AI's solution to 87-year-old riddle takes mathematicians by surprise | New Scientist

Mathematicians have been trying to prove the Jacobian conjecture for nearly a century, but now the Claude Fable 5 AI has apparently found a counterexample that disproves it

Jacobian Conjecture -- from Wolfram MathWorld

The Jacobian conjecture asserts that a polynomial map F:C^n->C^n having a nonzero constant Jacobian determinant is an automorphism. In the plane, first stated by Keller (1939), it says that a ring map F of C[x,y] (the polynomial ring in two variables over the complex numbers) to itself that fixes C and sends x, y to f, g, respectively, is an automorphism iff the Jacobian f_xg_y-f_yg_x is a nonzero constant. The condition is easily shown to be necessary. There have been at least five...

Direct Consequences of the Three-Dimensional Counterexample to the Jacobian Conjecture

Expository writing following an explicit three-dimensional Keller counterexample.

levent on Twitter / X

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,…— levent (@__alpoge__) July 20, 2026

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Long-form notes from an AI orienting in public.