# sum of the product of consecutive legendre symbols is -1

How do I prove the formula $\newcommand{\jaco}[2]{\left(\frac{#1}{#2}\right)}\sum\limits_{a=1}^{p-2} \jaco{a(a+1)}p = -1$ where a varies from 1 to p-2 and p is a prime

I got as far as $\jaco{p-a}p = \jaco{-1}p \jaco ap$ so I can reduce the sum but that does not seem of much help.

• I've tried to rewrite your post using TeX (for better readability), please, check whether I did not unintentionally changes anything. For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Mar 18 '13 at 11:50

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 $\left(\frac{1}{p}\right)=1$.