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Going over a past exam in my elementary number theory course, I noticed this question that caught my attention. The question asked for the conditions that allowed $-3$ to be a quadratic residue mod $p$. Doing some experimentation, I found that this was possible when $p \equiv 1 \pmod 3$. So I guess I have answered part of the question. But the proof is obviously nagging me:

Prove $-3$ is a quadratic residue in $\Bbb Z_p$ if and only if $p \equiv 1\pmod 3$.

I've done a bit of work on this, but haven't been able to come up with anything close to elegant nor conclusive. Any help would be appreciated.

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    $\begingroup$ Just to nitpick, $-3$ is a quadratic residue mod $2$ and $3$, as well. So for the iff to be true, you should probably assume $p \geq 5$. $\endgroup$ – Viktor Vaughn Aug 1 '20 at 3:32
  • $\begingroup$ -3 mod 7 = 2^2 mod 7? $\endgroup$ – Charlie Chang Aug 1 '20 at 4:12
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Here’s an argument that doesn’t depend on Quadratic Reciprocity.

First, for $-3$ to be a quadratic residue modulo $p$, that’s the same as having a $\sqrt{-3}$ in the field $\Bbb F_p$ with $p$ elements, i.e. $\Bbb Z/(p)$.

Second, $\sqrt{-3}\in\Bbb F_p$ if and only if $\Bbb F_p$ has all three cube roots of unity, since the nontrivial ones are $\frac{-1\pm\sqrt{-3}}2$.

Third, $\Bbb F_p$ has all three cube roots of unity if and only if $3$ divides $|\Bbb F_p^\times|=p-1$.

And finally, $3\mid(p-1)$ if and only if $p\equiv1\pmod3$.

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By quadratic reciprocity law we have $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p}{3}\right)\\=\left(\frac{p}{3}\right)$$

Therefore $$\left(\frac{-3}{p}\right)=1\iff \left(\frac{p}{3}\right)=1\iff p\equiv1\pmod{3}$$

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  • $\begingroup$ @JyrkiLahtonen That's not really a hint. I'm very sorry for this. I am changing it. $\endgroup$ – Shubhrajit Bhattacharya Aug 1 '20 at 5:17
  • $\begingroup$ Well I suppose OP still has to prove the reverse direction and deduce that $(\frac{p}{3}) $implies that $p$ is 1 mod 3. $\endgroup$ – NotSoTrivial Aug 1 '20 at 5:19
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    $\begingroup$ NVM, you edited it so that it is the full solution now $\endgroup$ – NotSoTrivial Aug 1 '20 at 5:20
  • $\begingroup$ My same nitpick as above: your last $\iff$ isn't really true, since $p$ could be $0$ mod $3$, i.e., $p = 3$. $\endgroup$ – Viktor Vaughn Aug 6 '20 at 7:19
  • $\begingroup$ @RichardD.James I avoided the case $p=3$. See the definition of Legendre Symbol. $\left(\frac{a}{p}\right)$ is defined to equal $0$ when $a\equiv0\pmod{p}$. See the definition here en.m.wikipedia.org/wiki/Legendre_symbol $\endgroup$ – Shubhrajit Bhattacharya Aug 6 '20 at 7:55

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