How many solutions do $x^{p-1} \equiv 1 \pmod p$ and $x^{p-1} \equiv 2 \pmod p$ have?

Question

This is my first post so I apologize for any kind of error.

I'm preparing a magistral degree exam in number theory, and I'm performing some exercise. I'm asking here this question: how can I prove how many solutions there are for $x^{p-1} \equiv 1\pmod p$ and $x^{p-1} \equiv 2 \pmod p$?

Edit: $p$ is an odd prime.

Answer 1

Do you know Fermat‘s little theorem?

Consider the multiplicative group $\Bbb Z^\times_p$.