Let $p$ be an odd prime. Prove that
$$ \sum_{a=1}^{p-1} (\frac{a}{p}) = 0 $$
where $(\frac{a}{p})$ is the Legendre symbol.
We use Proposition (7.4).
Let $p$ be an odd prime. Then there are exactly $\frac{p- 1}{2}$ quadratic residues and $\frac{p- 1}{2}$ quadratic non-residues of $p$.
This means that of the $p-1$ residues of $p$, there are $\frac{p- 1}{2}$ quadratic residues and $\frac{p- 1}{2}$ quadratic non-residues. The Legendre symbol of the quadratic residues has value 1, and has value -1 for the non-residues.
And so
$$ \sum_{a=1}^{p-1} (\frac{a}{p}) = (\frac{p-1}{2} \times 1) + (\frac{p-1}{2} \times -1) = 0$$