Fermat's 'proof' of his Last Theorem

The definition of a unique factorisation domain came up in my rings lecture about a week ago, and my lecturer mentioned that Fermat's 'proof' of his Last Theorem probably relied on the (false) assumption that all subrings of $\mathbb{C}$ are unique factorisation domains. Does anyone know what this 'proof' would have looked like?

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Essentially, $a^p+b^p$ factors easily in ring of algebraic integers in $\mathbb Q[\zeta_{p}]$ where $\zeta_p$ is a (non-trivial) $p$th root of unity. I don't really have much better memory than that. – Thomas Andrews Mar 8 '13 at 14:46
I seem to remember Kummer "proved" the theorem this way, and that when somebody pointed out there wasn't unique factorization he was led to ideals. – vonbrand Mar 8 '13 at 15:08
@vonbrand: I think it was Gabriel Lame and somebody was Joseph Liouville. – P.. Mar 8 '13 at 15:11
I think we need to issue preemptive protection to every question about crank-magnet topics... (And this one, being interesting, is a magnet for cranks who think they proved FLT using very short and elementary arguments.) – Asaf Karagila Dec 31 '13 at 10:39

4 Answers

Unless new historical documents are discovered, we can never know for certain what Fermat had in mind when he made his famous FLT remark. Most number-theorists probably share the same opinion as Weil (quoted below), that he made an elementary mistake, e.g. thinking that results for smaller exponents would generalize. Nowadays it is known that Fermat's Last Theorem cannot be proved by certain types of descent proofs similar to the classical simple proofs known for small exponents (search for "Tate Shafarevich obstruction").

Below is Andre Weil's opinion on this matter, from his historical treatise Number Theory, p.104.

As we have observed in Chap. I, S.X, the most significant problems in Diophantus are concerned with curves of genus 0 or 1. With Fermat this turns into an almost exclusive concentration on such curves. Only on one ill-fated occasion did Fermat ever mention a curve of higher genus, and there can hardly remain any doubt that this was due to some misapprehension on his part, even though, by a curious twist of fate, his reputation in the eyes of the ignorant came to rest chiefly upon it. By this we refer of course to the incautious words "et generaliter nullam in infinitum potestatem" in his statement of "Fermat's last theorem" as it came to be vulgarly called: "No cube can be split into two cubes, nor any biquadrate into two biquadrates, nor generally any power beyond the second into two of the same kind" is what he wrote into the margin of an early section of his Diophantus (Fe.I.291, Obs.II), adding that he had discovered a truly remarkable proof for this "which this margin is too narrow to hold". How could he have guessed that he was writing for eternity? We know his proof for biquadrates (cf. above, S.X); he may well have constructed a proof for cubes, similar to the one which Euler discovered in 1753 (cf. infra, S.XVI); he frequently repeated those two statements (e.g. Fe.II.65,376,433), but never the more general one. For a brief moment perhaps, and perhaps in his younger days (cf. above, S.III), he must have deluded himself into thinking that he had the principle of a general proof; what he had in mind on that day can never be known.

Remark  It is a common inaccurate hunch that shortly-stated theorems should have short proofs. However, this is easily disproved. For any formal system that has nontrivial power (e.g. Peano arithmetic) there is no recursive algorithm that decides theoremhood. Suppose that there existed a recursive bound $\rm\ L(n)\$ on the length of proofs of a statement of length $\rm\:n.\:$ Then we could test for theoremhood simply be enumerating and testing all possible proofs of length $\rm\le L(n).\,$ Therefore there can be no such recursive bound on the length of proofs. Therefore there exist short-stated theorems with arbitrarily long proofs -- proofs so long that they are probably not amenable to human comprehension (this was observed by Goedel in his 1936 paper on speedup theorems).

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your remark is the coolest thing I have heard in a long, long time. Thank you... but how do we know there there is no bound? How do we know that we just haven't checked all of the proofs yet? – user27182 Mar 8 '13 at 22:10
@user27182 I think that MathGems is trying to say that, if such a bound exists, then there is an algorithm which can determine the theoremhood of statements in a non-trivial formal system. But such an algorithm cannot exist, for reasons that I think are related to the theorem of Gödel. Regards. – awllower May 2 '13 at 3:35
@awllower. Thank you, that cleared things up – user27182 May 2 '13 at 11:52
@user27182 NP. Glad to help! – awllower May 2 '13 at 12:57
Weil's book mentioned here is most probably Number Theory: An Approach Through History From Hammurapi to Legendre (1984). – lhf May 14 '13 at 12:19

It's hard to tell what Fermat had in mind.

Maybe the source for your rumor is the following: About 1850, the two mathematicians Gabriel Lamé and Augustin Louis Cauchy had announced a proof for Fermat's Last Theorem. However, the mathematician Ernst Eduard Kummer realized and pointed out that both proofs wronly assumed the UFD property.

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Read the first chapter of Washington's book on cyclotomic fields for a proof of Fermat's last theorem assuming a certain subring of $\mathbb{C}$ is a UFD (this subring is a UFD for certain values of the exponent $p$ in $x^p + y^p = z^p$, so the result is actually a valid proof for those values of $p$). This was indeed Lame's approach as was mentioned in a comment.

However, my understanding was that people don't generally think this was what Fermat had in mind. Maybe I am wrong.

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It is anybody's guess. Many of Fermat's results he just announced without proof, most of them were much later proved by Euler. And the whole area of abstract algebra (rings and such) were far in the future then.

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Yes, the areas of mathematics were ahead of his time, but he may well have still assumed that you can factorise uniquely in $\mathbb{C}$ and some of it's subsets. I'm interested in how a proof assuming that 'fact' might look. – user27182 Mar 8 '13 at 14:59
It was Euler, some century after Fermat, who made complex numbers into full mathematical citizens. So I doubt Fermat would have used them much. – vonbrand Mar 8 '13 at 15:18

protected by Asaf KaragilaDec 31 '13 at 10:38

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