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

## Solution 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.