜\\beta\\in\\mathcal\{A\}the hypergraphJβJ\_\{\\beta\}is not33\-colorable\.

###### Proof\.

Supposeψ\\psiis a proper33\-coloring ofJβJ\_\{\\beta\}\. The copies ofSSgiveψ⁡\(wj\)=ψ⁡\(wj\+1\)\\psi\(w\_\{j\}\)=\\psi\(w\_\{j\+1\}\)for everyj∈\[t−1\]j\\in\[t\-1\]by Lemma[C\.12\.2](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem2), so

ψ⁡\(w1\)=ψ⁡\(w2\)=⋯=ψ⁡\(wt\),\\psi\(w\_\{1\}\)=\\psi\(w\_\{2\}\)=\\cdots=\\psi\(w\_\{t\}\),and after relabeling the colors we may assume this common value is11\. For eachv∈V⁡\(G\)v\\in V\(G\)the copy ofDDattached tov1,v2,v3v^\{1\},v^\{2\},v^\{3\}forces those three vertices to receive three distinct colors, so there is a unique indexϕ⁡\(v\)∈\[3\]\\phi\(v\)\\in\[3\]withψ⁡\(vϕ⁡\(v\)\)=1\\psi\(v^\{\\phi\(v\)\}\)=1\.

Letu​v∈E⁡\(G\)uv\\in E\(G\)and supposeϕ⁡\(u\)=ϕ⁡\(v\)=i\\phi\(u\)=\\phi\(v\)=i\. Then the lifted triple\{ui,vi,wβ⁡\(u​v\)\}\\\{u^\{i\},v^\{i\},w\_\{\\beta\(uv\)\}\\\}is monochromatic of color11, a contradiction\. Henceϕ\\phiis a proper33\-coloring ofGG, which is impossible becauseGGis44\-chromatic\. ∎

#### C\.12\.3Recovering the edge coloring

Foru​v∈E⁡\(G\)uv\\in E\(G\)write

Lu​v=\{\{ui,vi,wβ⁡\(u​v\)\}:i∈\[3\]\}L\_\{uv\}=\\bigl\\\{\\\{u^\{i\},v^\{i\},w\_\{\\beta\(uv\)\}\\\}:i\\in\[3\]\\bigr\\\}for the set of three lifted triples overu​vuv\. These sets are pairwise disjoint asu​vuvranges overE⁡\(G\)E\(G\)\.

###### Lemma C\.12\.6\.

LetK⊆JβK\\subseteq J\_\{\\beta\}be a subhypergraph that is not33\-colorable\. ThenKKcontains a triple fromLu​vL\_\{uv\}for everyu​v∈E⁡\(G\)uv\\in E\(G\)\.

###### Proof\.

Suppose instead thatKKcontains no triple ofLu​vL\_\{uv\}for someu​v∈E⁡\(G\)uv\\in E\(G\), so thatKKis a subhypergraph ofJβ−Lu​vJ\_\{\\beta\}\-L\_\{uv\}\. We exhibit a proper33\-coloring ofJβ−Lu​vJ\_\{\\beta\}\-L\_\{uv\}, which restricts to one ofKK\. SinceGGis44\-critical,G−u​vG\-uvhas a proper33\-coloringϕ:V⁡\(G\)→\[3\]\\phi\\colon V\(G\)\\to\[3\]\. Give every palette vertexwjw\_\{j\}the color11, and for eachz∈V⁡\(G\)z\\in V\(G\)colorz1,z2,z3z^\{1\},z^\{2\},z^\{3\}with the three colors in such a way thatzϕ⁡\(z\)z^\{\\phi\(z\)\}receives the color11\.

Let\{xi,yi,wβ⁡\(x​y\)\}\\\{x^\{i\},y^\{i\},w\_\{\\beta\(xy\)\}\\\}be a retained lifted triple, sox​y∈E⁡\(G\)xy\\in E\(G\)andx​y≠u​vxy\\neq uv\. If this triple were monochromatic then, aswβ⁡\(x​y\)w\_\{\\beta\(xy\)\}has color11, we would haveϕ⁡\(x\)=ϕ⁡\(y\)=i\\phi\(x\)=\\phi\(y\)=i, contradictingϕ⁡\(x\)≠ϕ⁡\(y\)\\phi\(x\)\\neq\\phi\(y\)\. So no retained lifted triple is monochromatic\. On each copy ofSSthe two designated vertices have received the same color, and on each copy ofDDthe three designated vertices have received three distinct colors, so by Lemma[C\.12\.2](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem2)the coloring extends over the fresh internal vertices of every gadget copy\. ThusJβ−Lu​vJ\_\{\\beta\}\-L\_\{uv\}is33\-colorable, contradicting the hypothesis onKK\. ∎

For eachβ∈𝒜\\beta\\in\\mathcal\{A\}choose an edge\-minimal non\-33\-colorable subhypergraph ofJβJ\_\{\\beta\}and delete its isolated vertices; call the resultKβK\_\{\\beta\}\. Such a subhypergraph exists by Lemma[C\.12\.5](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem5)\.

###### Lemma C\.12\.7\.

EachKβK\_\{\\beta\}is a44\-critical linear triple system with at leaste⁡\(G\)e\(G\)edges\. Moreover the mapβ↦Kβ\\beta\\mapsto K\_\{\\beta\}is injective\.

###### Proof\.

Linearity is inherited fromJβJ\_\{\\beta\}by Lemma[C\.12\.4](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem4)\. By the choice ofKβK\_\{\\beta\}it is not33\-colorable whileKβ−eK\_\{\\beta\}\-eis33\-colorable for every edgeee, and it has no isolated vertices\. It is44\-colorable: delete an edgeee, take a proper33\-coloring ofKβ−eK\_\{\\beta\}\-e, and ifeeis monochromatic recolor one vertex ofeewith a fourth color\. The recolored vertex then lies in no monochromatic edge, since it is the only vertex of that color, and every other edge is unaffected\. HenceKβK\_\{\\beta\}is44\-chromatic and44\-critical\. By Lemma[C\.12\.6](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem6)it contains a triple ofLu​vL\_\{uv\}for each of thee⁡\(G\)e\(G\)edgesu​vuvofGG, and these sets are pairwise disjoint, soKβK\_\{\\beta\}has at leaste⁡\(G\)e\(G\)edges\.

For injectivity, fixu​v∈E⁡\(G\)uv\\in E\(G\)and read offβ⁡\(u​v\)\\beta\(uv\)from the labeled hypergraphKβK\_\{\\beta\}as follows\. By Lemma[C\.12\.6](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem6)some triple of the form\{ui,vi,wj\}\\\{u^\{i\},v^\{i\},w\_\{j\}\\\}belongs toKβK\_\{\\beta\}, and as observed in the proof of Lemma[C\.12\.4](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem4)the only edges ofJβJ\_\{\\beta\}containinguiu^\{i\}andviv^\{i\}are the lifted triples overu​vuv, all of which use the palette vertexwβ⁡\(u​v\)w\_\{\\beta\(uv\)\}\. Soj=β⁡\(u​v\)j=\\beta\(uv\)\. ThereforeKβK\_\{\\beta\}determinesβ\\beta, and distinct edge colorings give distinct labeled hypergraphs onV∗V^\{\\ast\}\. ∎

#### C\.12\.4Proof of the main theorem

###### Proof of Theorem[C\.12\.1](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem1)\.

Letmmbe large, letGG,𝒜\\mathcal\{A\}and\{Kβ:β∈𝒜\}\\\{K\_\{\\beta\}:\\beta\\in\\mathcal\{A\}\\\}be as above, and putM=\|V∗\|M=\|V^\{\\ast\}\|, so thatM≤C0​mM\\leq C\_\{0\}mby Lemma[C\.12\.4](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem4)\. By Lemmas[C\.12\.3](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem3)and[C\.12\.7](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem7)the family\{Kβ:β∈𝒜\}\\\{K\_\{\\beta\}:\\beta\\in\\mathcal\{A\}\\\}consists of at least\(2​m\)e⁡\(G\)\(2m\)^\{e\(G\)\}distinct44\-critical linear triple systems, all labeled inside the single vertex setV∗V^\{\\ast\}\. Each has between11andMMvertices, so there is an integerq≤Mq\\leq Msuch that at least

\(2​m\)e⁡\(G\)M\\frac\{\(2m\)^\{e\(G\)\}\}\{M\}of them have exactlyqqvertices\. A labeled system insideV∗V^\{\\ast\}isomorphic to a fixedqq\-vertex system is determined by an injection of the latter’s vertex set intoV∗V^\{\\ast\}, so each isomorphism class accounts for at mostM\(M−1\)⋯\(M−q\+1\)≤MMM\(M\-1\)\\cdots\(M\-q\+1\)\\leq M^\{M\}of them\. Therefore

T⁡\(4,3,2,q\)≥\(2​m\)e⁡\(G\)M⋅MM\.T\(4,3,2,q\)\\ \\geq\\ \\frac\{\(2m\)^\{e\(G\)\}\}\{M\\cdot M^\{M\}\}\.Usinge⁡\(G\)≥a​m2e\(G\)\\geq am^\{2\}andM≤C0​mM\\leq C\_\{0\}m,

log⁡\(\(2​m\)e⁡\(G\)M⋅MM\)≥a​m2​log⁡\(2​m\)−\(C0​m\+1\)​log⁡\(C0​m\)≥a2​m2​log​m\\log\\left\(\\frac\{\(2m\)^\{e\(G\)\}\}\{M\\cdot M^\{M\}\}\\right\)\\ \\geq\\ a\\,m^\{2\}\\log\(2m\)\-\(C\_\{0\}m\+1\)\\log\(C\_\{0\}m\)\\ \\geq\\ \\frac\{a\}\{2\}\\,m^\{2\}\\log mfor all sufficiently largemm\.

It remains to compareqqwithmm\. On one handq≤M≤C0​mq\\leq M\\leq C\_\{0\}m\. On the other hand a linear triple system onqqvertices has at most\(q2\)/3\\binom\{q\}\{2\}\\big/3edges, since its edges contain three vertex pairs each and no pair is repeated, so Lemma[C\.12\.7](https://arxiv.org/html/2608.16977#A3.SS12.Thmtheorem7)gives

q26≥13​\(q2\)≥e⁡\(G\)≥a​m2,\\frac\{q^\{2\}\}\{6\}\\ \\geq\\ \\frac\{1\}\{3\}\\binom\{q\}\{2\}\\ \\geq\\ e\(G\)\\ \\geq\\ a\\,m^\{2\},whenceq≥6​a​mq\\geq\\sqrt\{6a\}\\,m\. Thusq=q⁡\(m\)=Θ⁡\(m\)q=q\(m\)=\\Theta\(m\), and in particularq⁡\(m\)→∞q\(m\)\\to\\inftyasm→∞m\\to\\infty, so the integersq⁡\(m\)q\(m\)take infinitely many distinct values\. Finallym≥q/C0m\\geq q/C\_\{0\}andlog⁡m≥log⁡q−log⁡C0≥12​log​q\\log m\\geq\\log q\-\\log C\_\{0\}\\geq\\tfrac\{1\}\{2\}\\log qonceq≥C02q\\geq C\_\{0\}^\{2\}, so

T⁡\(4,3,2,q\)≥exp⁡\(a2​m2​log⁡m\)≥exp⁡\(a4​C02​q2​log⁡q\)\.T\(4,3,2,q\)\\ \\geq\\ \\exp\\left\(\\frac\{a\}\{2\}m^\{2\}\\log m\\right\)\\ \\geq\\ \\exp\\left\(\\frac\{a\}\{4C\_\{0\}^\{2\}\}\\,q^\{2\}\\log q\\right\)\.Writingn=q⁡\(m\)n=q\(m\)andc=a/\(4​C02\)c=a/\(4C\_\{0\}^\{2\}\)completes the proof\. ∎

## References\.

- Abbott & Liu \(1978\)Harvey L Abbott and AC Liu\.The existence problem for colour critical linear hypergraphs\.*Acta Mathematica Academiae Scientiarum Hungarica*, 32\(3\):273–282, 1978\.
- Abbott et al\. \(1980\)HL Abbott, A Liu, and Bjarne Toft\.The enumeration problem for color critical linear hypergraphs\.*Journal of Combinatorial Theory, Series B*, 29\(1\):106–115, 1980\.
- Kostochka \(2006\)Alexandr Kostochka\.Color\-critical graphs and hypergraphs with few edges: a survey\.In*More Sets, Graphs and Numbers: A Salute to Vera Sós and András Hajnal*, pp\. 175–197\. Springer, 2006\.
- Rödl & Siggers \(2006\)Vojtech Rödl and M Siggers\.Color critical hypergraphs with many edges\.*Journal of Graph Theory*, 53\(1\):56–74, 2006\.
- Toft \(1970\)Bjarne Toft\.On the maximal number of edges of critical k\-chromatic graphs\.*\(No Title\)*, 1970\.

### C\.13Strongly connected digraphs with no smallkk\-kernel

This problem asks how small akk\-kernel a strongly connected digraph is guaranteed to have\. Spiro asked whether\|V⁡\(G\)\|/\(k\+1\)\+Ok​\(1\)\|V\(G\)\|/\(k\+1\)\+O\_\{k\}\(1\)is always enough oncek≥3k\\geq 3; we show that it is not, and that the coefficient1/\(k\+1\)1/\(k\+1\)is itself wrong, the truth lying between1/k1/kand1/\(k−1\)1/\(k\-1\)\. The digraphs witnessing this are a hub feedingmmparallel directed paths that all return through one common tail, long enough to force the hub out of everykk\-kernel\. Two refutations of the same question were published after our run, and the construction below was obtained independently of both\.

