# Quadratic Reciprocity Summation Proof [duplicate]

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!

## 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.

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