sum of the product of consecutive legendre symbols is -1

summation elementary-number-theory legendre-symbol

Note that for $1 \leq a \leq p-2$, $a$ has a unique inverse between $1$ and $p-2$. $$\sum_{a=1}^{p-2}{\left(\frac{a(a+1)}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{(\frac{a+1}{a})}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{1+a^{-1}}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{1+a}{p}\right)}=\sum_{a=2}^{p-1}{\left(\frac{a}{p}\right)}=-1$$

The first equality holds since $\left(\frac{a(a+1)}{p}\right)=\left(\frac{(\frac{a+1}{a})}{p}\right)\left(\frac{a^2}{p}\right)=\left(\frac{(\frac{a+1}{a})}{p}\right)$.

The last equality holds because $\sum\limits_{a=1}^{p-1}{\left(\frac{a}{p}\right)}=0$ and $\left(\frac{1}{p}\right)=1$.

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