I need to prove that -1 is a quadratic residue of an odd prime p iff p = 1 (mod 4)
Any Ideas? Thanks
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The ammount of times this question's been asked in this site rapidly tends to infinity. Anyway, what have you tried so far? – Timbuc May 19 '15 at 16:21
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Not much to be honest. I have a feeling that i will have to use a consequence from eulers criterion and possibly Wilsons Theorem. – keith nolan May 19 '15 at 16:25
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If you are allowed to use Euler's Criterion, then it is one line since $(-1)^{(p-1)/2}=1$ if $p$ is of the form $4k+1$, and is $-1$ otherwise. – André Nicolas May 19 '15 at 16:29
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How come? I cant see it? – keith nolan May 19 '15 at 16:30
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If $-1$ is a quadratic residue modulo $p$, then we have $-1\equiv a^2\pmod p$, hence $$ (-1)^{\frac{p-1}2}\equiv (a^2)^{\frac{p-1}2}=a^{p-1}\equiv1\pmod p $$ Hence $p\equiv1\pmod4$ . Now if $p\equiv1\pmod4$, then $$ -1\equiv \left((\frac{p-1}2)!\right)^2 $$ by Wilson's theorem.
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