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    Let \displaystyle p be an odd prime.

    A primitive root mod \displaystyle p is an integer \displaystyle \omega with the order of \displaystyle \omega =p-1

    I have shown

    \displaystyle \left(\frac{\omega}{p}\right)=-1 (This is the Legendre symbol)

    Here is the question:

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

    I am aware of the "standard" proof of this result which uses Lagrange's theorem (for polynomials) but I don't know how to prove it using primitive roots.

    Obviously since \displaystyle \left(\frac{\omega}{p}\right)=-1 then any primitive root is a quadratic non residue but this is as far as I have got.

    I tried to use Euler's criterion somehow but couldn't see a way forward.
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    I would think you're supposed to do something very "low level", based on the fact that every element of Z_p* can be written as \omega^k for some integer k, and then the quadratic residue status can be found based on whether k is odd or even.
 
 
 
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