Show that if if $x^3+y^3=z^3$ for some $x,y,z$ $ \in \mathbb{Z}$ then one of $x,y,z$ is divisible by $7$.
I'm stuck on this problem. I know that for any integer $n\in \mathbb{Z}$, $[n^3 \pmod{7}]$ $\in\{0,1,6\}$. So I think of breaking it into cases:
Case 1) $n^3 \equiv 0 \pmod{7}$
Case 2) $n^3 \equiv 1 \pmod{7}$
Case 3) $n^3 \equiv 6 \pmod{7}$
I'm not sure how to proceed with any of the above cases.
Edit:
In order to proceed, one needs to know the following proposition:
Let a,b be integers. Let p be a prime. If p|ab then p|a or p|b. If b=a then p|ab implies p|a.
Therefore for the first case, if $7|z^3$ then $7|z$ as 7 is prime. The other two cases follow similarly
The proof of the proposition is the following:
Assume $p|ab$ and p does not divide a. Then if I assume $gcd(p,a)=d>1$ Then it follows that d |a and d|p. As p is prime, d =1 or p. If d =p then p|a, which is a contradiction so $gcd(p,b)=1$ and so $p|b$
You've done all the work... simply observe that $$x^3+y^3=z^3\implies x^3+y^3\equiv z^3\pmod 3$$ Thus, the sum of two residues has to be another residue. This only works for $0+1=1, 0+6=6, 0+0=0$ and $6+1\equiv 0$. Notice that every possibility contains a number $\equiv 0\bmod 7$.
Sidenote $\ $ This is a nice exercise in order to practice modular arithmetic, however Fermat's Last Theorem tells us that there is no solution for $z\not= 0\implies 7\mid z$