$\sum_{n=1}^{n=p}\left(\frac{n^2+a}{p}\right)=-1$.

Question

Show that if $(a,p)=1$, $p$ an odd prime then, $\sum_{n=1}^{n=p}\left(\frac{n^2+a}{p}\right)=-1$, where $\left(\frac{n^2+a}{p}\right)$ is the Jacobi symbol.

Answer 1

Here are some hints to get started

• Let $N(a)=\{ \exists\,, x \in \mathbb{Z}_{p} \ : \ x^2\equiv a \pmod{p}\}$, then it is easy to see that $|N(a)|=1+\left(\frac{a}{p}\right)$

• Count the number of solutions to $x^{2}-n^2\equiv a\pmod{p}$ for a fixed $a$.