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Prove that if $p \equiv 3 \pmod {4}$, then $x \equiv \pm a^{(p+1)/4}\pmod{p}$ are the solutions of the congruence $x^2 \equiv a\pmod{p}$ if they exist.


I know this has to do with Euler's criterian. But other than that I'm not sure how to start the proof. Any help is appreciated thanks!

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We have: $$ a^{\frac{p-1}{2}}\equiv 1\pmod{p}\Longleftrightarrow \left(\frac{a}{p}\right)=+1 $$

and:

$$ \left(a^{\frac{p+1}{4}}\right)^2 = a \cdot a^{\frac{p-1}{2}}, $$ hence if $a$ is a quadratic residue, $\pm a^{\frac{p+1}{4}}$ are its square roots in $\mathbb{F}_p$.

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