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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

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

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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  • $\begingroup$ 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? $\endgroup$ – MS93 Jan 29 '15 at 18:23
  • $\begingroup$ @MS93, Please find the edited vesrion $\endgroup$ – lab bhattacharjee Jan 29 '15 at 18:27
  • $\begingroup$ thank you very much! I got it! $\endgroup$ – MS93 Jan 29 '15 at 18:27

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