Suppose $p\in\mathbb{N}$ is an odd prime. I wish to show that in any reduced system of residues modulo $p$, we have that the number of quadratic residues is the same as the number of quadratic nonresidues. I further want to show that for $(a,p)=1$ we have $\sum_{x=0}^{p-1}(\frac{ax+b}{p})=0$ for $a,b\in\mathbb{Z}$.
I feel as though the second part follows from the first in that if we have equal number of quadratic residues as nonresidues then the sum of the Legendre symbols will be $1+(-1)=0$. But I'm not sure how to do the first part - in a complete system of residues there are $\frac{p-1}{2}$ nonresidues and $\frac{p+1}{2}$ residues, but for a reduced system I'm not sure.
