3
$\begingroup$

How do I prove the formula $\newcommand{\jaco}[2]{\left(\frac{#1}{#2}\right)}\sum\limits_{a=1}^{p-2} \jaco{a(a+1)}p = -1$ where a varies from 1 to p-2 and p is a prime

I got as far as $\jaco{p-a}p = \jaco{-1}p \jaco ap$ so I can reduce the sum but that does not seem of much help.

$\endgroup$
  • $\begingroup$ I've tried to rewrite your post using TeX (for better readability), please, check whether I did not unintentionally changes anything. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Mar 18 '13 at 11:50
10
$\begingroup$

Note that for $1 \leq a \leq p-2$, $a$ has a unique inverse between $1$ and $p-2$. $$\sum_{a=1}^{p-2}{\left(\frac{a(a+1)}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{(\frac{a+1}{a})}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{1+a^{-1}}{p}\right)}=\sum_{a=1}^{p-2}{\left(\frac{1+a}{p}\right)}=\sum_{a=2}^{p-1}{\left(\frac{a}{p}\right)}=-1$$

The first equality holds since $\left(\frac{a(a+1)}{p}\right)=\left(\frac{(\frac{a+1}{a})}{p}\right)\left(\frac{a^2}{p}\right)=\left(\frac{(\frac{a+1}{a})}{p}\right)$.

The last equality holds because $\sum\limits_{a=1}^{p-1}{\left(\frac{a}{p}\right)}=0$ and $\left(\frac{1}{p}\right)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.