Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2 =1$  $y=3, x=4$, so form Pell I get that $\alpha= (4+3\sqrt{2})^n$ for every integer $n$, and a private solution for $-\frac{1}{2}x^2+ y^2 =-1$  is  $y=1, x=2$, so the total solution is $\alpha= (1+\sqrt{2}) \cdot  (+/- (4+3\sqrt{2})^n)$. Are all these steps correct? and if not- how should I solve this one?
Thank you!

The standard way is to find one solution to $x^2-2y^2=2$, e.g., $x=2$, $y=1$, and find the fundamental solution to $x^2-2y^2=1$, which is $x=3$, $y=2$, and then go $(2+\sqrt2)(3+2\sqrt2)^n$, etc., etc.
