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.
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What is $\alpha$ ? – dmtri Oct 16 '19 at 10:04
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It's $a$ and $a$ is any integer co prime to $p$. – Nitish Kumar Oct 16 '19 at 10:07
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@KaviRamaMurthy the question has been taken from Niven Zuckerman, section 3.3 question 20. Please help me to solve it. – Nitish Kumar Oct 16 '19 at 10:07
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An introduction to theory of numbers by Ivan Niven , Zuckerman. – Nitish Kumar Oct 16 '19 at 10:08
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@nicomezi i hope this helps now. – Nitish Kumar Oct 16 '19 at 10:15
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Yes much better thank you. – nicomezi Oct 16 '19 at 10:15
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Here are some hints to get started
Let $N(a)=\{ \exists\,, x \in \mathbb{Z}_{p} \ : \ x^2\equiv a \pmod{p}\}$, then it is easy to see that $|N(a)|=1+\left(\frac{a}{p}\right)$
Count the number of solutions to $x^{2}-n^2\equiv a\pmod{p}$ for a fixed $a$.
C.S.
- 5,528
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Can you please elaborate your answer. Sorry for the delayed message. – Nitish Kumar Oct 16 '19 at 13:59
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