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Given prime number $p\equiv 1 \pmod 4$. Prove if $a∈F_p^×$ is a quadratic residue then the congruence $$x^4 ≡ a \pmod p$$ has either no solutions or four solutions. Give examples of each case.

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  • $\begingroup$ You can manage an example with for example $p=13$, surely? Trial and error will do! $\endgroup$ – Jyrki Lahtonen Feb 11 '15 at 11:20
  • $\begingroup$ Yeah I've found the examples thanks, still not sure how to tackle the proof though @JyrkiLahtonen $\endgroup$ – The Problem Feb 11 '15 at 11:43
  • $\begingroup$ I've started with x^4=a and a is quadratic residue, then simplify to x^2=q or x^2=-q where q,-q are square roots of a. Then we can see if q is a quadratic residue so is -q. So we'll get four solutions or no solutions, but I'm not sure how to conclude the 4 solutions are unique @JyrkiLahtonen $\endgroup$ – The Problem Feb 11 '15 at 11:53
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Let us put $a = b^2 \pmod p$. So $x^4 -b^2 = 0\mod p$, or $(x^2 - b)(x^2+b)=0\pmod p$. Since $p = 1\mod 4$, $-1$ is a quadratic residue (known theorem). Hence the previous equation can be written $(x^2-b)(x^2-\alpha^2 b) = 0\pmod p$ with $\alpha^2 = -1 \pmod p$. It is clear from this that if $b$ is a quadratic residue, say $b=c^2\pmod p$, the equation has four solutions $x = c, -c, \alpha c, -\alpha c$, otherwise it has no solution.

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