#### C\.13\.1Introduction

All digraphs here are finite and have no loops or parallel arcs; directed cycles of length two are allowed\. A setX⊆V⁡\(G\)X\\subseteq V\(G\)is*stable*if no arc ofGGhas both ends inXX, and for an integerk≥1k\\geq 1a*kk\-kernel*ofGGis a stable setX⊆V⁡\(G\)X\\subseteq V\(G\)such that every vertex ofGGcan be joined fromXXby a directed path of length at mostkk\. Paths of length00are allowed, so akk\-kernel covers itself\. A11\-kernel is a kernel and a22\-kernel is a quasikernel\. In a tournament every stable set is a single vertex, so a quasikernel is exactly a king\([Post & Zheng 2023](https://arxiv.org/html/2608.16977#bibm.bib7)\)\. We writeκk​\(G\)\\kappa\_\{k\}\(G\)for the minimum size of akk\-kernel ofGG\.

Not every digraph has a kernel, but every digraph has a quasikernel by the theorem of[Chvátal & Lovász 2006](https://arxiv.org/html/2608.16977#bibm.bib1), soκk​\(G\)\\kappa\_\{k\}\(G\)is defined for everyk≥2k\\geq 2\. How small a quasikernel one is entitled to expect is the subject of the*small quasi\-kernel conjecture*, thatκ2​\(G\)≤\|V⁡\(G\)\|/2\\kappa\_\{2\}\(G\)\\leq\|V\(G\)\|/2wheneverGGhas no source\. It was posed by P\. L\. Erdős and Székely in 1976 and stated in print by[Erdős & Székely 2010](https://arxiv.org/html/2608.16977#bibm.bib2);[Erdős et al\. 2023](https://arxiv.org/html/2608.16977#bibm.bib3)survey what is known\. After reversing every arc,kk\-kernels become the\(2,k\)\(2,k\)\-kernels of[Kwaśnik 2006](https://arxiv.org/html/2608.16977#bibm.bib4), that is, the stable sets that absorb every vertex within distancekk\.[Spiro 2026](https://arxiv.org/html/2608.16977#bibm.bib8)initiated their extremal study and asked what the source\-free hypothesis buys when it is strengthened to strong connectivity\.

*IfDDis a strongly connected digraph andq≥3q\\geq 3is an integer, does there exist aqq\-kernelQQofDDsuch that\|Q\|≤\|V⁡\(D\)\|q\+1\+Oq​\(1\)\|Q\|\\leq\\frac\{\|V\(D\)\|\}\{q\+1\}\+O\_\{q\}\(1\)?*

This is Question 7\.7 of[Spiro 2026](https://arxiv.org/html/2608.16977#bibm.bib8), and it is recorded as Conjecture 1\.4 by[Nguyen et al\. 2024](https://arxiv.org/html/2608.16977#bibm.bib5), who attribute it to Spiro and leave it open\. We writekkthroughout for the integer calledqqthere\. Spiro observes that the coefficient1/\(k\+1\)1/\(k\+1\)would be best possible, by considering a collection of directed cyclesCk\+2C\_\{k\+2\}all sharing a single vertex\. The restriction tok≥3k\\geq 3is necessary, and is what distinguishes the published form of the question from the first preprint version, which asked it fork≥2k\\geq 2: Example 17 of[Erdős et al\. 2023](https://arxiv.org/html/2608.16977#bibm.bib3)exhibits strongly connected digraphs whose smallest quasikernel has size\(12−o⁡\(1\)\)​\|V⁡\(G\)\|\(\\tfrac\{1\}\{2\}\-o\(1\)\)\|V\(G\)\|, which is far above\|V⁡\(G\)\|/3\|V\(G\)\|/3\.

On the positive side,[Spiro 2026](https://arxiv.org/html/2608.16977#bibm.bib8)showed that under the hypotheses of the question there is always akk\-kernel of size at most about\|V⁡\(G\)\|/log⁡k\|V\(G\)\|/\\log k, by producing aboutlog⁡k\\log kpairwise disjointkk\-kernels\. This was improved by[Nguyen et al\. 2024](https://arxiv.org/html/2608.16977#bibm.bib5), who proved that every digraphGGwith\|V⁡\(G\)\|\>1\|V\(G\)\|\>1admitting a spanning out\-arborescence, and in particular every strongly connected digraph, satisfies

κk​\(G\)≤1\+\|V⁡\(G\)\|−2k−1\(k≥2\)\.\\kappa\_\{k\}\(G\)\\ \\leq\\ 1\+\\frac\{\|V\(G\)\|\-2\}\{k\-1\}\\qquad\(k\\geq 2\)\.\(34\)They deduce equation[34](https://arxiv.org/html/2608.16977#A3.E34)from the acyclic case, where the coefficient improves to1/k1/k: an acyclic digraph on at least two vertices with only one source has akk\-kernel of size at most1\+\(\|V⁡\(G\)\|−2\)/k1\+\(\|V\(G\)\|\-2\)/k\. The digraph of their Figure 1, a sourceyywith a single arc to a hubxxtogether withmmdisjoint directed paths onkkvertices leavingxx, shows that this is tight\. We show that the answer to Question 7\.7 is negative for everyk≥3k\\geq 3, and that it fails by a linear rather than a constant margin\.

###### Theorem C\.13\.1\.

For every integerk≥2k\\geq 2and every integerm≥1m\\geq 1there is a strongly connected digraphGk,mG\_\{k,m\}onm​k\+k\+2mk\+k\+2vertices whose smallestkk\-kernel has size exactly

κk​\(Gk,m\)=m\+1=\|V⁡\(Gk,m\)\|−2k\.\\kappa\_\{k\}\(G\_\{k,m\}\)\\ =\\ m\+1\\ =\\ \\frac\{\|V\(G\_\{k,m\}\)\|\-2\}\{k\}\.Consequently, for everyk≥2k\\geq 2and every constantccthere is a strongly connected digraphGGwithκk​\(G\)\>\|V⁡\(G\)\|/\(k\+1\)\+c\\kappa\_\{k\}\(G\)\>\|V\(G\)\|/\(k\+1\)\+c\.

Theorem[C\.13\.1](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem1)does more than defeat the additive termOk​\(1\)O\_\{k\}\(1\): it shows that the coefficient1/\(k\+1\)1/\(k\+1\)is itself wrong\. Writeck∗c\_\{k\}^\{\*\}for the infimum of the constantsccfor whichκk​\(G\)≤c​\|V⁡\(G\)\|\+Ok​\(1\)\\kappa\_\{k\}\(G\)\\leq c\|V\(G\)\|\+O\_\{k\}\(1\)holds for all strongly connectedGG\. Theorem[C\.13\.1](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem1)givesck∗≥1/kc\_\{k\}^\{\*\}\\geq 1/kand equation[34](https://arxiv.org/html/2608.16977#A3.E34)givesck∗≤1/\(k−1\)c\_\{k\}^\{\*\}\\leq 1/\(k\-1\)\. Fork≥3k\\geq 3, whether the value1/k1/kis itself admissible is asked by[Wang et al\. 2026](https://arxiv.org/html/2608.16977#bibm.bib9); fork=2k=2the corresponding statementc2∗=1/2c\_\{2\}^\{\*\}=1/2would follow from the small quasi\-kernel conjecture, since every strongly connected digraph is source\-free\. The excess is already substantial for moderate parameters:κ3​\(G3,50\)=51\\kappa\_\{3\}\(G\_\{3,50\}\)=51on155155vertices against155/4=38\.75155/4=38\.75, andκ10​\(G10,50\)=51\\kappa\_\{10\}\(G\_\{10,50\}\)=51on512512vertices against512/11≈46\.55512/11\\approx 46\.55\. Takingk=2k=2gives strongly connected digraphs whose smallest quasikernel has size exactly12​\|V⁡\(G\)\|−1\\tfrac\{1\}\{2\}\|V\(G\)\|\-1, one less than the bound of the small quasi\-kernel conjecture; digraphs achieving this up too⁡\(\|V⁡\(G\)\|\)o\(\|V\(G\)\|\)were already given by[Erdős et al\. 2023](https://arxiv.org/html/2608.16977#bibm.bib3)\.

Two refutations have since appeared\.[Wang et al\. 2026](https://arxiv.org/html/2608.16977#bibm.bib9)give for eachk≥3k\\geq 3strongly connected digraphs everykk\-kernel of which has size at least\(\|V⁡\(G\)\|−2\)/k\(\|V\(G\)\|\-2\)/k; this is the bound of Theorem[C\.13\.1](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem1), which also coversk=2k=2\.[Penev et al\. 2026](https://arxiv.org/html/2608.16977#bibm.bib6)obtain the weaker boundκk​\(G\)\>\(1/k−ε\)​\|V⁡\(G\)\|\\kappa\_\{k\}\(G\)\>\(1/k\-\\varepsilon\)\|V\(G\)\|, but with the extra feature that the digraphs may be taken of arbitrarily large directed girth\. By contrast, every directed cycle ofGk,mG\_\{k,m\}passes through the hubxxand hence traverses a whole arm, so these cycles have exactly the lengthsk\+2,k\+3,…,2​k\+2k\+2,k\+3,\\dots,2k\+2andGk,mG\_\{k,m\}has directed girth onlyk\+2k\+2\. The same paper shows that for oddk≥3k\\geq 3every source\-free bipartite digraph with a spanning out\-arborescence, and in particular every strongly connected bipartite digraph, has akk\-kernel of size at most\|V⁡\(G\)\|/\(k\+1\)\+1\|V\(G\)\|/\(k\+1\)\+1\. For oddkk, then, no bipartite digraph can refute Question 7\.7, and indeed the lengths just listed include odd ones, soGk,mG\_\{k,m\}is not bipartite\.

The construction is the acyclic example of Figure 1 of[Nguyen et al\. 2024](https://arxiv.org/html/2608.16977#bibm.bib5)with a return path attached: themmparallel directed paths leaving the hubxx, each onkkvertices, are brought back toxxthrough one common tail\. In the acyclic example the hub is unusable because the sourceyylies in everykk\-kernel and sends an arc toxx; hereyyis no longer a source, and the tail takes over that role\. The tail is long enough that some vertex of it must be used to cover its own far end, and every vertex of the tail sends an arc to the hub; this forces the hub out of everykk\-kernel and thereby forces each of themmpaths to pay for itself\.

#### C\.13\.2The construction

Fix integersk≥2k\\geq 2andm≥1m\\geq 1\. The digraphGk,mG\_\{k,m\}has vertex set

V\(Gk,m\)=\{x,y,p1,…,pk\}∪\{ai,j:1≤i≤m,1≤j≤k\}V\(G\_\{k,m\}\)=\\\{x,y,p\_\{1\},\\dots,p\_\{k\}\\\}\\ \\cup\\ \\\{a\_\{i,j\}:1\\leq i\\leq m,\\ 1\\leq j\\leq k\\\}and arc set consisting of

y→x,pj→x\(1≤j≤k\),pj→pj\+1\(1≤j<k\),pk→y,y\\to x,\\qquad p\_\{j\}\\to x\\ \\ \(1\\leq j\\leq k\),\\qquad p\_\{j\}\\to p\_\{j\+1\}\\ \\ \(1\\leq j<k\),\\qquad p\_\{k\}\\to y,together with, for eachi∈\{1,…,m\}i\\in\\\{1,\\dots,m\\\}, the*arm*

x→ai,1→ai,2→⋯→ai,k→p1\.x\\to a\_\{i,1\}\\to a\_\{i,2\}\\to\\cdots\\to a\_\{i,k\}\\to p\_\{1\}\.There are no other arcs\. WriteAi:=\{ai,1,…,ai,k\}A\_\{i\}:=\\\{a\_\{i,1\},\\dots,a\_\{i,k\}\\\}for theiith arm andP:=\{y,p1,…,pk\}P:=\\\{y,p\_\{1\},\\dots,p\_\{k\}\\\}for the*tail*\. ThusV⁡\(Gk,m\)V\(G\_\{k,m\}\)is the disjoint union of\{x\}\\\{x\\\},PPandA1,…,AmA\_\{1\},\\dots,A\_\{m\}, and

\|V⁡\(Gk,m\)\|=1\+\(k\+1\)\+m​k=m​k\+k\+2\.\|V\(G\_\{k,m\}\)\|=1\+\(k\+1\)\+mk=mk\+k\+2\.\(35\)Figure[14](https://arxiv.org/html/2608.16977#A3.F14)shows the digraph\.

xxa1,1a\_\{1,1\}a1,2a\_\{1,2\}⋯\\cdotsa1,ka\_\{1,k\}⋮\\vdotsam,1a\_\{m,1\}am,2a\_\{m,2\}⋯\\cdotsam,ka\_\{m,k\}p1p\_\{1\}p2p\_\{2\}⋯\\cdotspkp\_\{k\}yyFigure 14:The digraphGk,mG\_\{k,m\}\. The hubxxfeedsmmarms, each a directed path withkkinternal vertices running fromxxtop1p\_\{1\}, and the tailp1→p2→⋯→pk→y→xp\_\{1\}\\to p\_\{2\}\\to\\cdots\\to p\_\{k\}\\to y\\to xcloses the digraph up\. The dashed arcspj→xp\_\{j\}\\to x, present for everyjj, are what keepxxout of everykk\-kernel\.###### Lemma C\.13\.2\.

Gk,mG\_\{k,m\}is strongly connected\.

###### Proof\.

Every vertex reachesxx: the vertexyyand eachpjp\_\{j\}do so along a single arc, andai,ja\_\{i,j\}reachesp1p\_\{1\}along its own arm and then usesp1→xp\_\{1\}\\to x\. Converselyxxreaches every vertex: it reaches all ofAiA\_\{i\}along theiith arm and thenp1p\_\{1\}, thenp2,…,pkp\_\{2\},\\dots,p\_\{k\}along the tail, and finallyyy\. Hence any vertex reaches any other throughxx\. ∎

#### C\.13\.3A smallkk\-kernel

We first exhibit akk\-kernel of the size claimed in Theorem[C\.13\.1](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem1)\.

