Question

Find the general solution in positive integers of $x^2-17y^2=-1$.

Answer 1

Given equation is $x^2-17y^2=-1$.

$\sqrt{17}=4+\frac{1}{8+}\frac{1}{8+}....$

$\therefore x=4,y=1$ is a solution.

($\because$ Penultimate convergent is $\frac{4}{1},\because$ RHS$=-1$)

Thus, $(x+y\sqrt{17})(x-y\sqrt{17})={(4+\sqrt{17})}^n{(4-\sqrt{17})}^n$

$n$ $ϵ$ odd positive integer

Thus $2x={(4+\sqrt{17})}^n+{(4-\sqrt{17})}^n$

$\Rightarrow 2\sqrt{17}y={(4+\sqrt{17})}^n-{(4-\sqrt{17})}^n$