# How can I prove these summations for the legendre symbol

How can I prove for the Legendre symbol that:

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

This is essentially what I prove here, after you note that $\left(\dfrac{(p-1)p}{p}\right)=\left(\dfrac{p}{p}\right)=0$. The main idea is to note that $\left(\dfrac{a(a+1)}{p}\right)=\left(\dfrac{\dfrac{a+1}{a}}{p}\right)$. Of course, you need to be familar with the fact that $\sum\limits_{a=1}^{p-1}{\left(\dfrac{a}{p}\right)}=0$ (which follows from the fact that there are exactly $\dfrac{p-1}{2}$ non-zero quadratic residues $\pmod{p}$ (for $p>2$))