###### Lemma C\.13\.3\.

The setK0:=\{y\}∪\{ai,k:1≤i≤m\}K\_\{0\}:=\\\{y\\\}\\cup\\\{a\_\{i,k\}:1\\leq i\\leq m\\\}is akk\-kernel ofGk,mG\_\{k,m\}, and\|K0\|=m\+1\|K\_\{0\}\|=m\+1\.

###### Proof\.

First,K0K\_\{0\}is stable\. Indeed the only arcs incident withyyarey→xy\\to xandpk→yp\_\{k\}\\to y, and the only arcs incident withai,ka\_\{i,k\}areai,k−1→ai,ka\_\{i,k\-1\}\\to a\_\{i,k\}andai,k→p1a\_\{i,k\}\\to p\_\{1\}; none ofxx,pkp\_\{k\},p1p\_\{1\},ai,k−1a\_\{i,k\-1\}lies inK0K\_\{0\}\. In particular distinct verticesai,ka\_\{i,k\}andai′,ka\_\{i^\{\\prime\},k\}are nonadjacent, as they lie on different arms\.

For the covering condition,y→xy\\to xgivesdist⁡\(y,x\)=1\\mathrm\{dist\}\(y,x\)=1, andy→x→ai,1→⋯→ai,jy\\to x\\to a\_\{i,1\}\\to\\cdots\\to a\_\{i,j\}is a directed path of lengthj\+1j\+1, sodist⁡\(y,ai,j\)≤k\\mathrm\{dist\}\(y,a\_\{i,j\}\)\\leq kfor everyj≤k−1j\\leq k\-1\. Eachai,ka\_\{i,k\}covers itself, andai,k→p1→p2→⋯→pja\_\{i,k\}\\to p\_\{1\}\\to p\_\{2\}\\to\\cdots\\to p\_\{j\}has lengthj≤kj\\leq k, so any oneai,ka\_\{i,k\}covers all ofp1,…,pkp\_\{1\},\\dots,p\_\{k\}\. Every vertex ofGk,mG\_\{k,m\}is therefore joined fromK0K\_\{0\}by a directed path of length at mostkk\. ∎

#### C\.13\.4Everykk\-kernel is large

We now show that nokk\-kernel ofGk,mG\_\{k,m\}can be smaller\.

###### Lemma C\.13\.4\.

Everykk\-kernelKKofGk,mG\_\{k,m\}satisfies\|K\|≥m\+1\|K\|\\geq m\+1\.

###### Proof\.

The crucial observation is that the far end of the tail and the far end of each arm can each be covered from only one place\.

First consideryy\. The unique in\-neighbor ofyyispkp\_\{k\}, the unique in\-neighbor ofpjp\_\{j\}ispj−1p\_\{j\-1\}for2≤j≤k2\\leq j\\leq k, and the in\-neighbors ofp1p\_\{1\}are themmverticesai,ka\_\{i,k\}\. Walking backwards fromyytherefore givesdist⁡\(pk\+1−ℓ,y\)=ℓ\\mathrm\{dist\}\(p\_\{k\+1\-\\ell\},y\)=\\ellfor1≤ℓ≤k1\\leq\\ell\\leq k, while every remaining vertex is at distance at leastk\+1k\+1fromyy\. So the verticesuuwithdist⁡\(u,y\)≤k\\mathrm\{dist\}\(u,y\)\\leq kare exactly thek\+1k\+1vertices ofPP, and henceK∩P≠∅K\\cap P\\neq\\varnothing\.

Next, every vertex ofPPhas an arc toxx, byy→xy\\to xandpj→xp\_\{j\}\\to x\. Pickingu∈K∩Pu\\in K\\cap P, the pairu,xu,xis joined by an arc, so stability ofKKforcesx∉Kx\\notin K\.

Now fixiiand considerai,ka\_\{i,k\}\. The unique in\-neighbor ofai,ja\_\{i,j\}isai,j−1a\_\{i,j\-1\}for2≤j≤k2\\leq j\\leq k, and the unique in\-neighbor ofai,1a\_\{i,1\}isxx\. Walking backwards fromai,ka\_\{i,k\}therefore givesdist⁡\(ai,k−ℓ,ai,k\)=ℓ\\mathrm\{dist\}\(a\_\{i,k\-\\ell\},a\_\{i,k\}\)=\\ellfor1≤ℓ≤k−11\\leq\\ell\\leq k\-1anddist⁡\(x,ai,k\)=k\\mathrm\{dist\}\(x,a\_\{i,k\}\)=k, while every remaining vertex is at distance at leastk\+1k\+1fromai,ka\_\{i,k\}\. So the verticesuuwithdist⁡\(u,ai,k\)≤k\\mathrm\{dist\}\(u,a\_\{i,k\}\)\\leq kare exactly thek\+1k\+1vertices ofAi∪\{x\}A\_\{i\}\\cup\\\{x\\\}, and sincex∉Kx\\notin Kwe getK∩Ai≠∅K\\cap A\_\{i\}\\neq\\varnothing\.

The setsP,A1,…,AmP,A\_\{1\},\\dots,A\_\{m\}are pairwise disjoint, so\|K\|≥m\+1\|K\|\\geq m\+1\. ∎

#### C\.13\.5Proof of the main theorem

###### Proof of Theorem[C\.13\.1](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem1)\.

The digraphGk,mG\_\{k,m\}is strongly connected by Lemma[C\.13\.2](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem2)and hasm​k\+k\+2mk\+k\+2vertices by equation[35](https://arxiv.org/html/2608.16977#A3.E35)\. Lemmas[C\.13\.3](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem3)and[C\.13\.4](https://arxiv.org/html/2608.16977#A3.SS13.Thmtheorem4)giveκk​\(Gk,m\)=m\+1\\kappa\_\{k\}\(G\_\{k,m\}\)=m\+1, and

m\+1=m​k\+kk=\|V⁡\(Gk,m\)\|−2km\+1=\\frac\{mk\+k\}\{k\}=\\frac\{\|V\(G\_\{k,m\}\)\|\-2\}\{k\}by equation[35](https://arxiv.org/html/2608.16977#A3.E35)\. For the last assertion, fixk≥2k\\geq 2and compute

κk​\(Gk,m\)−\|V⁡\(Gk,m\)\|k\+1=\(m\+1\)−m​k\+k\+2k\+1=\(m\+1\)​\(k\+1\)−m​k−k−2k\+1=m−1k\+1\.\\kappa\_\{k\}\(G\_\{k,m\}\)\-\\frac\{\|V\(G\_\{k,m\}\)\|\}\{k\+1\}=\(m\+1\)\-\\frac\{mk\+k\+2\}\{k\+1\}=\\frac\{\(m\+1\)\(k\+1\)\-mk\-k\-2\}\{k\+1\}=\\frac\{m\-1\}\{k\+1\}\.This tends to infinity asm→∞m\\to\\infty, so givenccwe may choosemmwith\(m−1\)/\(k\+1\)\>c\(m\-1\)/\(k\+1\)\>cand takeG=Gk,mG=G\_\{k,m\}\. ∎

## References\.

- Chvátal & Lovász \(2006\)Vašek Chvátal and László Lovász\.Every directed graph has a semi\-kernel\.In*Hypergraph Seminar: Ohio State University 1972*, pp\. 175–175\. Springer, 2006\.
- Erdős & Székely \(2010\)Péter L Erdős and László A Székely\.Two conjectures on quasi\-kernels\.In*Fete of Combinatorics and Computer Science*, volume 20 of*Bolyai Society Mathematical Studies*, pp\. 357–358\. Springer, 2010\.Open problems no\. 4\.
- Erdős et al\. \(2023\)Péter L Erdős, Ervin Győri, Tamás Róbert Mezei, Nika Salia, and Mykhaylo Tyomkyn\.On the small quasi\-kernel conjecture\.*arXiv preprint arXiv:2307\.04112*, 2023\.
- Kwaśnik \(2006\)Maria Kwaśnik\.On the \(k; l\)\-kernels\.In*Graph Theory: Proceedings of a Conference held in Łagów, Poland, February 10–13, 1981*, pp\. 114–121\. Springer, 2006\.
- Nguyen et al\. \(2024\)Tung Nguyen, Alex Scott, and Paul Seymour\.Distant digraph domination\.*arXiv preprint arXiv:2409\.05039*, 2024\.
- Penev et al\. \(2026\)Irena Penev, Maya Stein, and Ana Trujillo\-Negrete\.Smallqq\-kernels in digraphs\.*arXiv preprint arXiv:2608\.00825*, 2026\.
- Post & Zheng \(2023\)Logan Post and Zeyu Zheng\.Common kings of a chain of cycles in a strong tournament\.*Graphs and Combinatorics*, 39\(4\):71, 2023\.
- Spiro \(2026\)Sam Spiro\.Generalized quasikernels in digraphs\.*European Journal of Combinatorics*, 133:104307, 2026\.
- Wang et al\. \(2026\)Xiaoyi Wang, Bo Deng, and Bin Chen\.Counterexamples to a problem of spiro on k\-kernels\.*Discrete Mathematics*, 349\(12\):115303, 2026\.

### C\.14FF\-positivity of chromatic symmetric functions of hypertrees

The chromatic symmetric function of a hypergraph is the generating function for the colorings under which no hyperedge is monochromatic\. Unlike its graph analogue, it need not expand nonnegatively in Gessel’s fundamental quasisymmetric basis, but[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)proved that it does whenever the hypergraph is a hypertree all of whose hyperedges have prime size, and conjectured that primality is superfluous\. We prove the conjecture\. The prime case rests on partitioning the nonconstant colorings of anrr\-element hyperedge intor\!r\!blocks indexed by the linear orders of that hyperedge, and no general construction of such a partition is known\. We replace the partition by a decomposition of the indicator function of “nonconstant” into chain conditions with nonnegative rational weights; such a decomposition exists for everyrr, and it glues across a hypertree exactly as a partition would\.

#### C\.14\.1Introduction

A*hypergraph*is a pairH=\(V,E\)H=\(V,E\)withVVfinite andEEa family of subsets ofVV, each of size at least22, called*hyperedges*\. A*coloring*ofHHis a mapκ:V→ℕ\\kappa:V\\to\\mathbb\{N\}, whereℕ=\{1,2,…\}\\mathbb\{N\}=\\\{1,2,\\dots\\\}, andκ\\kappais*proper*if no hyperedge is monochromatic, that is, ifκ\\kappais nonconstant on everye∈Ee\\in E\. The*chromatic symmetric function*ofHHis

XH=∑κ​proper∏v∈Vxκ⁡\(v\),X\_\{H\}=\\sum\_\{\\kappa\\ \\text\{proper\}\}\\ \\prod\_\{v\\in V\}x\_\{\\kappa\(v\)\},introduced for ordinary graphs by[Stanley 1995](https://arxiv.org/html/2608.16977#bibn.bib6)and extended to hypergraphs by[Stanley 1998](https://arxiv.org/html/2608.16977#bibn.bib7)\.

Writen=\|V\|n=\|V\|and\[n\]=\{1,…,n\}\[n\]=\\\{1,\\dots,n\\\}\. ForS⊆\[n−1\]S\\subseteq\[n\-1\]the*fundamental quasisymmetric function*of[Gessel 1984](https://arxiv.org/html/2608.16977#bibn.bib2)is

FS\(n\)=∑i1≤⋯≤inj∈S⇒ij<ij\+1xi1⋯xin,F\_\{S\}^\{\(n\)\}=\\sum\_\{\\begin\{subarray\}\{c\}i\_\{1\}\\leq\\cdots\\leq i\_\{n\}\\\\ j\\in S\\Rightarrow i\_\{j\}<i\_\{j\+1\}\\end\{subarray\}\}x\_\{i\_\{1\}\}\\cdots x\_\{i\_\{n\}\},and theFS\(n\)F\_\{S\}^\{\(n\)\}withS⊆\[n−1\]S\\subseteq\[n\-1\]form a basis of the degree\-nncomponent of the ringQSym\\mathrm\{QSym\}of quasisymmetric functions\. A homogeneous quasisymmetric function of degreennis*FF\-positive*if all of its coefficients in this basis are nonnegative\.

A*path*inHHis a sequencev1,e1,v2,e2,…,em,vm\+1v\_\{1\},e\_\{1\},v\_\{2\},e\_\{2\},\\dots,e\_\{m\},v\_\{m\+1\}withvi,vi\+1∈eiv\_\{i\},v\_\{i\+1\}\\in e\_\{i\}for eachii, in which the hyperedgeseie\_\{i\}are distinct and the verticesviv\_\{i\}are distinct except thatv1=vm\+1v\_\{1\}=v\_\{m\+1\}is allowed\. It is a*cycle*ifv1=vm\+1v\_\{1\}=v\_\{m\+1\}andm≥2m\\geq 2\. The hypergraphHHis*connected*if any two vertices are joined by a path, and a*hypertree*is a connected hypergraph with no cycles; this is the convention of[Gessel & Kalikow 2005](https://arxiv.org/html/2608.16977#bibn.bib3)adopted by[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)\.

