Proving summations involving the Legendre symbol 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 $p \equiv 1 \pmod 4$.

 A: The first question is answered in several other posts:


*

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

*How can I prove these summations for the legendre symbol 

*How do you find $\sum_{j=0}^{p-1}\left(\frac{j(j+1)}p\right)$,p = prime, where $\left(\frac{j(j+1)}p\right)$ is the Legendre symbol?
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?)
A: Some hints:


*

*$a(a+1)$ is a quadratic residue if and only if $\frac{a+1}{a}=1+a^{-1}$ is quadratic residue and then how many quadratic residu of the form $1+x$ with $x$ invertible are there?

*How many quadratic residues are there in $\Bbb Z^*$ and how many non quadratic residue are there?

