0
$\begingroup$

This question already has an answer here:

Let $f(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{Z}$, and let $p$ be an odd prime that does not divide $a$. In addition, let $d = b^2 - 4ac$.

Show that, if $p$ does not divide $d$, then: \begin{align} \sum_{x=1}^p\left(\frac{f(x)}{p}\right) = -\left(\frac{a}{p}\right) \end{align}

I first started off with substituting $f(x)$ into the LHS of the equation: \begin{align} \sum_{x=1}^p\left(\frac{ax^2 + bx + c}{p}\right) \end{align}

But I'm not sure how to go from here. Any tips would be appreciated!

$\endgroup$

marked as duplicate by Lord Shark the Unknown, awllower, YuiTo Cheng, Aqua, Community Jul 19 at 6:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ The Legendre symbol is not a fraction, so you can’t do the step of “separating constant. $\endgroup$ – Thomas Andrews Jul 19 at 3:43
  • $\begingroup$ Specifically $\left(n\over p\right)$ is notation called a Legendre symbol, and the value always one of $-1,0,1.$ $\endgroup$ – Thomas Andrews Jul 19 at 3:45
  • $\begingroup$ Same question here: math.stackexchange.com/questions/1891644/… $\endgroup$ – Robert Z Jul 19 at 3:55