For an ordinary graph,XGX\_\{G\}is alwaysFF\-positive:[Stanley 1995](https://arxiv.org/html/2608.16977#bibn.bib6)splits the proper colorings according to the acyclic orientation each one induces and identifies every piece as a\(P,ω\)\(P,\\omega\)\-partition enumerator\. For hypergraphs this fails\.[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)records that the hypergraph onV=\[4\]V=\[4\]with hyperedges\{1,2,3\}\\\{1,2,3\\\}and\{2,3,4\}\\\{2,3,4\\\}has

XH=2​F\{1\}\+6​F\{2\}\+2​F\{3\}\+4​F\{1,2\}\+8​F\{1,3\}\+4​F\{2,3\}−2​F\{1,2,3\},X\_\{H\}=2F\_\{\\\{1\\\}\}\+6F\_\{\\\{2\\\}\}\+2F\_\{\\\{3\\\}\}\+4F\_\{\\\{1,2\\\}\}\+8F\_\{\\\{1,3\\\}\}\+4F\_\{\\\{2,3\\\}\}\-2F\_\{\\\{1,2,3\\\}\},and that the hypergraph with hyperedges\{1,2,3\},\{1,4\},\{2,4\},\{3,4,5\}\\\{1,2,3\\\},\\\{1,4\\\},\\\{2,4\\\},\\\{3,4,5\\\}is notFF\-positive either, although distinct hyperedges of it meet in at most one vertex\. Both of these contain cycles\. On the other hand[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)proves thatXHX\_\{H\}isFF\-positive wheneverHHis a hypertree each of whose hyperedges has prime size, and in that case obtains the explicit expansionXH=∑πFDesH⁡\(π\)\(n\)X\_\{H\}=\\sum\_\{\\pi\}F^\{\(n\)\}\_\{\\operatorname\{Des\}\_\{H\}\(\\pi\)\}over alln\!n\!bijectionsπ:V→\[n\]\\pi:V\\to\[n\], where theHH\-descentsDesH⁡\(π\)\\operatorname\{Des\}\_\{H\}\(\\pi\)are read off from the unique path between consecutively labeled vertices\.[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)asks for the removal of the primality hypothesis:

*LetHHbe a hypertree\. ThenXHX\_\{H\}isFF\-positive\.*

The role of primality is isolated by[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)\. Call an integerr≥2r\\geq 2*splittable*if the nonconstant colorings of anrr\-element set can be partitioned intor\!r\!blocks indexed by the linear orders of that set, the block of a linear order consisting of the colorings that are weakly increasing along it and strictly increasing at some prescribed set of steps; he proves thatXHX\_\{H\}isFF\-positive for every hypertree all of whose hyperedge sizes are splittable\. He shows that every prime is splittable, by the cyclic standardization of[Gessel & Reutenauer 1993](https://arxiv.org/html/2608.16977#bibn.bib4), and reports a splitting forr=4r=4found by computer search; he also observes that splittability ofrris equivalent to partitionability of the Coxeter complex of typeAr−1A\_\{r\-1\}with the empty face removed, a scheduling problem in the sense of[Breuer & Klivans 2016](https://arxiv.org/html/2608.16977#bibn.bib1)\. Conjecture A was therefore known for every hypertree whose hyperedge sizes are all prime or equal to44, and in particular for every hypertree with no hyperedge on more than55vertices\. Each further composite size requires a splitting of its own\. In a different direction,[Pawlowski 2018](https://arxiv.org/html/2608.16977#bibn.bib5)proves thatXHX\_\{H\}is Schur\-positive, henceFF\-positive, for every hyperforest whose line graph is bipartite, which settles Conjecture B of[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9); this does not reach all hypertrees, since three66\-element hyperedges through a common vertex form a hypertree whose line graph isK3K\_\{3\}and whose hyperedge sizes are neither prime nor44\.

###### Theorem C\.14\.1\.

LetH=\(V,E\)H=\(V,E\)be a finite hypertree andn=\|V\|n=\|V\|\. Then

XH=∑S⊆\[n−1\]aS​FS\(n\)with​aS∈ℤ≥0​for every​S⊆\[n−1\]\.X\_\{H\}=\\sum\_\{S\\subseteq\[n\-1\]\}a\_\{S\}\\,F\_\{S\}^\{\(n\)\}\\qquad\\text\{with \}a\_\{S\}\\in\\mathbb\{Z\}\_\{\\geq 0\}\\ \\text\{for every \}S\\subseteq\[n\-1\]\.

This is Conjecture A of[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)\. Every bijectionV→\[n\]V\\to\[n\]is a proper coloring, and the coefficient ofx1x2⋯xnx\_\{1\}x\_\{2\}\\cdots x\_\{n\}inFS\(n\)F\_\{S\}^\{\(n\)\}equals11for everySS, so the coefficients of Theorem[C\.14\.1](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem1)satisfy∑S⊆\[n−1\]aS=n\!\\sum\_\{S\\subseteq\[n\-1\]\}a\_\{S\}=n\!, as[Taylor 2015](https://arxiv.org/html/2608.16977#bibn.bib9)observes\. Theorem[C\.14\.1](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem1)therefore says thatXHX\_\{H\}is a sum ofn\!n\!fundamental quasisymmetric functions counted with multiplicity\.

Certain cases are immediate\. Ifn=1n=1thenE=∅E=\\varnothing, since every hyperedge has at least two vertices, andXH=x1\+x2\+⋯=F∅\(1\)X\_\{H\}=x\_\{1\}\+x\_\{2\}\+\\cdots=F\_\{\\varnothing\}^\{\(1\)\}\. We assumen≥2n\\geq 2from now on, so that every vertex ofHHlies in a hyperedge\.

We now summarize the proof\. Fix a hyperedgeeeof sizerrand, for a linear orderτ\\tauofeeand a setB⊆\[r−1\]B\\subseteq\[r\-1\], consider the colorings ofeethat are weakly increasing alongτ\\tauand strictly increasing at the steps inBB\. The key point is that the indicator function of “κ\\kappais nonconstant onee” is a nonnegative rational combination of the indicator functions of these conditions, taken over allr\!r\!ordersτ\\tauand all nonemptyBB, with weights depending only onrrandBB\. The weights are forced by the fundamental expansion of the chromatic symmetric function of a single hyperedge, and their nonnegativity is the elementary fact that every subset of\[r−1\]\[r\-1\]is the descent set of at least one permutation of\[r\]\[r\]\. Multiplying this local identity over the hyperedges expressesXHX\_\{H\}as a nonnegative rational combination of generating functions for systems of weak and strict inequalities, one inequality per consecutive pair inside each chosen order\. The hypertree hypothesis makes each such system a\(P,ω\)\(P,\\omega\)\-partition condition on an ordinary tree, so each of these generating functions isFF\-positive, and the fundamental coefficients ofXHX\_\{H\}are nonnegative rationals\. They are integers because the monomial quasisymmetric coefficients ofXHX\_\{H\}are integers and the two bases are related by an integral unitriangular matrix\.

#### C\.14\.2A weighted local decomposition for one hyperedge

Fixr≥2r\\geq 2\. Forπ∈𝔖r\\pi\\in\\mathfrak\{S\}\_\{r\}writeDes⁡\(π\)=\{j∈\[r−1\]:π⁡\(j\)\>π⁡\(j\+1\)\}\\operatorname\{Des\}\(\\pi\)=\\\{j\\in\[r\-1\]:\\pi\(j\)\>\\pi\(j\+1\)\\\}, and forB⊆\[r−1\]B\\subseteq\[r\-1\]put

Ar​\(B\)=\#⁡\{π∈𝔖r:Des⁡\(π\)=B\},wr​\(B\)=Ar​\(B\)−\(−1\)\|B\|r\!\.A\_\{r\}\(B\)=\\\#\\\{\\pi\\in\\mathfrak\{S\}\_\{r\}:\\operatorname\{Des\}\(\\pi\)=B\\\},\\qquad w\_\{r\}\(B\)=\\frac\{A\_\{r\}\(B\)\-\(\-1\)^\{\|B\|\}\}\{r\!\}\.SinceAr​\(∅\)=1A\_\{r\}\(\\varnothing\)=1we havewr​\(∅\)=0w\_\{r\}\(\\varnothing\)=0, so the empty set never contributes below\.

###### Lemma C\.14\.2\.

For everyB⊆\[r−1\]B\\subseteq\[r\-1\]one haswr​\(B\)≥0w\_\{r\}\(B\)\\geq 0\.

###### Proof\.

Every subset of\[r−1\]\[r\-1\]is the descent set of at least one permutation in𝔖r\\mathfrak\{S\}\_\{r\}\. Indeed, letBBcut\[r\]\[r\]into consecutive blocks, fill the first block increasingly with the largest available values of\[r\]\[r\], the second block increasingly with the next largest available values, and so on\. This permutation ascends inside each block and descends exactly at the cuts, so its descent set isBBandAr​\(B\)≥1A\_\{r\}\(B\)\\geq 1\. If\|B\|\|B\|is odd thenr\!​wr​\(B\)=Ar​\(B\)\+1\>0r\!\\,w\_\{r\}\(B\)=A\_\{r\}\(B\)\+1\>0, and if\|B\|\|B\|is even thenr\!​wr​\(B\)=Ar​\(B\)−1≥0r\!\\,w\_\{r\}\(B\)=A\_\{r\}\(B\)\-1\\geq 0\. ∎

ForP=\{p1<⋯<pk\}⊆\[r−1\]P=\\\{p\_\{1\}<\\cdots<p\_\{k\}\\\}\\subseteq\[r\-1\]let

α⁡\(P\)=\(α1​\(P\),…,αk\+1​\(P\)\)=\(p1,p2−p1,…,pk−pk−1,r−pk\)\\alpha\(P\)=\(\\alpha\_\{1\}\(P\),\\dots,\\alpha\_\{k\+1\}\(P\)\)=\(p\_\{1\},\\,p\_\{2\}\-p\_\{1\},\\,\\dots,\\,p\_\{k\}\-p\_\{k\-1\},\\,r\-p\_\{k\}\)be the composition ofrrwhose partial sums are the elements ofPP, withα⁡\(∅\)=\(r\)\\alpha\(\\varnothing\)=\(r\)\.

###### Lemma C\.14\.3\.

If∅≠P⊆\[r−1\]\\varnothing\\neq P\\subseteq\[r\-1\]then

∑∅≠B⊆Pwr​\(B\)=1∏iαi​\(P\)\!\.\\sum\_\{\\varnothing\\neq B\\subseteq P\}w\_\{r\}\(B\)=\\frac\{1\}\{\\prod\_\{i\}\\alpha\_\{i\}\(P\)\!\}\.

###### Proof\.

A permutation of\[r\]\[r\]has descent set contained inPPexactly when it is increasing on each of the blocks thatPPcuts\[r\]\[r\]into, and such a permutation is determined by the unordered choice of values placed in those blocks\. Hence

∑B⊆PAr​\(B\)=r\!∏iαi​\(P\)\!\.\\sum\_\{B\\subseteq P\}A\_\{r\}\(B\)=\\frac\{r\!\}\{\\prod\_\{i\}\\alpha\_\{i\}\(P\)\!\}\.SinceP≠∅P\\neq\\varnothingwe also have∑B⊆P\(−1\)\|B\|=0\\sum\_\{B\\subseteq P\}\(\-1\)^\{\|B\|\}=0, and therefore

∑B⊆Pwr​\(B\)=1r\!​\(r\!∏iαi​\(P\)\!−0\)=1∏iαi​\(P\)\!\.\\sum\_\{B\\subseteq P\}w\_\{r\}\(B\)=\\frac\{1\}\{r\!\}\\left\(\\frac\{r\!\}\{\\prod\_\{i\}\\alpha\_\{i\}\(P\)\!\}\-0\\right\)=\\frac\{1\}\{\\prod\_\{i\}\\alpha\_\{i\}\(P\)\!\}\.The summandwr​\(∅\)w\_\{r\}\(\\varnothing\)vanishes, so it may be omitted\. ∎

Leteebe a set of sizerr\. For a linear orderτ=\(u1,…,ur\)\\tau=\(u\_\{1\},\\dots,u\_\{r\}\)of the elements ofeeand a setB⊆\[r−1\]B\\subseteq\[r\-1\], letC⁡\(τ,B\)C\(\\tau,B\)be the set of coloringsκ:e→ℕ\\kappa:e\\to\\mathbb\{N\}with

κ⁡\(u1\)≤⋯≤κ⁡\(ur\),κ⁡\(uj\)<κ⁡\(uj\+1\)​for every​j∈B\.\\kappa\(u\_\{1\}\)\\leq\\cdots\\leq\\kappa\(u\_\{r\}\),\\qquad\\kappa\(u\_\{j\}\)<\\kappa\(u\_\{j\+1\}\)\\ \\text\{ for every \}j\\in B\.
###### Proposition C\.14\.4\.

For every coloringκ:e→ℕ\\kappa:e\\to\\mathbb\{N\},

𝟏\{κ​is nonconstant on​e\}=∑τ∑∅≠B⊆\[r−1\]wr\(B\)1\{κ∈C\(τ,B\)\},\\mathbf\{1\}\_\{\\\{\\kappa\\ \\text\{is nonconstant on\}\\ e\\\}\}=\\sum\_\{\\tau\}\\ \\sum\_\{\\varnothing\\neq B\\subseteq\[r\-1\]\}w\_\{r\}\(B\)\\,\\mathbf\{1\}\_\{\\\{\\kappa\\in C\(\\tau,B\)\\\}\},whereτ\\tauruns over allr\!r\!linear orders ofee\.

###### Proof\.

Suppose first thatκ\\kappais constant\. Then no strict inequality holds, soκ∉C⁡\(τ,B\)\\kappa\\notin C\(\\tau,B\)for everyτ\\tauand every nonemptyBB, and the right hand side is00\.

