Zeilberger's potential proof of Fermat's last theorem.

Question

Doron Zeilberger suggested the following potential proof for Fermat's last theorem:

Let's define: $$W(n,a,b,c) \equiv (a^n + b^n - c^n)^2$$ I am almost sure that there exists a polynomial, discoverable by computer, with positive coefficients such that: $$W(n,a,b,c) = P\left(W(n,a-1,b,c), W(n,a,b-1,c), \ldots W(n -1,a,b,c), \ldots\right)$$ for $n>3$.

Since $W > 0$ for $n= 3$, and $abc>0$ FLT would follow.

$$$$

Could someone explain how / why exactly "FLT would follow"?

Moreover, why wouldn't one have to find a separate polynomial for each (unbound) $n$?

I particularly enjoy the seven words starting the second line of text. I think that's what Fermat said, when someone had asked him about the wonderful proof. :-) — Asaf Karagila, Apr 22 '15 at 02:40
@TimRaczkowski - The question is not whether this polynomial exists, but rather why that would prove the theorem. — Nathaniel Bubis, Apr 22 '15 at 02:46
Can someone clarify what is suppposed to go in the two $\dots$? — Dan Brumleve, Apr 22 '15 at 02:53
@DanBrumleve - I assumed more elements of the form $W(a- A, b- B, c - C, n- N)$, hence my second question regarding the infinite nature of such an expression. — Nathaniel Bubis, Apr 22 '15 at 02:57
@DanielV - "This guy" is a rather distinguished mathematician who is in fact a strong supporter of the use of computer programming in mathematics. — Nathaniel Bubis, Apr 22 '15 at 03:47
@DanielV: How on Earth is maths related to computer programming? Do you think Riemann had a computer? — Sylvain Julien, Apr 22 '15 at 08:18
@SylvainJulien - Lot's of things can't be proven by hand, e.g. the four color theorem. — Nathaniel Bubis, Aug 24 '16 at 07:43

score 6 · Accepted Answer · answered Apr 22 '15 at 02:47

6

The proof is by infinite descent. Let $n,a,b,c$ be the smallest possible solution to $W(n,a,b,c) = 0$, where $n>3$. Since he is almost sure that there exists a polynomial, discoverable by computer, with positive coefficients such that: $$W(n,a,b,c) = P\left(W(n,a-1,b,c), W(n,a,b-1,c), \ldots W(n -1,a,b,c), \ldots\right)$$ for $n>3$. This would mean that $W(n,a,b,c) > 0$ contradicting the fact that $a,b,c$ is the smallest possible solution to $W(n,a,b,c) = 0$.

answered Apr 22 '15 at 02:47

Adhvaitha

20,657

This answer does not make a lot of sense (at least to me), and it does not explain the use or construction of the described polynomial. – Nathaniel Bubis Apr 22 '15 at 03:01
@nbubis Note that none of $W(n-1,a,b,c), W(n,a-1,b,c), W(n,a,b-1,c), W(n,a,b,c-1),...$ are zeros. Hence, if we look at a polynomial expression with positive coefficients with these terms then $W(n,a,b,c)$ would just be a sum of positive terms an hence will never be $0$. – Adhvaitha Apr 22 '15 at 03:04
But how do you know that they are not zero? e.g. $W(2, 3, 4,5) =0$ – Nathaniel Bubis Apr 22 '15 at 16:05
Also, wouldn't sch a polynomial be unique for every $n$? – Nathaniel Bubis Apr 22 '15 at 16:06
@nbubis n will be larger than 2, also he assumes there is some general formula valid for all $n$ that one could prove using e.g. induction. – Count Iblis Jun 29 '16 at 23:26

Zeilberger's potential proof of Fermat's last theorem.

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