Let $p$ be a prime number.
Let $S_p=\{1≤x≤p^2:p^2│x^{p-1}-1\}$.
Prove that $|S_p|=p-1$.
I managed to prove that $|S_p|\geq p-1$.
I took $g$, a primitive root modulo $p^2$, and then proved that $\langle g^p\rangle\subseteq S_p$ ...
Now, in order to finish, it is needed to show that $|S_p|\leq p-1$, for this I wanted to say something about number roots of the polynomial $x^{p-1}-1$ over $\mathbb{F}_p$, but it doesn't help since we are dealing with numbers in $\mathbb{Z}/p^2\mathbb{Z}$.
In a finite cyclic group of order $n$ and for a divisor $d \mid n$, there are exactly $d$ elements whose order divides $d$ (they form the unique cyclic subgroup of order $d$).
If we take a generator $g$, these statements follow from the fact that the order of $g^k$ is $n/\gcd(n,k)$: for $g^k$ to have order dividing $d$, we need $\frac nd\mid k$.