= Norm-six equation in the quadratic field of square root ten
{title2=$x^2-10y^2=6$}
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 solutions
$$
x+\sqrt{10}y=\pm\epsilon^n(\epsilon+(-1)^n),\qquad \epsilon=3+\sqrt{10},\quad n\in\mathbb Z.
$$
Restriction 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.
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