$x^4\equiv-1 \pmod p$ is solvable if and only if $p \equiv 1 \pmod 8$ or $p=2$

Question

Prove that $x^4\equiv-1 \pmod p$ is solvable if and only if $p\equiv1\pmod 8$ or $p=2$.

I was thinking of using cases like for $8m+1$, $8m+2$... in order to prove it but I didn't get anything satisfying.

Answer 1

The multiplicative group $\mathbb Z/p\mathbb Z$ is cyclic of order $p-1$. If a solution of $x^4\equiv -1\pmod p$ is also a solution of $x^8\equiv 1\pmod p$, hence an element of order dividing $8$. If $p\ne 2$, we also know that it is not an element of order dividing $4$ because $x^4\not\equiv 1\pmod p$, hence an element of order exactly $8$. Therefore $8$ must be a divisor of the group order $p-1$.

Answer 2

$\!\!\bmod p\!:\ x^4\equiv \color{}{\overbrace{-1\color{#c00}{\not\equiv 1}}^{\large p\ \neq\ 2}}\Rightarrow $ $\ x^8\equiv 1\Rightarrow $ $\,x\,$ has order $\,\color{#c00}8\,$ by Order Test, so $\,\overbrace{2^{p-1}\equiv 1}^{\rm Fermat}\Rightarrow \color{#c00}8\mid p\!-\!1$