• Jul 21st 2011, 12:48 AM
alexmahone
Prove that $\displaystyle \displaystyle\sum\limits_{j=1}^{p-1} \left(\frac{j}{p}\right)=0$, p an odd prime.

My attempt:

If $\displaystyle p \equiv 3\ (mod\ 4)$,

$\displaystyle \left(\frac{-j}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{j}{p}\right)=(-1)^{(p-1)/2}\left(\frac{j}{p}\right)=-\left(\frac{j}{p}\right)$ and the result follows immediately.

But what if $\displaystyle p \equiv 1\ (mod\ 4)$?
• Jul 21st 2011, 02:51 AM
melese
Quote:

Originally Posted by alexmahone
Prove that $\displaystyle \displaystyle\sum\limits_{j=1}^{p-1} \left(\frac{j}{p}\right)=0$, p an odd prime.

My attempt:

If $\displaystyle p \equiv 3\ (mod\ 4)$,

$\displaystyle \left(\frac{-j}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{j}{p}\right)=(-1)^{(p-1)/2}\left(\frac{j}{p}\right)=-\left(\frac{j}{p}\right)$ and the result follows immediately.

But what if $\displaystyle p \equiv 1\ (mod\ 4)$?

Theorem: For an odd prime $\displaystyle p$ there are exactly $\displaystyle (p-1)/2$ quadratic residues and exactly $\displaystyle (p-1)/2$ quadratic nonresidues.*

So restating the theorem, half of the terms $\displaystyle (1/p), (2/p), ..., ((p-1)/p)$ are $\displaystyle -1$ and half are $\displaystyle 1$.