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 $m^p+n^p\equiv 0 \pmod{p^2}$.
 A: From little Fermat, $m^p \equiv m \pmod p$ and $n^p \equiv n \pmod p$. Hence, $p$ divides $m+n$ i.e. $m+n = pk$.
$$m^p + n^p = (pk-n)^p + n^p = p^2 M + \dbinom{p}{1} (pk) (-n)^{p-1} + (-n)^p + n^p\\ = p^2 M + p^2k (-n)^{p-1} \equiv 0 \pmod {p^2}$$
A: I will prove first the following:
Suppose that $a$ and $b$ are integers and $p$ is prime, such that $a,b\nmid p$. 
If $a^p\equiv b^p\pmod{p}$, then $a\equiv b\pmod{p} $
Proof:
From Fermat's theorem, we know $a^p\equiv a\pmod{p}$ and $b^p\equiv b\pmod{p}$, therefor, if $a^p\equiv b^p\pmod{p}$, then $a\equiv b\pmod{p}$
We ready to prove our main problem,
From our lemma above, we conclude that, $a^k\equiv b^k\pmod{p}$.
Now,
$$a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+...+b^{p-1})=(a-b)\sum^{p}_{k=1}a^{p-k}b^{k-1}$$
but,
$$b^{k-1}\equiv a^{k-1}\pmod{p}$$
Therefor,
$$\sum^{p}_{k=1}a^{p-k}b^{k-1}\equiv\sum^{p}_{k=1} a^{p-k}a^{k-1}=\sum^{p}_{k=1}a^{p-1}=pa^{p-1}\equiv 0\pmod{p}$$
So, $p\mid\sum^{p}_{k=1}a^{p-k}b^{k-1}$, and $p\mid (a-b)$, we get $p^2\mid (a^p-b^p)$.
