 Norm-six equation in the quadratic field of square root ten

The ring of integers is $\mathbb{Z}[\sqrt{10}]$, its class number is two, and its units are $\pm (3+\sqrt{10}{)}^n$ for $n\in \mathbb{Z}$. Factoring a norm-six principal ideal above 2 and 3 gives all solutionsRestriction to nonnegative $n$ gives only the positive-coordinate branch up to simultaneous sign. For example, $4-\sqrt{10}$ is a norm-six element requiring $n=-1$; adding conjugates repairs a nonnegative-index presentation.