Suppose now thatκ\\kappais nonconstant, and let the fibers ofκ\\kappahave sizesα1,…,αk\\alpha\_\{1\},\\dots,\\alpha\_\{k\}listed in increasing order of color, so thatk≥2k\\geq 2andα1\+⋯\+αk=r\\alpha\_\{1\}\+\\cdots\+\\alpha\_\{k\}=r\. There are exactly∏iαi\!\\prod\_\{i\}\\alpha\_\{i\}\!linear ordersτ=\(u1,…,ur\)\\tau=\(u\_\{1\},\\dots,u\_\{r\}\)along whichκ\\kappais weakly increasing, namely those obtained by listing the fibers in increasing order of color and ordering each fiber arbitrarily\. For any other orderτ\\tauthe coloringκ\\kappalies in noC⁡\(τ,B\)C\(\\tau,B\)at all\. Fix one of the∏iαi\!\\prod\_\{i\}\\alpha\_\{i\}\!weakly increasing orders\. Along it the strict jumps occur exactly at the set

P=\{α1,α1\+α2,…,α1\+⋯\+αk−1\},P=\\\{\\alpha\_\{1\},\\ \\alpha\_\{1\}\+\\alpha\_\{2\},\\ \\dots,\\ \\alpha\_\{1\}\+\\cdots\+\\alpha\_\{k\-1\}\\\},which is nonempty becausek≥2k\\geq 2, andα⁡\(P\)=\(α1,…,αk\)\\alpha\(P\)=\(\\alpha\_\{1\},\\dots,\\alpha\_\{k\}\)\. Henceκ∈C⁡\(τ,B\)\\kappa\\in C\(\\tau,B\)if and only ifB⊆PB\\subseteq P, so the inner sum for thisτ\\tauequals

∑∅≠B⊆Pwr​\(B\)=1∏iαi\!\\sum\_\{\\varnothing\\neq B\\subseteq P\}w\_\{r\}\(B\)=\\frac\{1\}\{\\prod\_\{i\}\\alpha\_\{i\}\!\}by Lemma[C\.14\.3](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem3)\. Summing over the∏iαi\!\\prod\_\{i\}\\alpha\_\{i\}\!weakly increasing orders gives11, as desired\. ∎

###### Example C\.14\.5\.

Taker=4r=4, the smallest hyperedge size not covered by primality\. The descent\-set counts on𝔖4\\mathfrak\{S\}\_\{4\}areA4​\(∅\)=A4​\(\{1,2,3\}\)=1A\_\{4\}\(\\varnothing\)=A\_\{4\}\(\\\{1,2,3\\\}\)=1,A4​\(\{1\}\)=A4​\(\{3\}\)=A4​\(\{1,2\}\)=A4​\(\{2,3\}\)=3A\_\{4\}\(\\\{1\\\}\)=A\_\{4\}\(\\\{3\\\}\)=A\_\{4\}\(\\\{1,2\\\}\)=A\_\{4\}\(\\\{2,3\\\}\)=3andA4​\(\{2\}\)=A4​\(\{1,3\}\)=5A\_\{4\}\(\\\{2\\\}\)=A\_\{4\}\(\\\{1,3\\\}\)=5, so

w4​\(\{1\}\)=w4​\(\{3\}\)=w4​\(\{1,3\}\)=16,w4​\(\{2\}\)=14,w\_\{4\}\(\\\{1\\\}\)=w\_\{4\}\(\\\{3\\\}\)=w\_\{4\}\(\\\{1,3\\\}\)=\\tfrac\{1\}\{6\},\\qquad w\_\{4\}\(\\\{2\\\}\)=\\tfrac\{1\}\{4\},w4​\(\{1,2\}\)=w4​\(\{2,3\}\)=w4​\(\{1,2,3\}\)=112\.w\_\{4\}\(\\\{1,2\\\}\)=w\_\{4\}\(\\\{2,3\\\}\)=w\_\{4\}\(\\\{1,2,3\\\}\)=\\tfrac\{1\}\{12\}\.Ifκ\\kappatakes two distinct values onee, each twice, then∏iαi\!=4\\prod\_\{i\}\\alpha\_\{i\}\!=4orders are weakly increasing, each withP=\{2\}P=\\\{2\\\}, and the total contribution is4⋅w4​\(\{2\}\)=14\\cdot w\_\{4\}\(\\\{2\\\}\)=1\. Ifκ\\kappais injective then a single order is weakly increasing, withP=\{1,2,3\}P=\\\{1,2,3\\\}, and the total contribution is16\+14\+16\+112\+16\+112\+112=1\\tfrac\{1\}\{6\}\+\\tfrac\{1\}\{4\}\+\\tfrac\{1\}\{6\}\+\\tfrac\{1\}\{12\}\+\\tfrac\{1\}\{6\}\+\\tfrac\{1\}\{12\}\+\\tfrac\{1\}\{12\}=1\.

#### C\.14\.3Gluing the local identities over a hypertree

For each hyperedgeeechoose a linear orderτe=\(ue,1,…,ue,\|e\|\)\\tau\_\{e\}=\(u\_\{e,1\},\\dots,u\_\{e,\|e\|\}\)of its vertices together with a nonempty setBe⊆\[\|e\|−1\]B\_\{e\}\\subseteq\[\|e\|\-1\], and letΩ\\Omegadenote such a collection of choices\. Put

WΩ=∏e∈Ew\|e\|​\(Be\),W\_\{\\Omega\}=\\prod\_\{e\\in E\}w\_\{\|e\|\}\(B\_\{e\}\),which is nonnegative by Lemma[C\.14\.2](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem2), and letKΩK\_\{\\Omega\}be the generating function∑κ∏v∈Vxκ⁡\(v\)\\sum\_\{\\kappa\}\\prod\_\{v\\in V\}x\_\{\\kappa\(v\)\}over the coloringsκ:V→ℕ\\kappa:V\\to\\mathbb\{N\}satisfying, for everye∈Ee\\in E,

κ⁡\(ue,1\)≤⋯≤κ⁡\(ue,\|e\|\),κ⁡\(ue,j\)<κ⁡\(ue,j\+1\)​for every​j∈Be\.\\kappa\(u\_\{e,1\}\)\\leq\\cdots\\leq\\kappa\(u\_\{e,\|e\|\}\),\\qquad\\kappa\(u\_\{e,j\}\)<\\kappa\(u\_\{e,j\+1\}\)\\ \\text\{ for every \}j\\in B\_\{e\}\.\(36\)A coloring is proper exactly when it is nonconstant on every hyperedge, so multiplying the identity of Proposition[C\.14\.4](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem4)over the hyperedges ofHHgives the pointwise identity

𝟏\{κ​proper\}=∑ΩWΩ​1\{κ​satisfies the inequalities of​Ω\},\\mathbf\{1\}\_\{\\\{\\kappa\\ \\text\{proper\}\\\}\}=\\sum\_\{\\Omega\}W\_\{\\Omega\}\\,\\mathbf\{1\}\_\{\\\{\\kappa\\ \\text\{satisfies the inequalities of\}\\ \\Omega\\\}\},in which “the inequalities ofΩ\\Omega” abbreviates equation[36](https://arxiv.org/html/2608.16977#A3.E36)for everye∈Ee\\in Eand the sum overΩ\\Omegais finite\. Weighting by∏v∈Vxκ⁡\(v\)\\prod\_\{v\\in V\}x\_\{\\kappa\(v\)\}and summing over all colorings therefore gives

XH=∑ΩWΩ​KΩ\.X\_\{H\}=\\sum\_\{\\Omega\}W\_\{\\Omega\}K\_\{\\Omega\}\.\(37\)We now use the hypertree hypothesis to identify eachKΩK\_\{\\Omega\}as a\(P,ω\)\(P,\\omega\)\-partition enumerator\.

###### Lemma C\.14\.6\.

LetH=\(V,E\)H=\(V,E\)be a hypertree with\|V\|≥2\|V\|\\geq 2\. Then its incidence graph is a tree, distinct hyperedges ofHHmeet in at most one vertex, and

∑e∈E\(\|e\|−1\)=\|V\|−1\.\\sum\_\{e\\in E\}\(\|e\|\-1\)=\|V\|\-1\.

###### Proof\.

The incidence graph has vertex setV⊔EV\\sqcup Eand an edge joiningvvtoeewheneverv∈ev\\in e\. Since\|V\|≥2\|V\|\\geq 2andHHis connected, every vertex ofHHlies in a hyperedge, and every path ofHHfromvvtov′v^\{\\prime\}is a walk of the incidence graph fromvvtov′v^\{\\prime\}; as each hyperedge is adjacent to its own vertices, the incidence graph is connected\. It is bipartite and simple, so each of its cycles has even length2​m2mwithm≥2m\\geq 2and readsv1,e1,v2,e2,…,vm,em,v1v\_\{1\},e\_\{1\},v\_\{2\},e\_\{2\},\\dots,v\_\{m\},e\_\{m\},v\_\{1\}with theviv\_\{i\}distinct and theeie\_\{i\}distinct, which is precisely a cycle ofHH\. AsHHhas no cycles the incidence graph is acyclic, hence a tree\. A tree on\|V\|\+\|E\|\|V\|\+\|E\|vertices has\|V\|\+\|E\|−1\|V\|\+\|E\|\-1edges, and the incidence graph has∑e∈E\|e\|\\sum\_\{e\\in E\}\|e\|edges, so∑e∈E\|e\|=\|V\|\+\|E\|−1\\sum\_\{e\\in E\}\|e\|=\|V\|\+\|E\|\-1, which is the displayed identity\. Finally, if distinct hyperedgese,e′e,e^\{\\prime\}both contained distinct verticesu,vu,v, thenu,e,v,e′,uu,e,v,e^\{\\prime\},uwould be a cycle ofHH\. ∎

GivenΩ\\Omega, letTΩT\_\{\\Omega\}be the graph on vertex setVVwhose edges join consecutive vertices in the chosen order of each hyperedge,

E\(TΩ\)=\{\{ue,j,ue,j\+1\}:e∈E,1≤j<\|e\|\}\.E\(T\_\{\\Omega\}\)=\\bigl\\\{\\\{u\_\{e,j\},u\_\{e,j\+1\}\\\}\\ :\\ e\\in E,\\ 1\\leq j<\|e\|\\bigr\\\}\.
###### Lemma C\.14\.7\.

For everyΩ\\Omegathe graphTΩT\_\{\\Omega\}is a tree onVV\.

###### Proof\.

It is connected\. Indeed, letu,v∈Vu,v\\in Vbe distinct and letu=v1,e1,…,em,vm\+1=vu=v\_\{1\},e\_\{1\},\\dots,e\_\{m\},v\_\{m\+1\}=vbe a path ofHH\. For eachiithe verticesviv\_\{i\}andvi\+1v\_\{i\+1\}both lie ineie\_\{i\}, and the edges ofTΩT\_\{\\Omega\}contributed byeie\_\{i\}form a path through all ofeie\_\{i\}, soviv\_\{i\}andvi\+1v\_\{i\+1\}are joined inTΩT\_\{\\Omega\}\. Next, the∑e∈E\(\|e\|−1\)\\sum\_\{e\\in E\}\(\|e\|\-1\)listed pairs are pairwise distinct: consecutive pairs inside one linear order are distinct, and two pairs coming from different hyperedges are distinct because distinct hyperedges share at most one vertex by Lemma[C\.14\.6](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem6)\. HenceTΩT\_\{\\Omega\}is a connected graph on\|V\|\|V\|vertices with∑e∈E\(\|e\|−1\)=\|V\|−1\\sum\_\{e\\in E\}\(\|e\|\-1\)=\|V\|\-1edges, again by Lemma[C\.14\.6](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem6), and is therefore a tree\. ∎

Orient every edge ofTΩT\_\{\\Omega\}asue,j→ue,j\+1u\_\{e,j\}\\to u\_\{e,j\+1\}, and call this oriented edge*strict*ifj∈Bej\\in B\_\{e\}and*weak*otherwise\. By Lemma[C\.14\.7](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem7)the underlying graph is a tree, so this orientation is acyclic and the transitive closure of these relations is a partial order; letPΩP\_\{\\Omega\}be the resulting poset onVV, generated by the relationsue,j<PΩue,j\+1u\_\{e,j\}<\_\{P\_\{\\Omega\}\}u\_\{e,j\+1\}\. LetQΩQ\_\{\\Omega\}be the second orientation of the same tree obtained by keeping every weak edge and reversing every strict edge\. It is again acyclic, so we may choose a bijectionωΩ:V→\[n\]\\omega\_\{\\Omega\}:V\\to\[n\]that is increasing alongQΩQ\_\{\\Omega\}; equivalently, along an oriented edgeu→vu\\to vofTΩT\_\{\\Omega\},

ωΩ​\(u\)<ωΩ​\(v\)​if the edge is weak,ωΩ​\(u\)\>ωΩ​\(v\)​if the edge is strict\.\\omega\_\{\\Omega\}\(u\)<\\omega\_\{\\Omega\}\(v\)\\ \\text\{ if the edge is weak\},\\qquad\\omega\_\{\\Omega\}\(u\)\>\\omega\_\{\\Omega\}\(v\)\\ \\text\{ if the edge is strict\}\.\(38\)Figure[15](https://arxiv.org/html/2608.16977#A3.F15)shows the construction on a small hypertree\.

