Jacobi Symbol: $\sum_{n=1}^{n=p}\left(\frac{n^2+a}{p}\right)=-1$

Asked Oct 16 '19 at 14:07

Active Oct 17 '19 at 04:56

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Show that if $(a,p)=1$, $p$ an odd prime then, $\sum_{n=1}^{n=p}\left(\frac{n^2+a}{p}\right)=-1$, where $\left(\frac{n^2+a}{p}\right)$ is the Jacobi symbol.

This question has been taken from the book : An introduction to theory of numbers by Niven, Zuckerman, Montgomery. Section 3.3., question 20. Thanks in advance.

edited Oct 17 '19 at 04:56

asked Oct 16 '19 at 14:07

Could you write more than just" I couldn't understand that"? What exactly is the problem? Did you try to understand the other duplicates? – Dietrich Burde Oct 16 '19 at 19:14
I did't find any other duplicates. How to proceed using that hint is not at all clear to me. – Oct 16 '19 at 19:15
You did not search long, right? Start with this duplicate. – Dietrich Burde Oct 16 '19 at 19:22
Thanks for the help. Let me go through it. +1 – Oct 16 '19 at 19:26

Jacobi Symbol: $\sum_{n=1}^{n=p}\left(\frac{n^2+a}{p}\right)=-1$

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