Proving summations involving the Legendre symbol

Question

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 $p \equiv 1 \pmod 4$.

Answer 1

Some hints:

1. $a(a+1)$ 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?
2. How many quadratic residues are there in $\Bbb Z^*$ and how many non quadratic residue are there?

Answer 2

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?)