112233445566e1e\_\{1\}e2e\_\{2\}the hypertreeHH112233445566the oriented treeTΩT\_\{\\Omega\}Figure 15:On the left, the hypertreeHHonV=\[6\]V=\[6\]with hyperedgese1=\{1,2,3\}e\_\{1\}=\\\{1,2,3\\\}ande2=\{3,4,5,6\}e\_\{2\}=\\\{3,4,5,6\\\}drawn as gray ovals\. On the right, the treeTΩT\_\{\\Omega\}for the choiceτe1=\(1,2,3\)\\tau\_\{e\_\{1\}\}=\(1,2,3\),τe2=\(3,4,5,6\)\\tau\_\{e\_\{2\}\}=\(3,4,5,6\),Be1=\{2\}B\_\{e\_\{1\}\}=\\\{2\\\}andBe2=\{1,3\}B\_\{e\_\{2\}\}=\\\{1,3\\\}: each hyperedge is threaded along its chosen order, and the three bold arrows are the strict steps, the two thin gray arrows the weak ones\. Reversing the bold arrows gives the acyclic orientationQΩQ\_\{\\Omega\}, and the labelingωΩ\\omega\_\{\\Omega\}that assigns1,2,3,4,5,61,2,3,4,5,6to the vertices4,6,1,3,5,24,6,1,3,5,2in this order is increasing along it, as required by equation[38](https://arxiv.org/html/2608.16977#A3.E38)\.We use the order\-preserving convention forPP\-partitions\. Given a finite posetPPonnnelements and a bijectionω:P→\[n\]\\omega:P\\to\[n\], a*\(P,ω\)\(P,\\omega\)\-partition*is a mapσ:P→ℕ\\sigma:P\\to\\mathbb\{N\}such thatx<Pyx<\_\{P\}yimpliesσ⁡\(x\)≤σ⁡\(y\)\\sigma\(x\)\\leq\\sigma\(y\), and impliesσ⁡\(x\)<σ⁡\(y\)\\sigma\(x\)<\\sigma\(y\)when moreoverω⁡\(x\)\>ω⁡\(y\)\\omega\(x\)\>\\omega\(y\)\. WriteΓ⁡\(P,ω\)=∑σ∏v∈Pxσ⁡\(v\)\\Gamma\(P,\\omega\)=\\sum\_\{\\sigma\}\\prod\_\{v\\in P\}x\_\{\\sigma\(v\)\}for the generating function of the\(P,ω\)\(P,\\omega\)\-partitions\. The fundamental theorem of\(P,ω\)\(P,\\omega\)\-partitions, for which see[Gessel 1984](https://arxiv.org/html/2608.16977#bibn.bib2)or, in the order\-reversing convention,[Stanley 2011](https://arxiv.org/html/2608.16977#bibn.bib8), states that

Γ⁡\(P,ω\)=∑π∈ℒ⁡\(P\)FDω​\(π\)\(n\),Dω​\(π\)=\{i∈\[n−1\]:ω⁡\(πi\)\>ω⁡\(πi\+1\)\},\\Gamma\(P,\\omega\)=\\sum\_\{\\pi\\in\\mathcal\{L\}\(P\)\}F^\{\(n\)\}\_\{D\_\{\\omega\}\(\\pi\)\},\\qquad D\_\{\\omega\}\(\\pi\)=\\\{i\\in\[n\-1\]:\\omega\(\\pi\_\{i\}\)\>\\omega\(\\pi\_\{i\+1\}\)\\\},\(39\)whereℒ⁡\(P\)\\mathcal\{L\}\(P\)is the set of linear extensions ofPP, each written as the wordπ1π2⋯πn\\pi\_\{1\}\\pi\_\{2\}\\cdots\\pi\_\{n\}that lists the elements ofPPin the corresponding order\. In particularΓ⁡\(P,ω\)\\Gamma\(P,\\omega\)isFF\-positive with integer coefficients\.

###### Lemma C\.14\.9\.

For everyΩ\\Omegaone hasKΩ=Γ⁡\(PΩ,ωΩ\)K\_\{\\Omega\}=\\Gamma\(P\_\{\\Omega\},\\omega\_\{\\Omega\}\)\. In particularKΩK\_\{\\Omega\}is a nonnegative integral combination of fundamental quasisymmetric functions\.

###### Proof\.

Suppose first thatκ\\kappasatisfies equation[36](https://arxiv.org/html/2608.16977#A3.E36)\. Then along every oriented edgeu→vu\\to vofTΩT\_\{\\Omega\}one hasκ⁡\(u\)≤κ⁡\(v\)\\kappa\(u\)\\leq\\kappa\(v\), with strict inequality when the edge is strict\. Ifx<PΩyx<\_\{P\_\{\\Omega\}\}ythere is a directed path fromxxtoyyinTΩT\_\{\\Omega\}, and chaining the inequalities along it givesκ⁡\(x\)≤κ⁡\(y\)\\kappa\(x\)\\leq\\kappa\(y\)\. Suppose in addition thatωΩ​\(x\)\>ωΩ​\(y\)\\omega\_\{\\Omega\}\(x\)\>\\omega\_\{\\Omega\}\(y\)\. By equation[38](https://arxiv.org/html/2608.16977#A3.E38)the value ofωΩ\\omega\_\{\\Omega\}increases along every weak edge, so if all edges of that directed path were weak we would getωΩ​\(x\)<ωΩ​\(y\)\\omega\_\{\\Omega\}\(x\)<\\omega\_\{\\Omega\}\(y\)\. Hence the path contains a strict edge, so at least one of the chained inequalities is strict andκ⁡\(x\)<κ⁡\(y\)\\kappa\(x\)<\\kappa\(y\)\. Thusκ\\kappais a\(PΩ,ωΩ\)\(P\_\{\\Omega\},\\omega\_\{\\Omega\}\)\-partition\.

Conversely, supposeκ\\kappais a\(PΩ,ωΩ\)\(P\_\{\\Omega\},\\omega\_\{\\Omega\}\)\-partition and letu→vu\\to vbe an oriented edge ofTΩT\_\{\\Omega\}, so thatu<PΩvu<\_\{P\_\{\\Omega\}\}v\. If the edge is weak thenωΩ​\(u\)<ωΩ​\(v\)\\omega\_\{\\Omega\}\(u\)<\\omega\_\{\\Omega\}\(v\)by equation[38](https://arxiv.org/html/2608.16977#A3.E38)and the definition givesκ⁡\(u\)≤κ⁡\(v\)\\kappa\(u\)\\leq\\kappa\(v\)\. If it is strict thenωΩ​\(u\)\>ωΩ​\(v\)\\omega\_\{\\Omega\}\(u\)\>\\omega\_\{\\Omega\}\(v\)and the definition givesκ⁡\(u\)<κ⁡\(v\)\\kappa\(u\)<\\kappa\(v\)\. Ranging over the edges contributed by a hyperedgeeerecovers exactly the conditions equation[36](https://arxiv.org/html/2608.16977#A3.E36)foree\.

The colorings counted byKΩK\_\{\\Omega\}are therefore precisely the\(PΩ,ωΩ\)\(P\_\{\\Omega\},\\omega\_\{\\Omega\}\)\-partitions, and the last assertion follows from equation[39](https://arxiv.org/html/2608.16977#A3.E39)\. ∎

#### C\.14\.4Proof of the main theorem

###### Proof of Theorem[C\.14\.1](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem1)\.

We may assumen≥2n\\geq 2\. Every weightWΩW\_\{\\Omega\}in equation[37](https://arxiv.org/html/2608.16977#A3.E37)is nonnegative by Lemma[C\.14\.2](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem2), and everyKΩK\_\{\\Omega\}is a nonnegative integral combination of fundamental quasisymmetric functions by Lemma[C\.14\.9](https://arxiv.org/html/2608.16977#A3.SS14.Thmtheorem9)\. Hence equation[37](https://arxiv.org/html/2608.16977#A3.E37)exhibitsXHX\_\{H\}as a nonnegative rational combination of theFS\(n\)F\_\{S\}^\{\(n\)\}, so everyaSa\_\{S\}is a nonnegative rational number\.

It remains to see that theaSa\_\{S\}are integers\. ForT⊆\[n−1\]T\\subseteq\[n\-1\]letα⁡\(T\)=\(α1​\(T\),…,αm​\(T\)\)\\alpha\(T\)=\(\\alpha\_\{1\}\(T\),\\dots,\\alpha\_\{m\}\(T\)\)be the associated composition ofnn, defined as above withrrreplaced bynn, and let

MT\(n\)=∑i1<⋯<imxi1α1​\(T\)⋯ximαm​\(T\)M\_\{T\}^\{\(n\)\}=\\sum\_\{i\_\{1\}<\\cdots<i\_\{m\}\}x\_\{i\_\{1\}\}^\{\\alpha\_\{1\}\(T\)\}\\cdots x\_\{i\_\{m\}\}^\{\\alpha\_\{m\}\(T\)\}be the corresponding monomial quasisymmetric function\. Splitting the defining sum ofFS\(n\)F\_\{S\}^\{\(n\)\}according to the exact setTTof indicesjjwithij<ij\+1i\_\{j\}<i\_\{j\+1\}gives

FS\(n\)=∑T⊇SMT\(n\)\.F\_\{S\}^\{\(n\)\}=\\sum\_\{T\\supseteq S\}M\_\{T\}^\{\(n\)\}\.WriteXH=∑TcT​MT\(n\)X\_\{H\}=\\sum\_\{T\}c\_\{T\}M\_\{T\}^\{\(n\)\}\. ThencT=∑S⊆TaSc\_\{T\}=\\sum\_\{S\\subseteq T\}a\_\{S\}, so Möbius inversion in the Boolean lattice gives

aS=∑T⊆S\(−1\)\|S\|−\|T\|​cT\.a\_\{S\}=\\sum\_\{T\\subseteq S\}\(\-1\)^\{\|S\|\-\|T\|\}c\_\{T\}\.NowcTc\_\{T\}is the coefficient ofx1α1​\(T\)⋯xmαm​\(T\)x\_\{1\}^\{\\alpha\_\{1\}\(T\)\}\\cdots x\_\{m\}^\{\\alpha\_\{m\}\(T\)\}inXHX\_\{H\}, which is the number of proper coloringsκ:V→\[m\]\\kappa:V\\to\[m\]with\|κ−1​\(i\)\|=αi​\(T\)\|\\kappa^\{\-1\}\(i\)\|=\\alpha\_\{i\}\(T\)for everyi∈\[m\]i\\in\[m\], hence a nonnegative integer\. Therefore eachaSa\_\{S\}is an integer, and being also nonnegative it lies inℤ≥0\\mathbb\{Z\}\_\{\\geq 0\}\. ∎

## References\.

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- Stanley \(2011\)Richard P Stanley\.Enumerative combinatorics volume 1 second edition\.*Cambridge studies in advanced mathematics*, 2011\.
- Taylor \(2015\)Jair Taylor\.Chromatic symmetric functions of hypertrees\.*arXiv preprint arXiv:1506\.08262*, 2015\.

### C\.15Quartically many Fano subsquares in Latin squares

This problem asks whether a Latin square of ordernncan contain more than cubically many subsquares of order77\. We show that it can: for everyn≥56n\\geq 56there is a Latin square of ordernnwith more than\(n/28\)4\(n/28\)^\{4\}subsquares isotopic to the Fano squareS7S\_\{7\}\. A quartic upper bound is already due to[Browning et al\. 2015](https://arxiv.org/html/2608.16977#bibo.bib4), so it is the lower bound that settles the order of growth; we also prove the explicit upper boundn4n^\{4\}by a short self\-contained argument\. The construction is an affine lift of the Fano quasigroup over a finite field of characteristic22\. The upper bound comes from the observation that three rows and one column already generate the whole incidence structure ofS7S\_\{7\}\.

#### C\.15\.1Introduction

LetMMbe a Latin square of ordernn\. A*subsquare*of ordermminMMis a set ofmmrows together with a set ofmmcolumns whose inducedm×mm\\times msubarray contains onlymmdistinct symbols; that subarray is then itself a Latin square, and its symbol set is determined by the chosen rows and columns\. Two Latin squares are*isotopic*if one is carried to the other by relabeling rows, columns and symbols independently\. Following[Browning et al\. 2014](https://arxiv.org/html/2608.16977#bibo.bib3), writeζ⁡\(n,m\)\\zeta\(n,m\)for the largest number of subsquares of ordermmin a Latin square of ordernn, and, for a fixed Latin squareLL, writeζ∗​\(n,L\)\\zeta^\{\*\}\(n,L\)for the largest number of subsquares isotopic toLLin a Latin square of ordernn\.

LetV=𝔽23V=\\mathbb\{F\}\_\{2\}^\{3\}andP=V∖\{0\}P=V\\setminus\\\{0\\\}\. The*Fano square*S7S\_\{7\}is the Cayley table of the quasigroup\(P,∘\)\(P,\\circ\)defined by

x∘y=\{x,x=y,x\+y,x≠y,x\\circ y=\\begin\{cases\}x,&x=y,\\\\ x\+y,&x\\neq y,\\end\{cases\}\(40\)with addition inVV\. Equivalently,S7S\_\{7\}is the Steiner quasigroup of the Fano plane: the unordered triples\{x,y,x\+y\}\\\{x,y,x\+y\\\}withx≠yx\\neq yare exactly the seven lines ofPG⁡\(2,2\)\\mathrm\{PG\}\(2,2\)\. Writing1,…,71,\\dots,7for the nonzero vectors ofVVin binary notation,S7S\_\{7\}is the array

∘123456711325476232167453213765445674123547615326745236176543217\\begin\{array\}\[\]\{c\|ccccccc\}\\circ&1&2&3&4&5&6&7\\\\ \\hline\\cr 1&1&3&2&5&4&7&6\\\\ 2&3&2&1&6&7&4&5\\\\ 3&2&1&3&7&6&5&4\\\\ 4&5&6&7&4&1&2&3\\\\ 5&4&7&6&1&5&3&2\\\\ 6&7&4&5&2&3&6&1\\\\ 7&6&5&4&3&2&1&7\\end\{array\}
Wanless asked the following at the LOOPS ’11 open problem session\([LOOPS ’11 2011](https://arxiv.org/html/2608.16977#bibo.bib6), Problem 2\.8\)\.

