quadratic character of 2

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

$\left({\displaystyle \frac{2}{p}}\right)$$=$$1\text{ if }p\equiv \pm 1(mod8)$(1)

$\left({\displaystyle \frac{2}{p}}\right)$$=$$-1\text{ if }p\equiv \pm 3(mod8)$(2)

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

Let $\sigma$ be a primitive eighth root of unity in an algebraic closureof $\mathbb{Z}/p\mathbb{Z}$, 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 $p\equiv \pm 1(mod8)$, this implies ${\tau }^p=\tau$.If $p\equiv \pm 3(mod8)$, 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

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

which is enough.