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

 $\displaystyle\left(\frac{2}{p}\right)$ $\displaystyle=$ $\displaystyle 1\text{ if }p\equiv\pm 1\pmod{8}$ (1) $\displaystyle\left(\frac{2}{p}\right)$ $\displaystyle=$ $\displaystyle-1\text{ if }p\equiv\pm 3\pmod{8}$ (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 arguments about character sums.

Let $\sigma$ be a primitive eighth root of unity in an algebraic closure of $\mathbb{Z}/p\mathbb{Z}$, and write $\tau=\sigma+\sigma^{-1}$. We have $\sigma^{4}=-1$, whence $\sigma^{2}+\sigma^{-2}=0$, whence

 $\tau^{2}=2\;.$

By the binomial formula, we have

 $\tau^{p}=\sigma^{p}+\sigma^{-p}\;.$

If $p\equiv\pm 1\pmod{8}$, this implies $\tau^{p}=\tau$. If $p\equiv\pm 3\pmod{8}$, we get instead $\tau^{p}=\sigma^{5}+\sigma^{-5}=-\sigma^{-1}-\sigma=-\tau$. In both cases, we get $\tau^{p-1}=\left(\frac{2}{p}\right)$, proving (1) and (2).

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

 $\sigma=\exp(2\pi i/8)$
 $\tau=\sigma+\sigma^{-1}$

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

 $\tau^{p-1}\equiv\left(\frac{2}{p}\right)\pmod{p}$

which is enough.

Title quadratic character of 2 QuadraticCharacterOf2 2013-03-22 13:58:03 2013-03-22 13:58:03 mathcam (2727) mathcam (2727) 5 mathcam (2727) Theorem msc 11A15 ValuesOfTheLegendreSymbol