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I am trying to solve the following problems ($p$ is an odd prime).

  1. Find the sum $$\sum_{a=1}^{p-1}a \cdot \left (\frac{a}{p} \right),$$
  2. Find the sum $$\sum_{a=1}^{p-1} 2^a \cdot \left (\frac{a}{p} \right).$$

Some thoughts :

  1. I reduced the sum to $2S_P-\frac{p(p-1)}{2}$ where $S_p$ is the sum of the quadratic residues modulo $p$ but I don't know how to evaluate it .
  2. Nothing so far but more generally what can we say about the polynomial :

$$f(x)=\sum_{a=1}^{p-1} x^a \cdot \left (\frac{a}{p} \right)$$

Is this polynomial interesting in any way ?

Thanks for all the help .

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  • $\begingroup$ How did you reduce such sum? $\endgroup$ – Paolo Leonetti Aug 24 '15 at 21:47
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    $\begingroup$ Are you interested in computing the sums $\pmod{p}$ or in finding their exact values? In the first case, notice that $$\sum a\left(\frac{a}{p}\right) = \sum a^{\frac{p+1}{2}}.$$ $\endgroup$ – Jack D'Aurizio Aug 24 '15 at 21:56
  • $\begingroup$ For the case when $p \equiv 1 \pmod{4}$ it's easy to get (using a little symmetry) $S_p=\frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p \equiv 3 \pmod{4} $ . $\endgroup$ – user252450 Aug 24 '15 at 21:57
  • $\begingroup$ @ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ? $\endgroup$ – user252450 Aug 24 '15 at 22:01
  • $\begingroup$ and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity $\endgroup$ – reuns Mar 31 '16 at 16:57

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