(i) Check that $r = 17 + 12\sqrt{2}$ yields a solution of $x^2 − 2y^2 = 1$.
(ii) Hence or otherwise find a solution of $u^2− 8v^2 = 1$.
(iii) Hence or otherwise find a solution of $u^2 − 32v^2 = 1$.
(iv) Determine the condition on $m$ for $u^2− 2 (mv)^2 = 1$ to have a solution.
(i) From $r=17 + 12\sqrt{2}$ we have $x=17$ and $y=12$.
$$ x^2 -2y^2 = 17^2 -2(12^2) = 289 - 288 = 1 $$
So $r=17 + 12\sqrt{2}$ is a solution to $x^2 − 2y^2 = 1$.
(ii) We write $u^2− 8v^2 = 1$ as $u^2− 2(2v)^2 = 1$.
We have the solution $u=17, 2v=12$, which is $u=17, v=6$.
(iii) We write $u^2− 32v^2 = 1$ as $u^2− 2(4v)^2 = 1$.
We have the solution $u=17, 4v=12$, which is $u=17, v=3$.
(iv) From the solution $(17,12)$ we have $mv=12$, so the condition on $m$ is that it is a factor of 12.