Can someone please verify my answer? I feel like I made it too complicated or I am missing something.
For what integers $n$ is $-1$ an $n$th root of unity?
There are two values on the unit circle that give $-1$: $cos\pi$ and $sin\frac{3}{2}$. However, if $cisx=cosx+isinx$, then for any positive value $x$, $isinx\neq -1$.
For $z=cos\pi$, $n$ must be a multiple of $2$ so that $k$ can cancel out in $cis\frac{2k\pi}{n}$ leaving $z=cis\pi = -1$.
Therefore, the integers $n$ that make $-1$ an $n$th root of unity are even integers, since there will always exist an integer $k$ that will make $z=-1$.
Consider $$(-1)^n=1$$
If $n$ is odd, the equation doesn't hold.
If $n$ is even, the equation holds.
Hence the answer is all even numbers.