Exercise 9.4

Proof. $\lambda =1-\omega$, so $\omega \equiv 1(\mathit{mod}\lambda )$. Thus

Moreover $9=3\lambda \bar{\lambda }\equiv 0(\mathit{mod}3\lambda )$, thus $\gamma$ is congruent modulo $3\lambda$ to an integer between $0$ and $8$ of the form $3k-1$ : $\gamma \equiv 8,2$ or $5(\mathit{mod}3\lambda )$.

By Ex. 9.3, ${\chi }_{\gamma }(\omega )=1⟺m+n\equiv 0[3]$, and $m+n\equiv 0[3]$ implies $m+n=3k,k\in \mathbb{Z}$, so $\gamma \equiv -1+9k\equiv -1\equiv 8[3\lambda ]$.

Conversely, if $\gamma \equiv 8\equiv -1[3\lambda ]$, then $3\lambda \mid 3(m+n)$, so $\lambda \mid m+n$, and $N(\lambda )\mid N(m+n)$, $3\mid {(m+n)}^2$, thus $3\mid m+n$, $m+n\equiv 0[3]$, and so ${\chi }_{\gamma }(\omega )=1$. The two other cases are similar, so we obtain

If $\gamma =q$ is a rational prime, $q\equiv 8[9]$ implies $q\equiv 8[3\lambda ]$, since $3\lambda \mid 9=3\lambda \bar{\lambda }$, thus ${\chi }_q(\omega )=1$.

Conversely, if ${\chi }_q(\omega )=1$, then $q\equiv 8[3\lambda ]$, $q-8=\mu (3\lambda ),\mu \in D$, therefore

${(q-8)}^2=N(\mu )3^3,3^3\mid {(q-8)}^2$, thus $3^2\mid q-8$ and so $q\equiv 8[9]$. The two other cases are similar.