TITLE: Evaluate a character sum $\sum\limits_{r = 1}^{(p - 1)/2}r \left( \frac{r}{p} \right) = 0$ for a prime $p \equiv 7 \pmod 8$
QUESTION [5 upvotes]: I do not know how to prove the following statement:

Let $p \equiv 7 \pmod 8$ be a prime. Then
  $$\sum\limits_{r = 1}^{\frac{p - 1}{2}}r \left( \frac{r}{p} \right) = 0$$ where $\left( \dfrac{\cdot}{\cdot} \right)$ is the Legendre symbol. 

Could anybody help me to answer this question?

REPLY [3 votes]: Thanks for all the help. I have a proof now.
Since $p\equiv7\mod 8$, $(\frac{-1}{p}) = -1$ and $(\frac{2}{p}) = 1$. Thus we have $$\sum\limits_{{1\leq r\leq p - 1}}r(\frac{r}{p}) = \sum\limits_{1\leq r\leq \frac{p - 1}{2}}2r(\frac{r}{p}) - \sum\limits_{1\leq r\leq \frac{p - 1}{2}}(p - 2r)(\frac{r}{p}) = \sum\limits_{1\leq r\leq \frac{p - 1}{2}}(4r - p)(\frac{r}{p})$$ 
and 
$$\sum\limits_{1\leq r\leq p - 1}r(\frac{r}{p}) = \sum\limits_{1\leq r\leq\frac{p - 1}{2}}r(\frac{r}{p}) - \sum\limits_{1\leq r\leq\frac{p - 1}{2}}(p - r)(\frac{r}{p}) = \sum\limits_{1\leq r\leq \frac{p - 1}{2}}(2r - p)(\frac{r}{p})$$
hence $\sum\limits_{1\leq r\leq \frac{p - 1}{2}}r(\frac{r}{p}) = 0$.