Question

Find the general solution in positive integers of $x^2-3y^2=1$.

Answer 1

We know $\sqrt{3}=1+\sqrt{3}-1=1+\frac{2}{\sqrt{3}+1}$

$\frac{\sqrt{3}+1}{2}=1+\frac{\sqrt{3}-1}{2}=1+\frac{1}{\sqrt{3}+1}$

$\sqrt{3}+1=2+\sqrt{3}-1=2+\frac{2}{\sqrt{3}+1}$

Therefore, $\sqrt{3}=1+\frac{1}{1+}\frac{1}{2+}\frac{1}{1+}\frac{1}{2+}....$

Penultimate convergent$=2$

$x=2,y=1$ is a solution

Hence, $x^2-3y^2={(2^2-3)}^n;$

$(2+\sqrt{3}y)(x-\sqrt{3}y)={(2+\sqrt{3})}^n{(2-\sqrt{3})}^n$

$2x={(2+\sqrt{3})}^n+{(2-\sqrt{3})}^n$

$2\sqrt{3}y={(2+\sqrt{3})}^n-{(2-\sqrt{3})}^n$