Prove that $x^4\:\equiv\:-4\bmod\:p\:\iff\:p\equiv1\bmod\:4$

Question

I am trying to prove that

$x^4\:\equiv\:-4\:(\bmod\:p)\:\text{has a solution}\:\iff\:p\equiv1\:(\bmod\:4)$

However I do not know how to to go about this. I would appreciate if someone could show me the steps on how to solve this.

Answer 1

Note that $$x^4+4=[(x+1)^2+1][(x-1)^2+1]$$ So if there is a solution to $x^4=-4\pmod p$ then $-1$ is a square modulo $p$, and consequently $p=1\pmod4$ or $p=2$. Using the same identity, the converse is also as easy. Thus $$ x^4=-4\pmod p\quad\hbox{has a solution}\iff p=2 \hbox{ or } p=1 \pmod 4$$