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As above. I can show that this forms a group (under multiplication mod p) so clearly $ \dfrac{p-1}{2} $ is a possibility. But I must be missing something obvious as I don't see how to prove it.

Any help/ a solution would be appreciated.

Evgeny T
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    Many ways to proceed. Possibly the one putting least demands on your background is to first show that $x^2\equiv y^2$ if and only if $x\equiv-y$. If you can pull that off, then it follows that $x\mapsto x^2$ is a 2-1 mapping, and hence the image has half the size. – Jyrki Lahtonen Aug 29 '16 at 16:35
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    But, have you searched the site? It is quite llikely that very closely related questions have already been handled. Alas, the search engine sucks with formulas. – Jyrki Lahtonen Aug 29 '16 at 16:39
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    Have a look at the very pedagogical presentation of (https://en.wikipedia.org/wiki/Quadratic_residue). See on this site (http://math.stackexchange.com/q/151530) – Jean Marie Aug 29 '16 at 16:59
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    @JyrkiLahtonen You mean $x\equiv \pm y$. – user236182 Aug 29 '16 at 17:07
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    $x^2\equiv y^2\pmod{p}$ if and only if $p\mid x^2-y^2=(x+y)(x-y)$. By Euclid's lemma this is equivalent to either $p\mid x+y$ or $p\mid x-y$, i.e. $x\equiv \pm y\pmod{p}$. From this the fact that there are $\frac{p-1}{2}$ non-zero quadratic residues mod an odd prime $p$ follows. – user236182 Aug 29 '16 at 17:10
  • If you know about primitive roots, this is fairly straightforward, because squares are precisely even powers of your primitive root. – G Tony Jacobs Aug 30 '16 at 03:20

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