quadratic residue problem

**ezong** · Nov 1st 2009, 05:51 PM

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

**PaulRS** · Nov 2nd 2009, 03:43 AM

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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