Exercise 9.15

Proof. Since $\pi \mid p$, if $x^3\equiv a(\mathit{mod}p),x\in \mathbb{Z}$, then $x^3\equiv a(\mathit{mod}\pi )$, thus ${\chi }_{\pi }(a)=1$.

Conversely, suppose that ${\chi }_{\pi }(a)=1$. Then the equation $y^3\equiv a(\mathit{mod}\pi )$ has a solution $y=u+\mathit{v\omega },u,v\in \mathbb{Z}$. Moreover, the class of $y$ has a representative $x\in \mathbb{Z}$ modulo $\pi$ (see the proof of Proposition 9.2.1) :

So $x^3\equiv a(\mathit{mod}\pi )$ has a solution $x\in \mathbb{Z}$.

Thus $\pi \mid x^3-a,N(\pi )=p\mid {(x^3-a)}^2$, therefore $p\mid x^3-a$ in $\mathbb{Z}$, and so $x^3\equiv a(\mathit{mod}p)$.

Conclusion: if $p\equiv 1(\mathit{mod}3)$, $p=\pi \bar{\pi }$, where $\pi$ is a primary prime, and $a\in \mathbb{Z}$,

In other words, $x^3\equiv a(\mathit{mod}\pi )$ is solvable in $D$ iff it is solvable in $\mathbb{Z}$. □