Quadratic reciprocity

**alexmahone** · July 21st 2011, 12:48 AM

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

My attempt:

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

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

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

**melese** · July 21st 2011, 02:51 AM

Originally Posted by alexmahone

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

My attempt:

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

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

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

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

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

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