In the following, let $(\frac{a}{p})$ denote the Legendre symbol. Then
Show that $$\sum _{a=1}^{p-2} \left(\frac{a(a+1)}{p}\right)=-1$$ for an odd prime $p$.
I was thinking of factoring out $a^2$, but…
Show that $$\sum _{a=1}^{(p-1)/2} \left(\frac{a}{p}\right)=0$$ for a prime $p \equiv 1 \pmod 4$.
Best Answer
The first question is answered in several other posts:
For the second one, first you can notice that $$\newcommand\jaco[2]{\left(\frac{#1}{#2}\right)}\sum_{a=1}^{p-1} \jaco ap = 0,$$ since this sum contains the same number of $1$'s and $(-1)$'s.
Using the fact that $$\jaco{p-a}p = \jaco{-a}p = \jaco{-1}p \jaco ap \overset{(*)}= \jaco ap$$ you can divide the above sum into two sums which are equal to each other and therefore they are both zero.
(Can you say why the equation denoted by $(*)$ holds?)