Exercise 9.23

Proof. Let $p$ be a rational prime, $p\equiv 1(\mathit{mod}3)$, then $p=\pi \bar{\pi }$, where $\pi \in D$ is a primary prime : $\pi =a+\mathit{b\omega }=3m-1+3\mathit{n\omega }$.

$\text{∙}$

Suppose that there exists $x\in \mathbb{Z}$ such that $x^3\equiv 3(\mathit{mod}p)$. Then $x^3\equiv 3(\mathit{mod}\pi )$, so ${\chi }_{\pi }(3)=1$. By Exercise 9.5, ${\omega }^{2n}={\chi }_{\pi }(3)=1$, thus $3\mid n$, therefore $9\mid b=3n$, namely $b=9B,B\in \mathbb{Z}$.

$p=\mathit{N\pi }=a^2+b^2-\mathit{ab},4p={(2a-b)}^2+3b^2=C^2+243B^2$, where $C=2a-b,B=b∕9$. So there exists $C,B\in \mathbb{Z}$ such that $4p=C^2+243B^2$.

$\text{∙}$

Conversely, suppose that there exist $C,B\in \mathbb{Z}$ such that $4p=C^2+243B^2$.

As $4p={(2a-b)}^2+3b^2=C^2+3{(9B)}^2$, from the unicity proved in Exercise 8.13, we obtain $b=\pm 9B$, so $9\mid b=3n,3\mid n$, and ${\chi }_{\pi }(3)={\omega }^{2n}=1$.

Thus there exists $x\in D$ such that $x^3\equiv 3(\mathit{mod}\pi )$. As $p\equiv 1(\mathit{mod}3)$, $D∕\mathit{\pi D}=\{\bar{0},\dots ,\overline{p-1}\}$, so there exists $h\in \mathbb{Z}$ such that $x\equiv h(\mathit{mod}\pi )$, and $h^3\equiv 3(\mathit{mod}\pi )$.

Therefore $p=\mathit{N\pi }\mid N(h^3-3)$, namely $p\mid {(h^3-3)}^2$, where $p$ is a rational prime, thus $p\mid h^3-3$ : there exists $x\in \mathbb{Z}$ such that $x^3\equiv 3(\mathit{mod}p)$.

Moreover $4p=C^2+243B^2$ implies $p\equiv 1(\mathit{mod}3)$.

$$(p\equiv 1[3]\mathrm{and}\exists <!--\; FUNCTION\; APPLICATION\; -->x\in \mathbb{Z},x^3\equiv 3[p])⟺\exists <!--\; FUNCTION\; APPLICATION\; -->C\in \mathbb{Z},\exists <!--\; FUNCTION\; APPLICATION\; -->B\in \mathbb{Z},4p=C^2+243B^2.$$