# Quadratic Reciprocity problem.. help! [closed]

If $p$ is an odd prime, evaluate $\left(\frac{1\times2}{p}\right)+\left(\frac{2\times3}{p}\right)+\cdots+\left(\frac{(p-2)\times(p-1)}{p}\right)$

I don't know how I use properties of Legendre symbol. Please help!

## closed as off-topic by José Carlos Santos, Saad, Martin Sleziak, Jose Arnaldo Bebita-Dris, NamasteMay 27 '18 at 13:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Saad, Jose Arnaldo Bebita-Dris, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos May 27 '18 at 7:59
• umm.. I tried ((p-2)(p-1)/p)=(1x2/p), and thought there is (p-1)/2 quadratic residue and quadratic nonresidue.. but it didn't help to evaluate that. I think the answer is -1. – 박윤수 May 27 '18 at 8:05
• – Martin Sleziak May 27 '18 at 9:07
• – Martin Sleziak May 27 '18 at 9:07

This is $$S=\sum_{k=1}^{p-1}\left(\frac{k(k+1)}p\right) =\sum_{k=1}^{p-1}\left(\frac{k^*(k+1)}p\right) =\sum_{k=1}^{p-1}\left(\frac{k^*+1}p\right)$$ where $k^*$ denotes the modulo $p$ inverse of $k$.