Show $\sum\limits_{b=0}^{p-1}\left(\frac{b}{p}\right) = 0$ Show $\sum\limits_{b=0}^{p-1}\left(\frac{b}{p}\right) = 0$
The resources I have consulted said to use the fact that the number of quadratic residues $\text{mod } p$ is $\frac{p-1}{2}$ but I have no idea how to apply this to my problem. Any hints and suggestions are greatly appreciated. Also links to resources regarding this are also great
 A: If $\chi : \mathbb{Z}/n\mathbb{Z} \longrightarrow \mathbb{C}$ is a nontrivial Dirichlet character, meaning that if $\gcd(m,n) > 1$ then $\chi(m) = 0$, and otherwise $\chi$ is a homomorphism from $\left( \mathbb{Z}/ n\mathbb{Z} \right)^\times$ to $\mathbb{C}^\times$ which is not always $1$, then it is always true that
$$
S_n := \sum_{m \bmod n} \chi(m) = 0.
$$
So this is true with some generality. Perhaps the easiest way to show this is to take some $k$ relatively prime to $n$ with $\chi(k) \neq 1$ (which will exist because the character is nontrivial), and note that as $m$ varies through those numbers relatively prime to $n$, the  the $(mk \bmod n)$ also vary through numbers relatively prime to $n$. Thus
$$
\chi(k) S_n = \chi(k) \sum_{m \bmod n} \chi(m) = \sum_{m \bmod n} \chi(mk) = \sum_{l \bmod n} \chi(l) = S_n.
$$
But the only way for $\chi(k)S_n = S_n$ when $\chi(k) \neq 1$ is for $S_n = 0$. $\diamondsuit$
It turns out that this is only part of the story, and that there are a finite number of Dirichlet characters mod $n$ as well, which exhibit a similar sort of orthogonality relation.
For this question, the Legendre symbol is a nontrivial Dirichlet character, and is in fact the most friendly of the Dirichlet characters.
