Exercise 9.14

Proof. If $\pi$ is associate to 5, then $5^3\equiv 0\equiv 5(\mathit{mod}\pi )$, so $x^3\equiv 5(\mathit{mod}\pi )$ is solvable.

If $\pi$ is a primary prime not associate to 5, the Law of Cubic Reciprocity gives

(see Ex. 9.13)

Conclusion: the equation $5\equiv x^3[\pi ],x\in D$ is solvable iff the primary prime associate to $\pi$ is congruent modulo 5 to $1,2,3,4,1+2\omega ,2+4\omega ,3+\omega ,4+3\omega$ (or 0).

Examples:

$\text{∙}$ $q=23$ is a primary prime congruent to 3 modulo 5, thus the equation $x^3\equiv 5(\mathit{mod}23)$ has a solution $x\in D$ ( $x=19)$.

$\text{∙}$ $-4-3\omega$ is the primary prime associate to the prime $3-\omega$, and $-4-3\omega \equiv 1+2\omega (\mathit{mod}5)$, thus the equation $x^3\equiv 5(\mathit{mod}3-\omega )$ has a solution $a+\mathit{b\omega }\in \mathbb{Z}[\omega ]$.

Indeed, $7^3\equiv 8^3\equiv 11^3\equiv 5(\mathit{mod}13)$, and $3-\omega \mid 13$, so $7^3\equiv 8^3\equiv 11^3\equiv 5(\mathit{mod}3-\omega )$. □