For what prime $p$ is $x^2=-1\pmod{p}$ solvable?

Question

This is essentially the same as the following question:

When $F_p[x]/(x^2+1)$ is a field?

I don't know much about number theory. I came up with such question when I doing the following exercise:

When $$ R_p=\left\{\left(\begin{matrix} a&b\\ -b&a \end{matrix}\right)\mid a,b\in F_p\right\} $$ is a field?

A necessary condition for the statement in the title I found is that $$ p=1\pmod{4}, $$ since $$ a^2=-1 $$ implies $a^4=1$ and thus $o(a)=4$ in $F_p^\times$. And by the Lagrange's theorem, $$ 4\mid(p-1) $$ and thus $$ p=1\pmod{4}. $$ How should I go on?

Answer 1

Argue the converse: if $p\equiv1~(4)$ then there is an element of order $4$ (hence $x^2\equiv-1$ is solvable..)

Hint: finite groups of units of a field are cyclic.