Let $K$ denote the number field $\mathbb{Q}(\sqrt{15}).$ According to standard lore, we have that $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{15}]$. Moreover,
$2\mathcal{O}_{K} = \langle 2, 1+\sqrt{15}\rangle^{2}$
and
$3\mathcal{O}_{K} = \langle 3, \sqrt{15}\rangle^{2}.$
Resorting to the law of quadratic reciprocity, I have proven that the ideal $\langle 3, \sqrt{15}\rangle$ is non-principal. Since $h(\mathbb{Q}(\sqrt{15}))=2$, it must be the case that
$\langle 2, 1+\sqrt{15}\rangle \langle \alpha \rangle = \langle 3, \sqrt{15}\rangle \langle \beta \rangle$
for certain $\alpha, \beta \in \mathcal{O}_{K}\setminus \{0\}$.
Is there a quick way to determine such a pair $(\alpha, \beta)$?
Thanks!