quadratic character of 2

For any odd prime p, Gauss’s lemma quickly yields

(2p) = 1 if p±1(mod8) (1)
(2p) = -1 if p±3(mod8) (2)

But there is another way, which goes back to Euler, and is worth seeing, inasmuch as it is the prototype of certain more general argumentsMathworldPlanetmath about character sums.

Let σ be a primitive eighth root of unityMathworldPlanetmath in an algebraic closureMathworldPlanetmath of /p, and write τ=σ+σ-1. We have σ4=-1, whence σ2+σ-2=0, whence


By the binomial formula, we have


If p±1(mod8), this implies τp=τ. If p±3(mod8), we get instead τp=σ5+σ-5=-σ-1-σ=-τ. In both cases, we get τp-1=(2p), proving (1) and (2).

A variation of the argument, closer to Euler’s, goes as follows. Write


Both are algebraic integersMathworldPlanetmath. Arguing much as above, we end up with


which is enough.

Title quadratic character of 2
Canonical name QuadraticCharacterOf2
Date of creation 2013-03-22 13:58:03
Last modified on 2013-03-22 13:58:03
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 5
Author mathcam (2727)
Entry type Theorem
Classification msc 11A15
Related topic ValuesOfTheLegendreSymbol