*Fix a primepp\. Is there a family of latin squares with more than cubically many subsquares of orderpp? More precisely, is it true that for every constantccthere is a latin squareLLof ordernnsuch that there are more thanc​n3cn^\{3\}subsquares of orderppinLL?*

The note accompanying the problem records that the answer is negative forp∈\{2,3,5\}p\\in\\\{2,3,5\\\}, and that forp=7p=7the subsquares would have to be multiplication tables of a Steiner quasigroup\. Since the Steiner triple system of order77is unique up to isomorphism, the only Steiner quasigroup of order77is\(P,∘\)\(P,\\circ\)\. The casep=7p=7of the problem is therefore precisely the question, left open by[Browning et al\. 2014](https://arxiv.org/html/2608.16977#bibo.bib3), of whetherζ∗​\(n,S7\)\\zeta^\{\*\}\(n,S\_\{7\}\)grows faster than cubically\. We answer it affirmatively\.

The results of[Browning et al\. 2014](https://arxiv.org/html/2608.16977#bibo.bib3)give the exact order of growth ofζ⁡\(n,m\)\\zeta\(n,m\)for several smallmm: one hasζ⁡\(n,m\)=Θ⁡\(n3\)\\zeta\(n,m\)=\\Theta\(n^\{3\}\)form∈\{2,3,5\}m\\in\\\{2,3,5\\\}andζ⁡\(n,m\)=Θ⁡\(n4\)\\zeta\(n,m\)=\\Theta\(n^\{4\}\)form∈\{4,6,9,10\}m\\in\\\{4,6,9,10\\\}, with the sharper boundsn3/8\+O⁡\(n2\)≤ζ⁡\(n,2\)≤n3/4\+O⁡\(n2\)n^\{3\}/8\+O\(n^\{2\}\)\\leq\\zeta\(n,2\)\\leq n^\{3\}/4\+O\(n^\{2\}\)andn3/27\+O⁡\(n5/2\)≤ζ⁡\(n,3\)≤n3/18\+O⁡\(n2\)n^\{3\}/27\+O\(n^\{5/2\}\)\\leq\\zeta\(n,3\)\\leq n^\{3\}/18\+O\(n^\{2\}\)\. For a fixed square they show thatζ∗​\(n,L\)=Θ⁡\(n3\)\\zeta^\{\*\}\(n,L\)=\\Theta\(n^\{3\}\)whenLLis cyclic, thatζ∗​\(n,L\)=O⁡\(n3\)\\zeta^\{\*\}\(n,L\)=O\(n^\{3\}\)for a large class ofLL, and that everyLLadmits someε∈\(0,1\)\\varepsilon\\in\(0,1\)withζ∗​\(n,L\)=Ω⁡\(n2\+ε\)\\zeta^\{\*\}\(n,L\)=\\Omega\(n^\{2\+\\varepsilon\}\)\. The valuem=7m=7is absent from both lists, and by the note quoted above the only order\-77isotopy class that can pushζ⁡\(n,7\)\\zeta\(n,7\)past cubic growth isS7S\_\{7\}\. On the upper bound side,[Browning et al\. 2015](https://arxiv.org/html/2608.16977#bibo.bib4)show that a Latin square of ordernnhasO⁡\(nψ⁡\(m,t\)\+t\)O\(n^\{\\psi\(m,t\)\+t\}\)subsquares of ordermmfor all positive integerst≤m≤nt\\leq m\\leq n, whereψ⁡\(m,t\)=⌈12​⌊m/t⌋⌉\\psi\(m,t\)=\\lceil\\tfrac\{1\}\{2\}\\lfloor m/t\\rfloor\\rceilfor oddmm; withm=7m=7andt=2t=2this givesζ⁡\(n,7\)=O⁡\(n4\)\\zeta\(n,7\)=O\(n^\{4\}\)\. For subsquares of unbounded order,[Browning et al\. 2013](https://arxiv.org/html/2608.16977#bibo.bib2)prove that a Latin square of ordernnhas at mostnO⁡\(log⁡k\)n^\{O\(\\log k\)\}subsquares of orderkk\.

Extremal behavior here is very far from typical behavior\.[McKay & Wanless 1999](https://arxiv.org/html/2608.16977#bibo.bib7)show that for everyε\>0\\varepsilon\>0almost all Latin squares of ordernnhave at leastn3/2−εn^\{3/2\-\\varepsilon\}subsquares of order22, and[Allsop & Wanless 2025](https://arxiv.org/html/2608.16977#bibo.bib1)show that a uniformly randomk×nk\\times nLatin rectangle has no proper subsquare of order44or more with probability1−O⁡\(1/n\)1\-O\(1/n\)\. In particular a random Latin square of ordernnhas, with probability1−O⁡\(1/n\)1\-O\(1/n\), no proper subsquare of order77, so a positive answer must come from an explicit construction\. The main result of this section is the following\.

###### Theorem C\.15\.1\.

For everyn≥56n\\geq 56,

\(n28\)4<ζ∗​\(n,S7\)≤n4\.\\left\(\\frac\{n\}\{28\}\\right\)^\{4\}<\\zeta^\{\*\}\(n,S\_\{7\}\)\\leq n^\{4\}\.In particularζ∗​\(n,S7\)=Θ⁡\(n4\)\\zeta^\{\*\}\(n,S\_\{7\}\)=\\Theta\(n^\{4\}\)\.

Combined with the bound of[Browning et al\. 2015](https://arxiv.org/html/2608.16977#bibo.bib4)quoted above, this settles the order of growth ofζ⁡\(n,7\)\\zeta\(n,7\)as well, adding77to the list ofmmfor whichζ⁡\(n,m\)\\zeta\(n,m\)is known\.

###### Corollary C\.15\.2\.

ζ⁡\(n,7\)=Θ⁡\(n4\)\\zeta\(n,7\)=\\Theta\(n^\{4\}\)\.

We now summarize the construction\. The key point is thatS7S\_\{7\}is almost an𝔽2\\mathbb\{F\}\_\{2\}\-linear object: off the diagonal the rule equation[40](https://arxiv.org/html/2608.16977#A3.E40)is simply addition inVV, and the diagonal rulex∘x=xx\\circ x=xis the only obstruction to linearity\. We therefore work onP×𝔽qP\\times\\mathbb\{F\}\_\{q\}withqqa power of two, keep the additive rule off the diagonal, and replace the diagonal rule by a nontrivial affine combinationλ​u\+\(1\+λ\)​v\\lambda u\+\(1\+\\lambda\)vin the fiber coordinate\. The copies ofS7S\_\{7\}are then the translated graphs of the𝔽2\\mathbb\{F\}\_\{2\}\-linear mapsV→𝔽qV\\to\\mathbb\{F\}\_\{q\}: there areq3q^\{3\}such maps andqqtranslations, givingq4q^\{4\}subsquares in a Latin square of order7​q7q\. The affine parameterλ\\lambdais what forces the diagonal cells of each graph to close up correctly\.

#### C\.15\.2The construction

Fix a power of twoq≥4q\\geq 4and an elementλ∈𝔽q∖\{0,1\}\\lambda\\in\\mathbb\{F\}\_\{q\}\\setminus\\\{0,1\\\}\. Define a binary operation∗\*onP×𝔽qP\\times\\mathbb\{F\}\_\{q\}by

\(x,u\)∗\(y,v\)=\{\(x,λ​u\+\(1\+λ\)​v\),x=y,\(x\+y,u\+v\),x≠y,\(x,u\)\*\(y,v\)=\\begin\{cases\}\(x,\\ \\lambda u\+\(1\+\\lambda\)v\),&x=y,\\\\ \(x\+y,\\ u\+v\),&x\\neq y,\\end\{cases\}\(41\)where the first coordinate is computed inVVand the second in𝔽q\\mathbb\{F\}\_\{q\}\. WriteLqL\_\{q\}for the resulting array, with rows and columns indexed byP×𝔽qP\\times\\mathbb\{F\}\_\{q\}\.

###### Lemma C\.15\.3\.

LqL\_\{q\}is a Latin square of order7​q7q\.

###### Proof\.

Since\|P×𝔽q\|=7​q\|P\\times\\mathbb\{F\}\_\{q\}\|=7q, it suffices to check that∗\*is a quasigroup operation, that is, that for each row and each symbol there is a unique column producing that symbol, and likewise with the roles of rows and columns interchanged\.

Fix a row\(x,u\)\(x,u\)and a symbol\(z,w\)\(z,w\), and seek a column\(y,v\)\(y,v\)with\(x,u\)∗\(y,v\)=\(z,w\)\(x,u\)\*\(y,v\)=\(z,w\)\. Suppose first thatz=xz=x\. The second branch of equation[41](https://arxiv.org/html/2608.16977#A3.E41)would forcex\+y=z=xx\+y=z=x, hencey=0∉Py=0\\notin P, so the first branch applies andy=xy=x\. The remaining equationλ​u\+\(1\+λ\)​v=w\\lambda u\+\(1\+\\lambda\)v=whas the unique solution

v=\(1\+λ\)−1​\(w\+λ​u\),v=\(1\+\\lambda\)^\{\-1\}\(w\+\\lambda u\),because1\+λ≠01\+\\lambda\\neq 0\. Suppose next thatz≠xz\\neq x\. The first branch would forcez=xz=x, so the second branch applies andy=x\+zy=x\+z, which is nonzero and distinct fromxxbecausez≠xz\\neq xandz≠0z\\neq 0\. The remaining equationu\+v=wu\+v=whas the unique solutionv=w\+uv=w\+u\.

Now fix a column\(y,v\)\(y,v\)and a symbol\(z,w\)\(z,w\), and seek a row\(x,u\)\(x,u\)\. Ifz=yz=ythen exactly as beforex=yx=y, andλ​u\+\(1\+λ\)​v=w\\lambda u\+\(1\+\\lambda\)v=whas the unique solutionu=λ−1​\(w\+\(1\+λ\)​v\)u=\\lambda^\{\-1\}\(w\+\(1\+\\lambda\)v\)becauseλ≠0\\lambda\\neq 0\. Ifz≠yz\\neq ythenx=y\+z∈Px=y\+z\\in Pwithx≠yx\\neq y, andu=w\+vu=w\+v\. ∎

We next exhibit many Fano subsquares ofLqL\_\{q\}\. Letϕ:V→𝔽q\\phi\\colon V\\to\\mathbb\{F\}\_\{q\}be an𝔽2\\mathbb\{F\}\_\{2\}\-linear map and letr∈𝔽qr\\in\\mathbb\{F\}\_\{q\}\. Put

r~=λ−1​\(1\+λ\)​r,\\tilde\{r\}=\\lambda^\{\-1\}\(1\+\\lambda\)r,and define

Rϕ,r=\{\(x,ϕ⁡\(x\)\+r\):x∈P\},Cϕ,r=\{\(x,ϕ⁡\(x\)\+r~\):x∈P\},\\displaystyle R\_\{\\phi,r\}=\\\{\(x,\\phi\(x\)\+r\):x\\in P\\\},\\qquad C\_\{\\phi,r\}=\\\{\(x,\\phi\(x\)\+\\tilde\{r\}\):x\\in P\\\},Tϕ,r=\{\(x,ϕ⁡\(x\)\+r\+r~\):x∈P\}\.\\displaystyle T\_\{\\phi,r\}=\\\{\(x,\\phi\(x\)\+r\+\\tilde\{r\}\):x\\in P\\\}\.Each of these sets has exactly seven elements, since its members have distinct first coordinates\.

###### Lemma C\.15\.4\.

For every pair\(ϕ,r\)\(\\phi,r\)the subarray ofLqL\_\{q\}on the rowsRϕ,rR\_\{\\phi,r\}and the columnsCϕ,rC\_\{\\phi,r\}is a subsquare with symbol setTϕ,rT\_\{\\phi,r\}, and it is isotopic toS7S\_\{7\}\.

###### Proof\.

Index the row\(x,ϕ⁡\(x\)\+r\)\(x,\\phi\(x\)\+r\), the column\(y,ϕ⁡\(y\)\+r~\)\(y,\\phi\(y\)\+\\tilde\{r\}\)and the symbol\(z,ϕ⁡\(z\)\+r\+r~\)\(z,\\phi\(z\)\+r\+\\tilde\{r\}\)by their first coordinatesx,y,z∈Px,y,z\\in P\. We claim that the row indexed byxxtimes the column indexed byyyis the symbol indexed byx∘yx\\circ y\.

Supposex≠yx\\neq y\. The second branch of equation[41](https://arxiv.org/html/2608.16977#A3.E41)applies, and sinceϕ\\phiis𝔽2\\mathbb\{F\}\_\{2\}\-linear,

\(ϕ⁡\(x\)\+r\)\+\(ϕ⁡\(y\)\+r~\)=ϕ⁡\(x\+y\)\+r\+r~\.\(\\phi\(x\)\+r\)\+\(\\phi\(y\)\+\\tilde\{r\}\)=\\phi\(x\+y\)\+r\+\\tilde\{r\}\.The product is therefore the symbol indexed byx\+y=x∘yx\+y=x\\circ y\.

