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I know how to prove this the other way, but I don't see how the if and only if statement works in this direction. One thought I had was to try to show that the exponent was even as I know that this is the only way for $a$ to be a quadratic residue. I'm not really sure how to go about doing this though.

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  • $\begingroup$ The exponent will only be even if $p \equiv 1 \pmod{4}$, and that doesn't have to happen. You still need to handle the case $p \equiv 3 \pmod{4}$. $\endgroup$ – Robert Shore Mar 21 at 1:32
  • $\begingroup$ When $\ a^{(p-1)/2}\equiv 1 \pmod{p}\ $ the Tonelli-Shanks algorithm enables you to calculate the $\hspace{-0.2em}\pmod{p}\ $ square-root of $\ a\ $ explicitly. See the Wikipedia article at: en.wikipedia.org/wiki/Tonelli–Shanks_algorithm $\endgroup$ – lonza leggiera Mar 22 at 2:12
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If $a$ is a quadratic residue, there exists $x$ such that $$x^2\equiv a\pmod p$$

$$a^{(p-1)/2}\equiv x^{p-1}\equiv?\pmod p$$

Alternatively $g$ is a primitive root, $$(g^k)^{(p-1)}\equiv1\pmod p$$ will hold true iff $p-1$ divides $k(p-1)/2\iff2|k$

If $k=2m$ $$g^{2m}\equiv(g^m)^2\pmod p$$

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  • $\begingroup$ yes I know by Fermat's Little Theorem that this then is equal to 1. It is proving the other side of the if and only if statement where I am a little confused $\endgroup$ – joseph Mar 21 at 2:11
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    $\begingroup$ @josephF, Conversely if $x^2\equiv a,a^{(p-1)/2}\equiv -1,$ $$-1\equiv a^{(p-1)/2}\equiv x^{p-1}\equiv1$$ $\endgroup$ – lab bhattacharjee Mar 21 at 2:28
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The homomorphism $\phi: (\Bbb Z/p \Bbb Z)^* \to (\Bbb Z/ p \Bbb Z)^*$ defined by $\phi(x) = x^2$ has kernel $\{1, -1 \}$, so $|\operatorname{image}(\phi)|= \frac{p-1}{2}.$ Since $(\Bbb Z / p \Bbb Z)^*$ has $\frac{p-1}{2}$ elements satisfying $x^{\frac{p-1}{2}}=1$, $a$ must be in the image of $\phi$.

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