Under what $p$, $-1$ is a square in $\mathbf F_p$?

Question

Let $p$ be a prime number. If -1 is a square in $\mathbf F_p$, it means that there exists an element of order 4. So $p-1$ is divisible by 4 and therefore $p \equiv 1$ (mod 4). I wonder the converse is also true. Does $p \equiv 1$ (mod 4) mean the existence of an element whose square is $-1$ in $\mathbf F_p$?

Answer 1

Yes, the converse holds. Assume $p = 4k+1$ for some $k$. Every finite field has cyclic multiplicative group, so if we take $g$ to be a generator of this group, $g^{p-1} = g^{4k} = 1$. But then $g^{2k} = -1$ (since $(g^{2k})^2=1$ but $g^{2k}\neq 1$), and so $g^{k}$ is a square root of $-1$.

Answer 2

Assume $p>2$ prime.

$x^2=-1 \iff x^2+1=0 \iff \deg(x)=4$ (That means for is smaller for $n$ such taht $x^n-1$). This means

$x^2+1 | x^{p-1}-1 \iff 4|p-1$.