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?