$X^p + Y^p \equiv Z^p \pmod q$ Is it possible that although fermat's last theorem implies $x^p + y^p = z^p$ has no solution in integers $x,y,z$ and $p\geq3$, for some $q>p, x^p + y^p \equiv z^p \pmod q$. Like maybe $x^p \equiv 2 \pmod q$, $y^p$ is $1 \pmod q$ and $z^p \equiv 3 \pmod q$?

A. E. Pellet, Mémoire sur la théorie algébrique des équations, Bull. Soc. Math. France 15 (1887) 61-102, showed that for every prime $p$ there is a number $q_0(p)$ such that if $q$ is prime and $q\ge q_0(p)$ then $x^p+y^p+z^p\equiv0\bmod q$ has nontrivial solutions. (Later authors found explicit values for $q_0(p)$, and slicker proofs than that of Pellet, but it does appear that Pellet was the first one to get a result). 
