Exercise 6.8

Proof. As ${\omega }^2+\omega +1=0$, ${(2\omega +1)}^2=4{\omega }^2+4\omega +1=-4+1=-3$. Let $\alpha =2\omega +1$, so that ${\alpha }^2=-3$

From Prop. 6.1.6,

$\text{∙}$

If $p\equiv 0(\mathit{mod}3)$, $\left(\frac{-3}{p}\right)=0$.

$\text{∙}$

If $p\equiv 1(\mathit{mod}3)$, ${\omega }^p=\omega$, so ${\alpha }^p\equiv \alpha (\mathit{mod}p)$.

$\left(\frac{-3}{p}\right)\alpha \equiv \alpha (\mathit{mod}p)$, thus $\left(\frac{-3}{p}\right){\alpha }^2\equiv {\alpha }^2(\mathit{mod}p)$, $\left(\frac{-3}{p}\right)3\equiv 3(\mathit{mod}p)$. As $p\wedge 3=1$, $\left(\frac{-3}{p}\right)\equiv 1(\mathit{mod}p)$. Since $\left(\frac{-3}{p}\right)=\pm 1$, $\left(\frac{-3}{p}\right)=1$.

$\text{∙}$

If $p\equiv -1(\mathit{mod}3)$, ${\omega }^p={\omega }^{-1}={\omega }^2$, and $$\begin{array}{llll}\hfill {\alpha }^p& \equiv 2{\omega }^p+1(\mathit{mod}p)& \hfill & \\ \hfill & \equiv 2{\omega }^2+1=2(-1-\omega )+1=-2\omega -1=-\alpha (\mathit{mod}p).& \hfill & \\ \hfill & & \hfill \end{array}$$

$\left(\frac{-3}{p}\right)\alpha \equiv -\alpha (\mathit{mod}p)$, thus $\left(\frac{-3}{p}\right){\alpha }^2\equiv -{\alpha }^2(\mathit{mod}p)$, $\left(\frac{-3}{p}\right)3\equiv -3(\mathit{mod}p)$. As $p\wedge 3=1$, $\left(\frac{-3}{p}\right)\equiv -1(\mathit{mod}p)$. Since $\left(\frac{-3}{p}\right)=\pm 1$, $\left(\frac{-3}{p}\right)=-1$.

Conclusion :

In other words, $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$.

Note : $\alpha =2\omega +1=\omega -{\omega }^2=g$ is the quadratic Gauss sum for $p=3$. □