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!

Another way is to note that $x$ has to be even. Hence if $x=2x_1$, we get that
$$2x_1^2-y^2 = 1 \implies y^2 - 2x_1^2 = -1 \implies \left(y + \sqrt2 x_1\right)\left(y - \sqrt2 x_1\right) = -1$$
Now clearly, one solution is $(x_1,y) = (1,1)$.
Raise both sides to any odd power, i.e.,
$$\left(y + \sqrt2 x_1\right)^{2n-1}\left(y - \sqrt2 x_1\right)^{2n-1} = (-1)^{2n-1} = -1$$
and note that
$$\left(y + \sqrt2 x_1\right)^{2n-1} = Y_n(y,x_1) + \sqrt2 X_n(y,x_1)$$
$$\left(y - \sqrt2 x_1\right)^{2n-1} = Y_n(y,x_1) - \sqrt2 X_n(y,x_1)$$
which in turn gives us infinite other solutions since 
$$Y_n^2 -2 X_n^2 = -1$$
