Question

Find the general solution in positive integers of $x^2-5y^2=1$.

Answer 1

Given equation is $x^2-5y^2=1$.

$\sqrt{5}=2+\frac{1}{4+}\frac{1}{4+}....$

Penultimate convergent$=2+\frac{1}{4}=\frac{9}{4}$

$x=9,y=4$ is a solution.

$x^2+5y^2={(9^2-5.{,4}^2)}^n={(9+4\sqrt{5})}^n{(9-4\sqrt{5})}^n$

Therefore, $2x={(9+4\sqrt{5})}^n+{(9-4\sqrt{5})}^n$

$\Rightarrow 2\sqrt{5}y={(9+4\sqrt{5})}^n-{(9-4\sqrt{5})}^n$