Integer solutions of $x^3+y^3=z^3$ using methods of Algebraic Number Theory

Question

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

Answer 1

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.