# 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

$$x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)$$

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 $(x\pm1)^2\equiv-1\pmod p$

• o.k, I understood that for p=2 there is a solution (x=0), but what I need to do if p is odd prime as you mention? why it must divides exactly one of them? – MS93 Jan 29 '15 at 18:23
• @MS93, Please find the edited vesrion – lab bhattacharjee Jan 29 '15 at 18:27
• thank you very much! I got it! – MS93 Jan 29 '15 at 18:27