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nxtfariyesterday at 7:35 PM2 repliesview on HN

Everyone using Claude Fable to verify this proof is so funny. If you read the definition of the Jacobian Conjecture and (I am not exaggerating this) have passed a college Calc 3 class, you can just verify the proof yourself in 30 seconds. The problem was very hard to solve but the counterexample is very easy to verify!

--- edit, adding an explanation:

To summarize it, the conjecture says if you have any multi-variable polynomial function that maps an input to an output in the same dimensional space (take for example: F = (x+2, y+2), which maps 2D space into another 2D space), AND that function has a constant-valued non-zero Jacobian determinant, THEN the conjecture is that the polynomial has an inverse, meaning basically you can find a polynomial that turns the output space back into the input space.

Fable provided the example polynomial (which was very hard to do) and the coordinates which if you plug into it, results in two points being mapped to the same output point. This means that the polynomial can't be inverted, because if you have that output point, how do you know which input point it came from?

You can just plug in the two coordinates it gave into the equation and verify that you get the same output point from both. That's the contradiction of the conjecture and it takes 30 seconds.

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Something something outsourcing of thinking something.


Replies

mw888today at 9:24 AM

Something something pedantry.

mswphdyesterday at 10:43 PM

verification also requires computing the jacobian. an undergraduate-level exercise, but harder than verifying p(pt1) == p(pt2).