Sum of the Multiplicative Legendre's symbol Let $p$ a be odd prime show that $\displaystyle\sum_{a=1}^{p-2}\left(\frac{a(a+1)}{p}\right)=-1$.
Nota Bene : The  $(\frac{a}{p})$ is $\textbf{Legendre Symbol}$

For $a\in\mathbb F_p^*$ we have $(\frac{a(a+1)}{p})=(\frac{(a+1)/a}{p})$ (since $(\frac{a^2}p)=1$ and since the Legendre symbol is multiplicative), so
$$\sum\limits_{a\in\mathbb F_p^*}\left(\frac{a(a+1)}{p}\right)=\sum\limits_{a\in\mathbb F_p^*}\left(\frac{1+1/a}{p}\right)=\sum\limits_{b\in\mathbb F_p^*}\left(\frac{1+b}{p}\right)=-\left(\frac1p\right)+\sum\limits_{c\in\mathbb F_p}\left(\frac cp\right)$$
and since the sum over $c$ is $0$ (as many quadratic residues as non-residues) and since $\left(\frac1p\right)=1$, you get your result.
