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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?

  • 0
    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 $p$?2012-02-28
  • 1
    Didn't you - or someone - just ask this question yesterday?2012-02-28
  • 3
    Found it - http://math.stackexchange.com/questions/113954/the-bound-of-valuation. That's not the way to do things here.2012-02-28
  • 0
    It seemed that nobody noticed the second question I asked yesterday, so I put it here today. Sorry.2012-02-28

3 Answers 3