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Here is the setup:

Suppose $p$ is an odd prime.

A primitive root modulo $p$ is an integer with order $p-1$.

Suppose further $\omega$ is a primitive root modulo $p$.

It can be shown that $$\left(\frac{\omega}{p}\right)=-1$$

Here is the question:

Using primitive roots show there are the same number of quadratic residues as there are quadratic nonresidues modulo $p$.

My thoughts:

Any primitive root is a quadratic nonresidue by above.

For any primitive root we have $$\left(\frac{\omega}{p}\right)=-1$$ so by Euler's criterion we get $$\left(\frac{\omega}{p}\right)=-1 \equiv \omega^{(p-1)/2} \mod p$$

By Lagrange's theorem (for polynomials) we have the equation $$-1 \equiv \omega^{(p-1)/2} \mod p$$ has at most $(p-1)/2$ distinct solutions modulo $p$.

Here are just some things that spring to mind (I don't know if they are along the right tracks)

any help?

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  • $\begingroup$ There are $\frac {p-1}2$ squares...For every element $x^2\equiv (-x)^2 \pmod p$ and, since it's a field, there can't be any other square roots. Hence there are also $\frac {p-1}2$ non squares. $\endgroup$ – lulu Jan 12 '17 at 18:50
  • $\begingroup$ I don't understand how do you know that there are $(p-1)/2$ quadratic residues. It just seems like you state what I need to prove. I also need to use primitive roots in my answer. $\endgroup$ – Ryan S Jan 12 '17 at 18:52
  • $\begingroup$ I gave a proof! Group the residues as $\{\pm 1, \pm 2, \cdots, \pm \frac {p-1}2\}$. Squaring each term in that gives you all of the squares. $\endgroup$ – lulu Jan 12 '17 at 18:53
  • $\begingroup$ Yeah I get that proof but how does that use primitive roots? It says specifically that primitive roots must be used. $\endgroup$ – Ryan S Jan 12 '17 at 18:55
  • $\begingroup$ Ok, show that the squares are $g^{2i}$ and the non-squares are $g^{2i-1}$. $\endgroup$ – lulu Jan 12 '17 at 18:57
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An idea: take all the odd integers between $\;1\;$ and $\;p-1\;$, whose number is precisely $\;\cfrac{p-1}2\;$, and now observe that, for $\;n\in\Bbb N\;$ :

$$\left(\frac wp\right)=w^{(p-1)/2}\implies \left(\frac{w^{\frac{2n-1}2}}p\right)=\left(w^{\frac{2n-1}2}\right)^{\frac{p-1}2}=\left(w^{\frac{(p-1)}2}\right)^{\frac{2n-1}2}=(-1)^{\frac{2n-1}2}=-1$$

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  • $\begingroup$ What is $n$ here? $\endgroup$ – Ryan S Jan 12 '17 at 18:58
  • $\begingroup$ @RyanS A natural number. $\endgroup$ – DonAntonio Jan 12 '17 at 18:59

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