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Prove $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$, where $p$ is an odd prime, and the LHS is the legendre symbol.

I've got $-1 = x^2 \pmod p \implies (-1)^{\frac{p-1}{2}} = x^{p-1} = 1 = \left( \dfrac{-1}{p}\right)$. Now suppose that $-1$ is not a quadratic residue modulo $p$. Then if $(-1)^{\frac{p-1}{2}} = 1$ we have that $\frac{p-1}{2} = 2k$ for some $k$ as otherwise it would equal $-1$, but $p$ is an odd prime and $-1 \neq 1$. That doesn't seem to help us though. Any ideas?

Thanks.

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  • $\begingroup$ Do you know that $\;-1\;$ is a QR iff $\;p=1\pmod 4\;$ ? $\endgroup$ – Timbuc Apr 11 '15 at 19:15
  • $\begingroup$ I've seen that, I guess now to learn it. $\endgroup$ – Shine On You Crazy Diamond Apr 11 '15 at 19:19
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Say $(a,p) = 1$, where $p$ is odd prime. We have that, $$ (a/p) \equiv (-1)^{(p-1)/2} ~ (p) $$

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If you know what I asked in my comment, then it is trivial

$$(-1)^{\frac{p-1}2}=1\iff \frac{p-1}2\in2\Bbb Z\iff p=1\pmod 4$$

Otherwise we can try with some basic group theory: the multiplicative group $\;\Bbb F_p^*\;$ is cyclic of order $\;p-1\;$, thus it has one unique subgroup of each order dividing $\;p-1\;$ .

Thus, this group has a subgroup (cyclic, of course) of order four iff $\;4\mid (p-1)\;$ iff it has an element $\;x\;$ of order four, which means

$$(x^2)^2=1\implies x^2=-1\;\;\text{is the only element of order two in the group!}$$

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