Question

Solve in positive integers $x^2-13y^2=\pm 1$.

Answer 1

We can show that
$\sqrt{13}=3+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{6+\cdots }}}}}$
Here the number of quotients in the period is odd; the penultimate convergent in the first period is $\frac{18}{5}$; hence $x=18$, $y=5$ is a solution of $x^2-13y^2=-1$.
The penultimate convergent in the second recurring period is
$\frac{1}{2}(\frac{18}{5}+\frac{5}{18}\times 13)$, that is, $\frac{649}{180}$;
Hence $x=649$, $y=180$ is a solution of $x^2-13y^2=1$.
By forming the successive penultimate convergent of the recurring periods we can obtain any number of solutions of the equations
$x^2-13y^2=-1$, and $x^2-13y^2=+1$.