I'm asked to prove that the famous equation $$x^3+y^3=z^3$$ has no integer (non-trivial) solutions, i.e. FLT for $n=3$

I'm aware that on this website there are solutions using methods of Number Theory (the infinite descendant proof for example, or well, Wiles' Theorem) But my lecturer told us it can be done by methods of Algebraic Number Theory, i.e. using certain number fields and properties of them.

As an hint, he told us to consider the extension $$\mathbb{Q}(\sqrt{3})$$ and using the result that characterises ramified or non-ramified primes in quadratic fields.

Now I'd be lying saying that I have some idea on how to attack this problem.

I thought that something helpful would come using some analogue of the reasoning of finding roots of $x^2+y^2=z^2$, i.e. reasoning with the norm of a specific quadratic extension, but the norm gives a quadratic relation in this case, and not a cubic one. On the other hand I thought, ok let's consider cubic extension, but for $$\mathbb{Q}(\sqrt[3]{d})$$ the norm of $a+b\sqrt[3]{d}+c\sqrt[3]{d^2} $ is $$a^3+b^3d+c^3d^2-3abc$$ and so I have a kind of cubic relation, BUT I don't know how to get rid of the $abc$ term.

I'm aware that this is not a big effort, but this is what I'm able to think as a strategy to attack this problem.

Instead of full solutions I'd prefer suggestion and reasonings, otherwise I'll never learn how to proceed with these kind of problems :)

Thanks in advance

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    $\begingroup$ Personally I think it should be done in $\mathbb Q(\sqrt{-3})$, not $\mathbb Q(\sqrt3)$. Factor $a^3+b^3=(a+b)(a+b\zeta_3)(a+b\zeta_3^2)$. $\endgroup$ – Yai0Phah Nov 29 '15 at 11:52
  • $\begingroup$ @FrankScience it can be a typo from the lecturer for sure. I'll try working on that! $\endgroup$ – Luigi M Nov 29 '15 at 12:08
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    $\begingroup$ @PITTALUGA $\zeta_3 = \frac{1 + \sqrt{-3}}{2}$. $\endgroup$ – Brandon Carter Nov 29 '15 at 15:47
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    $\begingroup$ No, there is no easy way, because it is false: $1-\zeta_3$ may be a common factor, because $1-\zeta_3\mid 3\mid (a+b)$ for one case in question. $\endgroup$ – Dietrich Burde Nov 29 '15 at 16:52
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    $\begingroup$ You mean $\mathbb{Z}[\zeta_3]$ is a UFD. $\mathbb{Q}(\zeta_3)$ is a field. $\endgroup$ – Qiaochu Yuan Nov 29 '15 at 19:12

Fermat's equation for cubes is a common introduction to lecture notes on algebraic number theory, because it motivates to study rings of integers in a number field, and partly has been developed even for such Diophantine problems, e.g., Kummer's work concerning generalizing factorization to ideals. For the equation $x^3+y^3=z^3$ the number field is $\mathbb{Q}(\zeta)$ with a third primitive root of unity $\zeta=e^{2\pi i/3}$. Its ring of integers is given by $\mathbb{Z}[\zeta]$, which is indeed a factorial ring (because it is Euclidean). Its units are given by $\pm 1,\pm \zeta,\pm\zeta^{-1}$. This is crucial to prove Euler's result:

Theorem(Euler $1770$): The equation $x^3+y^3=z^3$ has no non-trivial integer solutions.

The proof uses divisibility properties of the ring $\mathbb{Z}[\zeta]$, starting from the equation $$ z^3=x^3+y^3=(x+y)(x+\zeta y)(x+\zeta^2y). $$ The first case is $p=3\nmid xyz$. We may suppose that $x,y,z$ are coprime. We have $z^3\equiv \pm 1\bmod 9$ and $x^3+y^3\equiv -2,0,2 \bmod 9$, so that $x^3+y^3\neq z^3$, a contradiction. Hence we suppose that $3\mid xyz$, i.e., say, $3\mid z$ and $3\nmid xy$. Now we reformulate the equation as $$ x^3+y^3=(3^mz)^3, $$ with $x,y,z$ pairwise coprime and $3\nmid xyz$, where we have solved the case $m=0$. The idea is now to use descent, i.e., to reduce it to the case $m=0$. The above equation becomes $$ (3^mz)^3=(x+y)(x+\zeta y)(x+\zeta^2y), $$ where the three factors are not coprime, because $1-\zeta$ is a common factor, because of $3=(1-\zeta)(1-\zeta^2)$, so that $(1-\zeta)\mid 3\mid (x+y)$. However, since $\mathbb{Z}[\zeta]$ is factorial, all factors are cubes, i.e. , $x+y=3^{3m-1}c^3$ with some $c\in \mathbb{Z}$, and so on. This finishes the proof, after some computations in this ring.

Unfortunately, this idea does not work for $x^p+y^p=z^p$ for primes $p$, except for $p\le 19$, because otherwise the ring of integers $\mathbb{Z}[\zeta_p]$ is no longer factorial.

  • $\begingroup$ Thanks for the answer! It will take some time trying to understand it, but I want to say thanks to you as soon as possible. I'll ask you here if there is something unclear to me (my knowledge in this field is very limited sadly) $\endgroup$ – Luigi M Nov 29 '15 at 20:23
  • $\begingroup$ Dear prof. Burde, the proof is clear up the the passage where you claim that the three factors are not coprimes, BUT still the three factors are cubes, and then $x+y=3^{3m-1}c^3$ which I don't get why it's a cube. Could you please clarify this passage for me? $\endgroup$ – Luigi M Nov 29 '15 at 21:07
  • $\begingroup$ Yes, I can clarify (although the answer is $0$-useful so far). The gcd is exactly $1-\zeta$, so that $z^3$, and hence all factors are cubes, up to units and powers of $(1-\zeta)$. This is only true because $\mathbb{Z}[\zeta]$is a UFD. $\endgroup$ – Dietrich Burde Nov 30 '15 at 9:28
  • $\begingroup$ Thanks, that was helpful. (Sorry for not upvoting and/or accepting the answer, but I usually do that after making sure I understand everything, if that was the concern). I have a last doubt before being done: if I understand correctly the mechanics of your proof, we want to show that if we have $x^3+y^3=(3^mz)^3$ then you can build a solution for $x^3+y^3=(3^nz)^3$ with $n<m$ and so find a absurd with the first case. So the fact that (up to $1-\xi$ and units, the factors are cubes, how help you to prove the claim? I hope the doubt is clear, o/w I'll try to reformulate $\endgroup$ – Luigi M Nov 30 '15 at 11:41
  • $\begingroup$ Yes, your doubt is clear. The missing step is not difficult, but really a bit long. I have more details in the introduction of my lecture notes on algebraic number theory. For the English version, see the book of K. Ireland and M. Rosen. $\endgroup$ – Dietrich Burde Nov 30 '15 at 12:17

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