Prove that for every prime number $|\{1\leq x \leq p^2 : p^2 | (x^{p-1}-1) \}|=p-1$

Question

Show that for every prime number $p$:

$$|\{1\leq x \leq p^2 : p^2 | (x^{p-1}-1) \}|=p-1$$

I'm not really sure how to approach this, any ideas?

Answer 1

Let $g$ be a generator of the cyclic group $(\mathbb Z/ p^2\mathbb Z)^\times$, whose order is $\varphi(p^2)=p(p-1)$, and define $\xi=g^{p-1}$. Then any invertible $1\leq x\leq p^2-1$ can be put in the form $x\equiv g^k\pmod{p^2}$ for some $1\leq k\leq p(p-1)$.

Now, if $x$ is such that $x^{p-1}\equiv1\pmod{p^2}$, then $x$ is invertible so $x\equiv g^k\pmod{p^2}$ and $\xi^{k}\equiv1\pmod{p^2}$, and since the order of $\xi$ is $p$, it follows that $k$ is multiple of $p$. There are exactly $p-1$ multiples of $p$ between $1$ and $p(p-1)$, hence the result.