<s>[INST] existence of solution to congruence $x^4 \equiv -4 \pmod p$ I stuck with the following question:
For which $p$ (prime numbers) there is a solution for the following congruence:
$x^4 \equiv -4 \pmod p$ 
I would greatly appreciate any help 
 [/INST] $$x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)$$
If $p$ divides both $(x^2+2x+2),(x^2-2x+2);$
$p$ must divide $(x^2+2x+2)-(x^2-2x+2)=4x$
If $p$ divides $4,p=2$
Else if $p|x\implies p$ must divide $-4\implies p=2$
So, for odd prime $p,$ it must divide exactly one of $x^2\pm2x+2=(x\pm1)^2+1$
So, we need $(x\pm1)^2\equiv-1\pmod p$
See $-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$
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