Variation of Fermat's little theorem of the form $a^{p-1} \equiv 1 \mod p^2$.

Question

If a natural number $a$ coprime to a prime $p$ satisfies $$a^{p-1} \equiv 1 \mod p^2$$ then, what can we say about $a$ and $p$ ?

Answer 1

There are primitive roots modulo $p^2$, so let $r$ be such a root, and let $a\equiv r^k$. Then

$$r^{k(p-1)}\equiv 1 \mod p^2$$

Which is equivalent to $k(p-1)$ being divisible by $p(p-1)$ (the order of $r$), which in turn is equivalent to $k$ being a multiple of $p$. Thus $a$ is congruent modulo $p^2$ to one of the numbers $1, r^p, r^{2p}, r^{3p} ...$