I am working on an exercise: If $x^5 +y^5 = z^5$ and $xyz \neq 0$, then $5$ divides at least one of $x$, $y$ or $z$.
I am thinking that the answer involves an application of Kummer's theorem, but I'm not sure if I am using it correctly. The theorem states that if $p$ is an odd regular prime, then $x^p + y^p = z^p$ has no integer solutions such that $xyz \neq 0$ and $p$ does not divide $xyz$.
My argument in the case $p=5$ is: If there are integers $x,y,z$ such that $x^5 +y^5 = z^5$ and $xyz \neq 0$, then since 5 is an odd regular prime, using Kummer's theorem it must be the case that 5 divides $xyz$. Then since 5 is a prime, 5 must divide at least one of $x, y, z$. Is this a correct argument?
The exercise then asks if the same argument works to prove the statement for $p=7$ (another odd regular prime); would the same argument work? The reason I am suspicious is I am sure the exercise would not ask you to do this unless something more involved was going on, but I can't see where the argument would fall down if 5 is replaced by 7.
Is there a more elementary way to do this?
Mod 25, the non-zero fifth powers are $1, 7, 18, 24$ (checked with the software pari). Now you see that the sum of any pair of these classes is never a fifth power mod 25.