## Solution of a Quadratic Diophantine Equation

\[x^2-15y^2=1\]

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 \[15^2-14 \times 4^2=1\]

.