Evaluate a character sum $\sum\limits_{r = 1}^{(p - 1)/2}r \left( \frac{r}{p} \right) = 0$ for a prime $p \equiv 7 \pmod 8$ 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?
 A: Since you note that you are doing this as an exercise in a book, I imagine you might like a hint more than a proof. Look at the following two formulas, both due to Dirichlet:
$$L(1) = - \frac{\pi}{\sqrt{q}} \sum_{n=1}^{q-1} \frac{n}{q} \left( \frac{n}{q} \right).$$
$$L(1) = \frac{\pi}{\sqrt{q} \left(2-\left( \frac{2}{q} \right) \right) } \sum_{n=1}^{(q-1)/2} \left( \frac{n}{q} \right)$$
Setting these two formulas equal to each other and doing some elementary manipluation will yield your result.
Proofs of both of these formulas can be found in Chapter 1 of Davenport's Multiplicative Number Theory, available online through google books. I would bet it is possible to eliminate the infite series and give a direct proof of the equality of the two sides, but I don't see it right now.
A: 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 $\sum\limits_{1\leq r\leq \frac{p - 1}{2}}r(\frac{r}{p}) = 0$.
A: I now have a completely elementary proof. I thought I'd say a little bit about how I found this. I suspected that the only relevant properties of primes which are $7 \mod 8$ are that $2$ is a QR and $-1$ is not. Just using these two facts gave me tons of relationships between sums of Legnedre symbols, but I was getting lost in a pile of relationships without being able to pick out the ones I needed. I cut down the clutter in two ways: (1) I was frequently breaking my sums up into parts. I decided that I was only going to break the set $\{ 1,2, \ldots, p-1 \}$ into four pieces, no more, and pursue that line to the end. If it failed, I'd go back and try more pieces. (2) Since all the relations I was finding were linear, I didn't have to try to fit them into a logical chain. I just had to write down everything I knew, and what I wanted to conclude; it was then a matter of mechanical linear algebra whether or not my givens implied my conclusion.
So, we will partition $\{ 1,2, \ldots, p-1 \}$ into $4$ sets. An element $r$ is in


*

*$A$ if $r$ is odd and $r < p/2$

*$B$ if $r$ is even and $r < p/2$

*$C$ if $r$ is odd and $r>p/2$

*$D$ if $r$ is even and $r>p/2$.


For $X$ one of the sets $A$, $B$, $C$, $D$, write $S(X)$ for $\sum_{r \in X} \left( \frac{r}{p} \right)$ and write $T(X)$ for $\sum_{r \in X} r \left( \frac{r}{p} \right)$. We have the following relations (exercise!)
$$
\begin{array}{r@{}c@{}lr@{}c@{}l}
S(D)&=&-S(A) & T(D) &=&- (p S(A) - T(A)) \\
S(C)&=&-S(B) & T(C)&=&- (p S(B) - T(B)) \\
S(A)+S(B)&=&S(B)+S(D) &2(T(A)+T(B)) &=& T(B)+T(D)
\end{array}$$
The left three equations imply that $(S(A), S(B), S(C), S(D)) = (0,h,-h,0)$ for some $h$. Then the right three imply that $(T(A),T(B),T(C),T(D)) = (x,-x,-x-ph,x)$ for some $x$. None of this required creative thought; I just found the kernel of a $6 \times 8$ matrix.
Your desired conclusion is $T(A)+T(B)=0$, which we see is true. The equality between Dirichlet's expressions is $S(A)+S(B) = -(1/p) (T(A)+T(B)+T(C)+T(D))$, which we also see is true.
