Claude Fable produced a counterexample to the Jacobian Conjecture

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Summary

A tweet claims that the AI model Claude Fable produced a counterexample to the Jacobian Conjecture, providing an explicit polynomial map that is not injective despite having constant nonzero Jacobian determinant.

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Cached at: 07/20/26, 09:47 AM

# levent (@__alpoge__) Source: [https://xcancel.com/__alpoge__/status/2079028340955197566](https://xcancel.com/__alpoge__/status/2079028340955197566) [![](https://pbs.twimg.com/profile_images/1824948681877155840/VqVHY3gF_bigger.jpg)](https://xcancel.com/__alpoge__) [levent](https://xcancel.com/__alpoge__) [@\_\_alpoge\_\_](https://xcancel.com/__alpoge__) [7h](https://xcancel.com/__alpoge__/status/2079028340955197566#m) hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final \(\(1\+xy\)^3 z \+ y^2 \(1\+xy\) \(4\+3xy\), y \+ 3 x \(1\+xy\)^2 z \+ 3 x y^2 \(4\+3xy\), 2 x \- 3 x^2 y \- x^3 z\): \\C^3\\to \\C^3, has jacobian determinant \-2, and sends \(0, 0, \-1/4\), \(1, \-3/2, 13/2\), and \(\-1, 3/2, 13/2\) to \(\-1/4, 0, 0\) Jul 20, 2026 · 2:19 AM UTC 514 968 6,805 2,416,605

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