Prove that all the positive solutions of $x^2 − 6y^2 = 1$ are produced by $r^n = (5 + 2\sqrt{6})^n$ where $n$ is a natural number.
We first find the least positive solution of $x^2 -6y^2=1$.
$y=1$ does not give an integer solution, but $y=2$ gives $x=5$. This is produced by $r=(5+2\sqrt{6})$.
By Theorem (8.21), all positive integer solutions are given by natural number powers of the least positive solution, that is,
$$ (5+2\sqrt{6})^n $$
where $n$ is a natural number.