Show $a^p \equiv b^p \mod p^2$ I am looking for a hint on this problem:

Suppose $a,b\in\mathbb{N}$ such that $\gcd\{ab,p\}=1$ for a prime $p$. Show that if $a^p\equiv b^p \pmod p$, then we have: $$a^p \equiv b^p \pmod {p^2}.$$

I have noted that $a,b$ are necessarily coprime to $p$ already, and Fermat's little theorem ($x^p\equiv x \pmod p$), but I do not see how I should apply it in this case if at all.
Any hints are appreciated!

Fermat's Little theorem should help you show $a\equiv b(\text{mod }p)$, at which point you have $a=b+pk$ for some $k\in\mathbb{Z}$. An application of the binomial theorem from here could give you the result you seek.
