Show $m^p+n^p\equiv 0 \mod p$ implies $m^p+n^p\equiv 0 \mod p^2$ Let $p$ an odd prime. Show that $m^p+n^p\equiv 0 \pmod p$ implies $m^p+n^p\equiv 0 \pmod{p^2}$.

From little Fermat, $m^p \equiv m \pmod p$ and $n^p \equiv n \pmod p$. Hence, $p$ divides $m+n$ i.e. $m+n = pk$.
$$m^p + n^p = (pk-n)^p + n^p = p^2 M + \dbinom{p}{1} (pk) (-n)^{p-1} + (-n)^p + n^p\\ = p^2 M + p^2k (-n)^{p-1} \equiv 0 \pmod {p^2}$$
