Number of Linear Factors of $x^{\frac{p-1}{2}}-1$ in $\mathbb{Z}_p$

Question

In my number theory class, I was given the following problem: How many linear factors does the polynomial $x^{\frac{p-1}{2}}-1$ have over $\mathbb{Z}_p$, where $p$ is an odd prime?

I'm not sure how to start this. Going just from looks, I might try to use Euler's criterion and then results about the amount of squares mod p, but I'm not really sure if that's applicable here because $x$ is indeterminate. What can I do to solve this?

Answer 1

$x^{\frac{p-1}{2}}-1 \equiv 0$(mod p)

$\iff (\frac{x}{p})=1$ which has exactly $\frac{p-1}{2}$ solutions[matches with degree of orginal polynomial] and are

$1^2,2^2,...,(\frac{p-1}{2})^2$ mod p

so $x^{\frac{p-1}{2}}-1=(x-1^2)...(x-(\frac{p-1}{2})^2)$ in $Z_p$

hence $\frac{p-1}{2}$ linear factors