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I have found various proofs of this question usig primitive roots, but I want to prove it without using primitive roots!

Here is my question again:

Let $p$ be prime. Prove that there are the same number of quadratic residues modulo p as there are non-residues. Clearly state where you have used that $p$ is prime.

Thanks! :)

Ineedhelppls
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    Hint: $x^2\equiv a\bmod p$ has exactly two solutions for each non-zero $a$. – Gerry Myerson Jan 29 '16 at 11:24
  • you should add the hypothesis of non-zero quadratic residues, otherwise you have $(p+1)/2$ residues and $(p-1)/2$ non-residues. – PITTALUGA Jan 29 '16 at 11:33
  • Would it be enough to say this?: Define a funtion $f: R -> R$ by $f(x)\equiv x^2\bmod p$ . Now $R$ is both the domain and codomain of $f$. Also, $Q$= image($f$) and $N$ = $R$/image($f$). Then we observe that each element of image($f$) is at the end of exactly two arrows of the mapping. Hence, |image($f$)| = $\frac{1}{2}$|$R$|, so |$Q$| = \frac{1}{2} |$R$|. But also, |$N$| = |$R$| - |$Q$| = \frac{1}{2} |$R$| = |$Q$|. We can see that the observation is true because if a is in the image($f$) then there exists x in $R$ such that $x^2\equiv a\bmod p$ as well as $(-x)^2\equiv a\bmod p$. – Ineedhelppls Jan 29 '16 at 13:08
  • So $a$ is at the end of at least 2 arrows of the mapping. Also, Lagrange's theorem says that there are exactly 2 solutions so $a$ is at the end of exactly 2 arrows of the mapping. – Ineedhelppls Jan 29 '16 at 13:10
  • $$x^2\equiv a^2\pmod{p}\iff p\mid x^2-a^2=(x+a)(x-a)$$

    $p$ is prime, so by Euclid's Lemma:

    $$\iff x\equiv \pm a\pmod{p}$$

    Therefore $(\pm 1)^2$, $(\pm 2)^2$, $((p-1)/2)^2$ are all the quadratic residues.

     – user236182 Feb 06 '16 at 02:31

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We are considering only invertible elements, and we have to assume $p>2$ since for $p=2$ the statament is false because there is just one invertible element. $x^2 \equiv y^2 \pmod p \iff x \equiv y \pmod p$ or $x \equiv -y \pmod p$ because $\mathbb Z_p$ is a field and thus an integral domain. So you reach exactly two times the same quadratic residue using the $p-1$ invertible elements. Thus you get exactly $\frac{p-1}{2}$ different values. The remaining are $(p-1) - \frac{p-1}{2}=\frac{p-1}{2}$.

Maffred
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