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Show that if $p$ is and odd prime and $h$ is an integer, $1\le h \le p$, then

$$\displaystyle\sum_{n=1}^{p}\left(\sum_{m=1}^{h}\left(\frac{m+n}{p}\right)\right)^2=h(p-h)$$ where $\left(\frac{m+n}{p}\right)$ denotes the Jacobi symbol.

My solution:

For $h=1$, we have $\left(\frac{1+n}{p}\right)^2$ is always 1 or 0. It is 0 only when $n=p-1$. So the sum comes out to be $p-1$, which is accordance with $h(p-h)$. Similarly, for $h=p$, and the sum is zero.

But I am having trouble when $ h \neq 1 or p$, how will I proceed in that case?

This question has been taken from the book : An introduction to theory of numbers by Niven, Zuckerman, Montgomery. Section 3.3., question 19. Thanks in advance.

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1 Answers1

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The square of the inner sum is $$\sum_{m_1,m_2=1}^h\left(\frac{(m_1+n)(m_2+n)}{p}\right).$$ The whole sum is $$\sum_{m_1,m_2=1}^h\sum_{n=1}^p\left(\frac{(m_1+n)(m_2+n)}{p}\right) =\sum_{m_1,m_2=1}^hS(m_1,m_2)$$ say. When $m_1=m_2$ then $S(m_1,m_2)=p-1$. When $m_1\ne m_2$ then $$S(m_1,m_2)=\sum_{n=1}^p\left(\frac{n(n+2m')}p\right)$$ where $2m'\equiv m_1-m_2\not\equiv0\pmod p$. We can drop the $n=p$ term, and then let $n'$ be the mod $p$ inverse of $n$ gives $$S(m_1,m_2)=\sum_{n=1}^{p-1}\left(\frac{n(n+2m')}p\right) =\sum_{n'=1}^{p-1}\left(\frac{1+2m'n'}p\right)=-1.$$ Therefore the original sum is $$h(p-1)-(h^2-h)=h(p-1-(h-1))=h(p-h).$$

Angina Seng
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  • I am having difficult in understanding the following:$$S(m_1,m_2)=\sum_{n=1}^p\left(\frac{n(n+2m')}p\right)$$ where $2m'\equiv m_1-m_2\not\equiv0\pmod p$. We can drop the $n=p$ term, and then let $n'$ be the mod $p$ inverse of $n$ gives $$S(m_1,m_2)=\sum_{n=1}^{p-1}\left(\frac{n(n+2m')}p\right) =\sum_{n'=1}^{p-1}\left(\frac{1+2m'n'}p\right)=-1.$$ can you please explain this part? Thanks for the solution. –  Oct 17 '19 at 06:47
  • @EpsilonDelta (i) is just a change of variable, (ii) is extracting a factor $(n^2/p)$. – Angina Seng Oct 17 '19 at 17:38
  • Thanks for the help –  Oct 21 '19 at 17:08