Pell's Equation 41

The continued fraction expansion of $\sqrt{41} = \big[6,\overline{2,2,12}\big]$ has period $3$ , which is odd. Thus the fundamental solution of the equation $p^2 - 41q^2 \; = \; -1$ is derived from the second convergent $\tfrac{32}{5}$ of the continued fraction expansion of $\sqrt{41}$ , so that $p=32$ and $q=5$ . Thus the fundamental solution of the equation $p^2 - 41q^2 \; = \; 1$ comes from the square $(32 + 5\sqrt{41})^2 \,=\, 2049 + 320\sqrt{41}$ , so that $p=2049$ and $q=320$ . Thus $p+q = \boxed{2369}$ .

$x^2-41y^2=1 \\ \frac{x^2}{y^2}-41=\frac{1}{y^2} \\ \frac{x^2}{y^2}=41+\frac{1}{y^2}$

As $y^2$ increases exponentially, the value of $\frac{1}{y^2}$ gets negligible. So we can write the equation as

$\frac{x^2}{y^2} \approx 41 \\ \frac{x}{y} \approx \sqrt{41}$

So we just need to check pairs of $x$ and $y$ such that $\frac{x}{y}$ converge to $\sqrt{41}$ .

To make the search easier, continued fraction of the square root can be found. Continued fraction appears to be $[6; \overline{2, 2, 12}]$ . On testing a few partial sums, it can be found that $x = 2049$ and $y = 320$ works.