Apart from Wiles'proof (which I didn't read), I know 4 proofs of the Fermat-Wiles theorem for exponent 3, i. e. impossibilty of $x^{3} + y^{3} + z^{3} = 0$ with x, y and z pairwise coprime nonzero rational integers :

1° Euler's proof (decomposition in the 3d cyclotomic field, which is a quadratic field, leads to an infinite descent); proves that there are no nontrivial solutions in rational numbers;

2° Kummer's proof; also a decomposition in the 3d cyclotomic field and an infinite descent, but the result is stronger than Euler's one : the equation has no nontrivial solution in the 3d cyclotomic field;

3° decomposition in $\mathbb{Q}(\sqrt[3]{2})$ : squaring $x^{3} + y^{3} = - z^{3}$ and substracting $4 x^{3} y^{3}$, we find that $z^{6} - 4 x^{3} y^{3}$ is a square in $\mathbb{Z}$. In view of arithmetic properties of the field $\mathbb{Q}(\sqrt[3]{2})$, this implies that $z^{2} - xy (\sqrt[3]{2})^{2}$ is a square in $\mathbb{Z}[\sqrt[3]{2}]$. This leads to an infinite descent;

4° decomposition in $\mathbb{Q}(\sqrt[3]{3})$ : we have for example $x + y = a^{3}, y + z = 9b^{3}, x + z = c^{3}$ with a, b, and c rational integers. This implies $a^{3} + 9b^{3} + c^{3} = 6 abc$.

(See for example Legendre, Second supplément, p. 7, section 9, online : https://books.google.fr/books?id=zHMs_R2SATYC&printsec=frontcover&hl=fr#v=onepage&q&f=false )

This can be written $(2c + 3b \sqrt[3]{3} + a (\sqrt[3]{3})^{2}) (9ab - 4c^{2} + (6bc - 3a^{2})\sqrt[3]{3} + (2ac - 9b^{2})(\sqrt[3]{3})^{2}) = c^{3}$

and also

$N_{\mathbb{Q}}^{\mathbb{Q}(\sqrt[3]{3})}(2c + 3b \sqrt[3]{3} + a (\sqrt[3]{3})^{2}) = - c^{3}$

and in view of arithmetic properties of $\mathbb{Q}(\sqrt[3]{3})$, this implies that there exist rational integers r, s, t such that

$2c + 3b \sqrt[3]{3} + a (\sqrt[3]{3})^{2} = (-2 + (\sqrt[3]{3})^{2}) (r + s \sqrt[3]{3} + t (\sqrt[3]{3})^{2})^{3}$

and $N_{\mathbb{Q}}^{\mathbb{Q}(\sqrt[3]{3})}(r + s \sqrt[3]{3} + t (\sqrt[3]{3})^{2}) = -c.$

This leads to an infinite descent. (I know, this fourth proof is ugly, but we are here for the fun.) The third proof was already published, but the fourth was perhaps not (I found it myself).

My question is : do you know other proofs of the Fermat-Wiles theorem for exponent 3 ? In particular, can anybody say if the paper of Z. Xu : "Another proof of Fermat's Last Theorem for Cubes", J. Southwest Teach. Univ., Ser. B (1987), No 1, 20-22, which claims to use no complex numbers, is accurate and different from proofs 3 and 4 above ? (This paper is in Chinese...)

  • 6
    $\begingroup$ The proof by Wiles doesn't work for $p = 3$ so that doesn't count. (Wiles' result implies that the Frey elliptic curve is isogenous to an elliptic curve $E$ over $\mathbf{Q}$ with torsion subgroup $\mathbf{Z}/2\mathbf{Z} \oplus \mathbf{Z}/2p \mathbf{Z}$. This is a contradiction for $p > 3$ by Mazur's theorem, but not a contradiction when $p = 3$. (Mazur's theorem is a crucial ingredient in Wiles' proof which even today cannot be removed.) $\endgroup$ Commented Jan 28, 2016 at 22:40

2 Answers 2


The fact that there is a proof using the cubic number field $K = {\mathbb Q}(\sqrt[3]{2})$ is related to the observation that the elliptic curve $x^3 + y^3 = z^3$ has an obvious $K$-rational point, namely $(1,1,\sqrt[3]{2})$. If I recall it correctly, this is a $2$-torsion point, which allows you to use a simple $2$-descent that is, in some ways, simpler than the usual proof by $3$-descent over ${\mathbb Q}(\sqrt{-3})$.

If you want, you can do a descent on this elliptic curve over any number field over which you have an $n$-torsion point for a small number $n$.

  • 1
    $\begingroup$ Thanks. I must confess that I don't know the theory of elliptic curves. Perhaps you know the reference for a proof using the number field $K = {\mathbb Q}(\sqrt[3]{2})$ and elliptic curves ? (The 3d proof I indicated uses only the usual arithmetic methods in a principal domain and the structure of the group of units.) $\endgroup$
    – Panurge
    Commented Jan 28, 2016 at 18:51
  • 2
    $\begingroup$ I don't know whether this is published anywhere; I used this as an example when thinking about 2-descent over number fields. Elliptic curves are a great tool for a deeper understanding of the "arithmetic" proofs concerning diophantine equations - their structure tells you essentially why these proofs work. $\endgroup$ Commented Jan 28, 2016 at 18:56
  • 2
    $\begingroup$ OK. By the way, there is also a nontrivial solution in $\mathbb{Q}(\sqrt[3]{3})$ (the field used in the fourth proof) : $1^{3} + 2^{3} = (\sqrt[3]{9})^{3}.$ Could it lead (by use of elliptic curves) to a simpler proof than he fourth proof above ? $\endgroup$
    – Panurge
    Commented Jan 28, 2016 at 19:05
  • $\begingroup$ While we are at it. It seems the question is equivalent to finding that the rank of the elliptic curve 27a3 is zero over $\mathbb{Q}$ as it is easy to find its torsion points. So for instance computing the modular symbol $[0]=1/9\neq0$ tells us that the $L$-series does not vanish at $s=1$ and hence by Gross-Zagier-Kolyvagin, we know that the rank is 0. Or by a theorem of Kato, too. $\endgroup$ Commented Jan 28, 2016 at 19:06

The Wikipedia article lists fourteen proofs, not counting Euler's.

  • 1
    $\begingroup$ Are you sure? Independent does not mean distinct, just that the proofs were given independently. $\endgroup$ Commented Jan 28, 2016 at 20:00
  • 4
    $\begingroup$ This article is essentially lifted from Ribenboim's books. The proofs up to Lame have serious gaps, the proofs by Gauss and Kummer are missing. $\endgroup$ Commented Jan 28, 2016 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.