From here, I understand how to find the number of $n$th roots of unity.
I would like to know how to find the number of solutions to this equation: $$ x^n = -1 $$ over the finite field $F_q$ with $q$ elements ($n$ is fixed).
My only thoughts on this is that it looks like finding $n$th roots of unity like in the other question. I can see that if $n$ is odd, then $-1$ is a solution. I can see that if $n$ is equal to $q-1$, then there are no solutions since the order of the group $F_q^X$ is $q-1$.
The solutions to $x^n=-1$ are precisely the primitive $(2n)$th roots of unity.
For any $m$ and any field $F$, if $F$ contains a primitive $m$th root of unity then the exact number of them it contains is $\varphi(m)$. The field $\Bbb F_q$ contains a primitive $m$th root of unity iff $m\mid(q-1)$.