TITLE: Showing that there are infinitely many integer solutions for the hyperbolic formula $|a^2 - 26 b^2| = 1$
QUESTION [5 upvotes]: I want to show that the formula 
$$ | a^2 - 26\cdot b^2| = 1$$
has infinitely many solutions $(a, b) \in \mathbb{Z}^2$.
First I tried to solve the formula for one of the two variables, to get something like $a = ±\sqrt{±1 + 26 b^2}$ or $b = \sqrt{-\frac{±1 - a^2}{26}} $, but both doesn't look very promising, seeing as I would now need to prove that there are infinitely many integer solutions for $b$ (or $a$) so that the respective other one is also an integer.
I did, however, get some solutions. I could (by more or less trying out values) find out that $(5, 1), (51, 10)$ are two solutions. (And therefore also $(-5, 1)$ and so on.) So maybe there is a recursive way to get new values, based on the ones I already have, i.e. a formula $\phi(a, b)$ so that, when $(a, b)$ is an integer solution, $\phi(a, b)$ is aswell? (We would probably need an extra condition like that the components of $\phi(a, b)$ are strictly larger than $a, b$, so that we really get new solutions and not the ones we already had before.)
Seeing as I don't actually want to find a formula or so for the actual solutions, it would also be sufficient if there's a proof that there are infinitely many solutions that doesn't involve actually finding any of the solutions. But so far, I haven't had any luck with that.
(I wasn't entirely sure what tags to choose; the problem seemed analytical to me if other tags are seen as more suiting, please feel free to correct them.)

REPLY [3 votes]: If $\alpha=a+b\sqrt{26}$, let $N(\alpha)=a^2 - 26b^2$. Then $N(\alpha\beta)=N(\alpha)N(\beta)$.
Now, by inspection, if $\theta = 5+\sqrt{26}$,then $N(5+\sqrt{26})=-1$.
Therefore, $|N(\theta^n)|=|N(\theta)^n|=1$ and so $\theta^n$ gives you an infinitude of solutions.
Note that $\theta^n=a_n + b_n \sqrt{26}$, with $a_n,b_n \in \mathbb Z$. In fact,
$$
\begin{array}%
a_{n+1} =& 5a_n+26b_n, && a_0 =& 1 \\
b_{n+1} =& a_n + 5b_n, && b_0 =& 0
\end{array}
$$