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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?

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We know that $\sum\limits_{r = 1}^{\frac{p - 1}{2}}r \left( \frac{r}{p} \right) \equiv0\mod p$, can we show this sum is bounded by $p$? – ksj03 Feb 28 '12 at 5:52
Didn't you - or someone - just ask this question yesterday? – Gerry Myerson Feb 28 '12 at 6:42
Found it - That's not the way to do things here. – Gerry Myerson Feb 28 '12 at 6:56
It seemed that nobody noticed the second question I asked yesterday, so I put it here today. Sorry. – ksj03 Feb 28 '12 at 8:34

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.

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Oh, okay. Now I see why they're relevant to the problem but I'll have to think some about all those relations. – anon Feb 29 '12 at 1:04
I think the big display is cut off at the left margin and needs editing. – Gerry Myerson Feb 29 '12 at 3:26
I asked a question related to this answer at – joriki Jun 28 at 10:16

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.

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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 $\sum\limits_{1\leq r\leq \frac{p - 1}{2}}r(\frac{r}{p}) = 0$.

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I added double dollar signs in order to make the formulas more readable. You should consider doing this in the future anytime you have a bulky mathematical expression. – Alex Becker Feb 29 '12 at 6:52

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