Given an odd prime $p$, I want to prove that $x^4 \equiv -4$ (mod $p$) is solvable if and only if $p \equiv 1$ (mod $4$). More specifically, I want to prove this using a hint, which says to first factorize $x^4+4$.
I proved that whenever the congruence is solvable, we have $p \equiv 1$ (mod $4$), however, I didn't manage to do that using the factorization but rather using Legendre symbols $(\cdot / \cdot)$. But I would like to prove the converse using the factorization.
This is the way I have tried to solve it but I'm unsure of whether it is correct or not:
So first we factorize $x^4+4$ as $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) = ((x+1)^2 + 1)((x-1)^2 + 1), $$ which can be written as $$ = (a^2 + 1)(b^2 + 1). $$
Now we want to show that there exists $a, b$ such that $(a^2 + 1)(b^2 + 1) \equiv 0$ (mod $p$). So in modulo $p$, we have that $$a^2 + 1 \equiv 0 \iff a^2 \equiv -1 $$ has a solution iff $(-1/p) = 1$. We have that $(-1/p) = (-1)^{(p-1)/2} = 1$ since $(p-1)/2$ is even whenever $p \equiv 1$ (mod $4$). The same goes for $b^2 + 1$, and we are done.
Was this correct? If not, I could use some advice on how to solve this.
The factorization you have found shows that $x^4 \equiv -4$ has a solution iff $y^2 \equiv -1$ has a solution: $$ x^4 + 4 = ((x+1)^2 + 1)((x-1)^2 + 1) $$
Now $y^2 \equiv -1$ has a solution iff there is an element of order $4$ mod $p$. This happens iff $4$ divides $p-1$ because the group of units mod $p$ is cyclic.
So you don't need Legendre symbols.