# First correct proof of FLT for exponent 3?

It is well known that Euler gave the first proof of FLT ($x^n + y^n = z^n$ has no nontrivial integral solutions for $n > 2$) for exponent $n=3$, but that his proof had gaps (which are not as easily closed as Weil seems to suggest in his excellent Number Theory - An Approach through History). Later proofs by Legendre and Kausler had the same gap, and in fact I do not know any correct proof published before Kummer's proof for all regular primes. Gauss had a beautiful proof, with the 3-isogeny clearly visible, which was published posthumously by Dedekind, and of course Dirichlet could have given a correct proof (he gave one for $n = 5$ in his very first article but apparently did not dare to provoke Legendre by suggesting his proof in Theorie des Nombres was incomplete) but did not.

The problem in the early proofs is this: if $p^2 + 3q^2 = z^3$, one has to show that $p$ and $q$ can be read off from $p + q \sqrt{-3} = (a + b\sqrt{-3})^3$. The standard proofs use unique factorization in ${\mathbb Z}[\zeta_3]$ or the equivalent fact that there is one class of binary quadratic forms with discriminant $-3$; Weil uses a (sophisticated, but elementary) counting argument.

I wonder whether there is any correct proof for the cubic Fermat equation before Kummer's proof for all regular prime exponents (1847-1850)?

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For the benefit of those of us who aren't familiar with the argument, what exactly is the "same gap" that is not "easily closed"? –  Timothy Chow Dec 14 '10 at 18:04
@Timothy: better now? –  Franz Lemmermeyer Dec 14 '10 at 18:10
@Franz: Yes, excellent...thanks. @Peter: Fermat produced an argument only for $n=4$. For other values of $n$, the question of whether Fermat had a proof is a matter of faith. –  Timothy Chow Dec 15 '10 at 2:49
@Peter: Yes, I do. In one of his last letters (to Carcavi), Fermat gives four theorems which he says can be proved by descent; he does admit, however, (depending on your interpretation) that there remain a few "details" to be supplied. One of these "theorems" is the primality of the Fermat numbers, another one the diophantine equations y^2 = x^3 - 2 and y^2 = x^3-4, and one is FLT for n=3. –  Franz Lemmermeyer Dec 15 '10 at 10:36
Weil says Fermat did have a proof for n = 3 in his book; strangely enough, this is preceded by a remark that Fermat's methods are based on 2-isogenies on elliptic curves. As he knew very well, the usual descent proof of FLT for n=3 is based on a 3-isogeny, which cannot be found in Diophantus or in any of Fermat's work on diophantine problems. –  Franz Lemmermeyer Dec 15 '10 at 10:37
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I had a look at Paulo Ribenboim's "13 lectures on Fermat's last Theorem" (Springer Verlag, 1979). In section 3 of Chapter III, he discusses (with full bibliographical details) the controversy around Euler's proof, and then provides a proof, using purely elementary number theory, which he attributes to Euler.

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Harold Edwards, in "Fermat's Last Theorem" (Spring-Verlag, 1977) also notes that Euler's original proof via descent fails for the above reasons, and gives a proof that cubes must factor into cubes by using methods known to Euler (arguments Euler used to show that primes congruent to 1 mod 3 can be written as $a^2 + 3b^2$), but also explicitly states that though Euler could have made this argument, he did not. –  Zack Wolske Dec 18 '11 at 1:36

Recently I have found two simple proofs of FLT for n=3. They use the ring $\mathbb{Z}[\sqrt[3]{2}]$. If $x^3+y^3+z^3=0$ then

(1) $x^6-4 (yz)^3=a^2$,

where $a=y^3-z^3$.

The idea is to deduce from (1) that the algebraic number $x^2-\sqrt[3]{4} yz$ is a square, more precisely:

(2) $x^2-\sqrt[3]{4} yz=u \beta^2$,

where $\beta \in \mathbb{Z}[\sqrt[3]{2}]$, and $u$ is a unit.

Once this is established, then it is not difficult to get a contradiction via infinite descent. It is similar to Euler's proof, but another number field is used.

The (2) follows from the unique factorization in the Euclidean ring $\mathbb{Z}[\sqrt[3]{2}]$

My second proof deduces (2) without using the unique factorization.

The idea is simple: find a number $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$, such that $\alpha^2$ is divided by the $x^2-\sqrt[3]{4} yz$ and $\alpha^2/(x^2-\sqrt[3]{4} yz)$ has a small norm. It turns out that the last number may have small norm indeed: less than 9. Then the norm of it can be either 1 or 4, and (2) easily follows. The method can be used to prove FLT for other concrete $n$ as well.

It is easy to get such $\alpha$ because $\alpha$ belongs to $I$, where $I$ is the ideal of norm $a$, generated by $a$ and $x^2-\sqrt[3]{4} yz$. We know from the theory of algebraic numbers, that there is an $\alpha \in I$, such that $N(\alpha)<m N(I)$, where $m$ is easily calculated constant, that turns to be less than 3. Then $N(\alpha^2/(x^2-\sqrt[3]{4} yz))<(3 a)^2/a^2=9$.

This proof is shorter and easier than that of Euler and Kummer.

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Unless you published this proof before 1847, it is not an answer to the question. –  Gerry Myerson Oct 10 '13 at 22:27
This is not an answer, but I think, it is a valuable comment. The proof is much simplier than that of Kummer, and Franz Lemmermeyer may like to know about it's existence. I hope he will include it in his next book. –  Felix Shmidel Oct 11 '13 at 10:29
Besides that this does not answer the question, did anyone understand how this proof is supposed to work? It seems implicitly to claim that descent shows that there are no non-trivial solutions of $x^2-\sqrt[3]{4}w=u\beta^2$ with $x,w\in\mathbb Z$ and $\beta\in\mathbb{Z}[\sqrt[3]{4}]$. –  Peter Mueller Oct 12 '13 at 17:17
I have explained it in math.stackexchange.com/questions/237556/… –  Felix Shmidel Oct 15 '13 at 15:00