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!
Asked
Active
Viewed 146 times
-2
-
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 07:59
-
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 08:05
-
Older posts about this: $\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$ and Proof Involving Legendre Symbol and Quadratic Residue Multiplication Found using Approach0. – Martin Sleziak May 27 '18 at 09:07
-
Possible duplicate of $\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$ – Martin Sleziak May 27 '18 at 09:07
1 Answers
0
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 $k$.
Angina Seng
- 158,341