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Math Help - quadratic residue problem

  1. #1
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    quadratic residue problem

    If a prime number p is in the form p=4k+1, then how do I prove that the sum of the quadratic residues mod p is equal to p(p-1)/4? I tried adding consecutive perfect squares up to [((p-1)/2)^2]^2 and then subtracting to get the congruences, but that didn't work too well. Any suggestions are welcome
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  2. #2
    Super Member PaulRS's Avatar
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    Note that, if p\equiv{1}(\bmod.4) then -1 is a qudratic residue, hence \left(\tfrac{k}{p}\right)_L=\left(\tfrac{p-k}{p}\right)_L ( Legendre's Symbol)

    Thus we find that 2S=\sum_{0\leq k<p;\left(\tfrac{k}{p}\right)_L=1} k+\sum_{0<k<p;\left(\tfrac{k}{p}\right)_L=1} (p-k) = \sum_{0<k<p;\left(\tfrac{k}{p}\right)_L=1} {p}. - the sum is taken over the quadratic residues-


    Now, how many quadratic residues are there?
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