Question

Find the smallest solution in positive integers of $x^2-19y^2=1$.

Answer 1

$\sqrt{19}=4+\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}......;$

Here the penultimate convergent is $\frac{170}{39}$;

Thus $x=170$ & $y=39$ is the smallest solution