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I want to prove $\newcommand{\jaco}[2]{\left(\frac{#1}{#2}\right)}\sum\limits_{n=1}^{p} \jaco{n^2+a}p = -1$, where $(a,p)=1$ and $p$ is an odd prime. I have seen similar problems like sum of the product of consecutive legendre symbols is -1, but I am not able to apply similar method to solve this one. Any ideas? Thank you.

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You can read the proof for the generalized result for $\displaystyle \sum_{k=1}^{p}\left( \frac{ak^2+bk+c}{p} \right)$ in here or here.

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