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What can I say about $x^4 \equiv -4 \mod p$ where $p$ is prime? In general what can I do with powers that are greater than $2$ and where I cannot use reciprocity, legendre/jacobi etc... In general what can I say about a quadratic polynomial modulo $p$: For instance $(x-1)^2 \equiv 1 \mod p$

By 'what can I say' I mean $p \equiv$ something $\mod 4$ or $8$

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    $\begingroup$ Is it $\,x^4=1\,\,\,or\,\,\,x^4=-4\,$? And what do you mean by "what can I say"? The equation $\,x^4=1\,\pmod p$ always has the solutions $\,-1,1\,$, which are different if $\,p\neq 2\,$... $\endgroup$
    – DonAntonio
    Nov 2, 2012 at 3:44
  • $\begingroup$ I mean $-4$. I just fixed my question. $\endgroup$
    – CodeKingPlusPlus
    Nov 2, 2012 at 3:47

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$$ x^4 + 4 = ((x-1)^2 + 1) ((x+1)^2 + 1) $$

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  • $\begingroup$ By Euclid's Lemma $p|((x-1)^2 + 1)$ or $p|((x+1)^2 +1)$. Then I get $(x-1)^2 \equiv 1 \mod p$ or $(x+1)^2 \equiv 1 \mod p$. I have learned about quadratic residues but not anything more complicated. Could you give me another hint? $\endgroup$
    – CodeKingPlusPlus
    Nov 2, 2012 at 3:56
  • $\begingroup$ Then you can multiply $-2^2$ to both sides of the congruence and then we have $-4$ is a quadratic residue $\mod p$ $\endgroup$
    – CodeKingPlusPlus
    Nov 2, 2012 at 4:06
  • $\begingroup$ artofproblemsolving.com/wiki/index.php/Sophie_Germain_Identity $\endgroup$
    – user236182
    Jan 31, 2017 at 0:57
  • $\begingroup$ @CodeKingPlusPlus . $\endgroup$
    – user236182
    Jan 31, 2017 at 0:59

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