Suppose insteadx=yx=y\. The first branch of equation[41](https://arxiv.org/html/2608.16977#A3.E41)applies, and in characteristic22,

λ⁡\(ϕ⁡\(x\)\+r\)\+\(1\+λ\)​\(ϕ⁡\(x\)\+r~\)=\(λ\+1\+λ\)​ϕ​\(x\)\+λ​r\+\(1\+λ\)​r~=ϕ⁡\(x\)\+λ​r\+\(1\+λ\)​r~\.\\lambda\\bigl\(\\phi\(x\)\+r\\bigr\)\+\(1\+\\lambda\)\\bigl\(\\phi\(x\)\+\\tilde\{r\}\\bigr\)=\(\\lambda\+1\+\\lambda\)\\phi\(x\)\+\\lambda r\+\(1\+\\lambda\)\\tilde\{r\}=\\phi\(x\)\+\\lambda r\+\(1\+\\lambda\)\\tilde\{r\}\.The definition ofr~\\tilde\{r\}says exactly thatλ​r~=\(1\+λ\)​r\\lambda\\tilde\{r\}=\(1\+\\lambda\)r, and addingλ​r\+r~\\lambda r\+\\tilde\{r\}to both sides of this identity turns it into

λ​r\+\(1\+λ\)​r~=r\+r~\.\\lambda r\+\(1\+\\lambda\)\\tilde\{r\}=r\+\\tilde\{r\}\.The product is therefore the symbol indexed byx=x∘xx=x\\circ x, which proves the claim\.

Consequently the subarray onRϕ,r×Cϕ,rR\_\{\\phi,r\}\\times C\_\{\\phi,r\}uses exactly the seven symbols ofTϕ,rT\_\{\\phi,r\}, so it is a subsquare of order77, and the three indexings by elements ofPPexhibit an isotopism fromS7S\_\{7\}onto it\. ∎

###### Lemma C\.15\.5\.

Theq4q^\{4\}subsquares produced by Lemma[C\.15\.4](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem4)are pairwise distinct\.

###### Proof\.

There areq3q^\{3\}𝔽2\\mathbb\{F\}\_\{2\}\-linear mapsV→𝔽qV\\to\\mathbb\{F\}\_\{q\}andqqchoices ofrr, so there areq4q^\{4\}pairs\(ϕ,r\)\(\\phi,r\); it suffices to show that distinct pairs give distinct row sets\. SupposeRϕ,r=Rψ,sR\_\{\\phi,r\}=R\_\{\\psi,s\}\. Each of the two sets contains exactly one element with first coordinatexx, for everyx∈Px\\in P, soϕ⁡\(x\)\+r=ψ⁡\(x\)\+s\\phi\(x\)\+r=\\psi\(x\)\+sfor everyx∈Px\\in P\. Thus the𝔽2\\mathbb\{F\}\_\{2\}\-linear maph=ϕ\+ψh=\\phi\+\\psitakes the constant valuer\+sr\+sonPP\. Choosing linearly independenta,b∈Va,b\\in Vgives

r\+s=h⁡\(a\+b\)=h⁡\(a\)\+h⁡\(b\)=\(r\+s\)\+\(r\+s\)=0\.r\+s=h\(a\+b\)=h\(a\)\+h\(b\)=\(r\+s\)\+\(r\+s\)=0\.Hencer=sr=s, andhhvanishes onPPand at00, soϕ=ψ\\phi=\\psi\. ∎

###### Corollary C\.15\.6\.

For every power of twoq≥4q\\geq 4one hasζ∗​\(7​q,S7\)≥q4\\zeta^\{\*\}\(7q,S\_\{7\}\)\\geq q^\{4\}\. Moreoverζ∗​\(n,S7\)\>\(n/28\)4\\zeta^\{\*\}\(n,S\_\{7\}\)\>\(n/28\)^\{4\}for everyn≥56n\\geq 56\.

###### Proof\.

The first assertion is immediate from Lemmas[C\.15\.3](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem3),[C\.15\.4](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem4)and[C\.15\.5](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem5)\.

For the second, letn≥56n\\geq 56and letqqbe the largest power of two with14​q≤n14q\\leq n\. Thenq≥4q\\geq 4, and maximality gives28​q\>n28q\>n, soq\>n/28q\>n/28\. By Evans’ embedding theorem, every partial Latin square of orderNNembeds in a Latin square of every order at least2​N2N\([Evans 1960](https://arxiv.org/html/2608.16977#bibo.bib5)\)\. Applying this toLqL\_\{q\}, which is a Latin square of order7​q7qand in particular a partial Latin square of order7​q7q, produces a Latin squareMMof ordern≥14​qn\\geq 14qwhose subarray on the first7​q7qrows and columns isLqL\_\{q\}\. Every subsquare ofLqL\_\{q\}is a subsquare ofMM, soMMcontains at leastq4\>\(n/28\)4q^\{4\}\>\(n/28\)^\{4\}subsquares isotopic toS7S\_\{7\}\. ∎

###### Example C\.15\.7\.

Takeq=4q=4, so that𝔽4=\{0,1,α,α2\}\\mathbb\{F\}\_\{4\}=\\\{0,1,\\alpha,\\alpha^\{2\}\\\}withα3=1\\alpha^\{3\}=1, andλ∈\{α,α2\}\\lambda\\in\\\{\\alpha,\\alpha^\{2\}\\\}\. ThenL4L\_\{4\}is a Latin square of order2828containing at least44=2564^\{4\}=256subsquares isotopic toS7S\_\{7\}\.

#### C\.15\.3The upper bound

It is convenient to view a Latin squareMMof ordernnas a tripartite incidence structure\. Its vertices are thennrows, thenncolumns and thennsymbols ofMM, and each of then2n^\{2\}cells contributes the triple consisting of its row, its column and the symbol it carries\. The defining property of a Latin square is exactly that any two vertices lying in different parts belong to a unique common triple\. Isotopisms are precisely the isomorphisms of these structures that respect the three parts\.

Observe that ifKKis a subsquare ofMMand two of the three vertices of some triple ofMMbelong toKK, then so does the third\. Indeed, if a row and a column ofKKare given then the symbol in that cell is a symbol ofKK; if a row and a symbol ofKKare given then the column in which that symbol occurs in that row ofKKis the unique such column inMM; and the third case is symmetric\.

Forx∈Px\\in PwriteRxR\_\{x\},CxC\_\{x\}andTxT\_\{x\}for the row, column and symbol vertices ofS7S\_\{7\}indexed byxx, so thatRxR\_\{x\},CyC\_\{y\}andTx∘yT\_\{x\\circ y\}form a triple for allx,y∈Px,y\\in P\.

###### Lemma C\.15\.8\.

Lete1,e2,e3e\_\{1\},e\_\{2\},e\_\{3\}be a basis ofV=𝔽23V=\\mathbb\{F\}\_\{2\}^\{3\}and pute4=e1\+e2\+e3e\_\{4\}=e\_\{1\}\+e\_\{2\}\+e\_\{3\}\. Then the four verticesRe1,Re2,Re3,Ce4R\_\{e\_\{1\}\},R\_\{e\_\{2\}\},R\_\{e\_\{3\}\},C\_\{e\_\{4\}\}generate all twenty\-one vertices ofS7S\_\{7\}under the rule that two known vertices in different parts determine the third vertex of their triple\.

###### Proof\.

Throughout we use equation[40](https://arxiv.org/html/2608.16977#A3.E40); note thatei≠e4e\_\{i\}\\neq e\_\{4\}fori∈\{1,2,3\}i\\in\\\{1,2,3\\\}, sincee4=eie\_\{4\}=e\_\{i\}would force the sum of the other two basis vectors to vanish\.

Pairing each ofRe1,Re2,Re3R\_\{e\_\{1\}\},R\_\{e\_\{2\}\},R\_\{e\_\{3\}\}withCe4C\_\{e\_\{4\}\}produces

Te2\+e3,Te1\+e3,Te1\+e2,T\_\{e\_\{2\}\+e\_\{3\}\},\\qquad T\_\{e\_\{1\}\+e\_\{3\}\},\\qquad T\_\{e\_\{1\}\+e\_\{2\}\},becauseei∘e4=ei\+e4e\_\{i\}\\circ e\_\{4\}=e\_\{i\}\+e\_\{4\}\. Next, ifx∈Px\\in Pandx∘y=wx\\circ y=wwithw≠xw\\neq x, theny≠xy\\neq xandy=x\+wy=x\+wis determined\. Applying this three times,

Re1,Te1\+e3​give​Ce3,Re1,Te1\+e2​give​Ce2,Re2,Te1\+e2​give​Ce1\.R\_\{e\_\{1\}\},T\_\{e\_\{1\}\+e\_\{3\}\}\\ \\text\{give\}\\ C\_\{e\_\{3\}\},\\qquad R\_\{e\_\{1\}\},T\_\{e\_\{1\}\+e\_\{2\}\}\\ \\text\{give\}\\ C\_\{e\_\{2\}\},\\qquad R\_\{e\_\{2\}\},T\_\{e\_\{1\}\+e\_\{2\}\}\\ \\text\{give\}\\ C\_\{e\_\{1\}\}\.The pairsRei,CeiR\_\{e\_\{i\}\},C\_\{e\_\{i\}\}then giveTeiT\_\{e\_\{i\}\}fori∈\{1,2,3\}i\\in\\\{1,2,3\\\}, sinceei∘ei=eie\_\{i\}\\circ e\_\{i\}=e\_\{i\}\. Applying the same rule as before,

Re1,Te2​give​Ce1\+e2,Re1,Te3​give​Ce1\+e3,Re2,Te3​give​Ce2\+e3\.R\_\{e\_\{1\}\},T\_\{e\_\{2\}\}\\ \\text\{give\}\\ C\_\{e\_\{1\}\+e\_\{2\}\},\\qquad R\_\{e\_\{1\}\},T\_\{e\_\{3\}\}\\ \\text\{give\}\\ C\_\{e\_\{1\}\+e\_\{3\}\},\\qquad R\_\{e\_\{2\}\},T\_\{e\_\{3\}\}\\ \\text\{give\}\\ C\_\{e\_\{2\}\+e\_\{3\}\}\.At this stage the columns

Ce1,Ce2,Ce3,Ce1\+e2,Ce1\+e3,Ce2\+e3,Ce4C\_\{e\_\{1\}\},\\ C\_\{e\_\{2\}\},\\ C\_\{e\_\{3\}\},\\ C\_\{e\_\{1\}\+e\_\{2\}\},\\ C\_\{e\_\{1\}\+e\_\{3\}\},\\ C\_\{e\_\{2\}\+e\_\{3\}\},\\ C\_\{e\_\{4\}\}are all known, and these are all seven columns ofS7S\_\{7\}\. PairingRe1R\_\{e\_\{1\}\}with each of them yields the symbolsTe1∘yT\_\{e\_\{1\}\\circ y\}fory∈Py\\in P; asyyruns overPPthe elemente1∘ye\_\{1\}\\circ yruns over all ofPP, so every symbol vertex is known\. PairingCe1C\_\{e\_\{1\}\}with each symbol vertex then yields every row vertex, since for eachz∈Pz\\in Pthere is a uniquex∈Px\\in Pwithx∘e1=zx\\circ e\_\{1\}=z\. ∎

###### Corollary C\.15\.9\.

For everynnone hasζ∗​\(n,S7\)≤n4\\zeta^\{\*\}\(n,S\_\{7\}\)\\leq n^\{4\}\.

###### Proof\.

LetMMbe a Latin square of ordernnand letKKbe a subsquare ofMMisotopic toS7S\_\{7\}, and lete1,e2,e3,e4e\_\{1\},e\_\{2\},e\_\{3\},e\_\{4\}be as in Lemma[C\.15\.8](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem8)\. Choose an isotopismθ\\thetafromS7S\_\{7\}ontoKKand record the quadruple

\(θ⁡\(Re1\),θ⁡\(Re2\),θ⁡\(Re3\),θ⁡\(Ce4\)\),\\bigl\(\\theta\(R\_\{e\_\{1\}\}\),\\ \\theta\(R\_\{e\_\{2\}\}\),\\ \\theta\(R\_\{e\_\{3\}\}\),\\ \\theta\(C\_\{e\_\{4\}\}\)\\bigr\),which consists of three rows and one column ofMM\. There are at mostn4n^\{4\}such quadruples, so it suffices to show that the quadruple determinesKK\.

Sinceθ\\thetais an isotopism ontoKK, it carries triples ofS7S\_\{7\}to triples ofMMthat lie inKK\. By the observation above, applying the rule “two vertices in different parts determine the third vertex of their triple” insideMMto vertices ofKKnever leavesKK, and it agrees with the corresponding rule inS7S\_\{7\}transported byθ\\theta\. By Lemma[C\.15\.8](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem8), starting from the recorded quadruple and iterating this rule insideMMtherefore produces exactly the twenty\-one vertices ofKK\. HenceKKis determined by the quadruple, as desired\. ∎

#### C\.15\.4Proof of the main theorem

###### Proof of Theorem[C\.15\.1](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem1)\.

Corollary[C\.15\.6](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem6)givesζ∗​\(n,S7\)\>\(n/28\)4\\zeta^\{\*\}\(n,S\_\{7\}\)\>\(n/28\)^\{4\}for everyn≥56n\\geq 56, and Corollary[C\.15\.9](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem9)givesζ∗​\(n,S7\)≤n4\\zeta^\{\*\}\(n,S\_\{7\}\)\\leq n^\{4\}for everynn\. ∎

###### Proof of Corollary[C\.15\.2](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem2)\.

Every subsquare isotopic toS7S\_\{7\}has order77, soζ⁡\(n,7\)≥ζ∗​\(n,S7\)=Ω⁡\(n4\)\\zeta\(n,7\)\\geq\\zeta^\{\*\}\(n,S\_\{7\}\)=\\Omega\(n^\{4\}\)by Theorem[C\.15\.1](https://arxiv.org/html/2608.16977#A3.SS15.Thmtheorem1)\. In the other direction, the bound of[Browning et al\. 2015](https://arxiv.org/html/2608.16977#bibo.bib4)withm=7m=7andt=2t=2givesζ⁡\(n,7\)=O⁡\(n4\)\\zeta\(n,7\)=O\(n^\{4\}\)\. ∎

## References\.

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