\[x^2-15y^2=1\]
using Continued fractions - specifically the convergents of the continued fraction representing \[\sqrt{14}\]
.\[\sqrt{14}\]
is represented by the continued fraction \[[3,1,2,1,6,1,2,1,6,1,2,1,6 ] = [3 \lt 1,2,1,6 \gt ]\]
. The first few convergents are
\[3, \; 3+ \frac{1}{1}=4, \; 3+ \frac{1}{1+ \frac{1}{2}} = \frac{11}{3}, \; 3+\frac{1}{1+ \frac{1}{2+ \frac{1}{1}}}= \frac{15}{4}\]
.Convergents up to the the last but one part of the recurring part - the second one here - gives us solutions of the equation.
We have
\[x=15, \; y=4\]
and \[15^2-14 \times 4^2=1\]
.