<s>[INST] 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!
 [/INST] Let's say $\alpha_n$ and $\beta_n$ the $n$-solution of the equation $x^2 - 2y^2 = 2$. We have: $$\left\{\begin{matrix}
\alpha_0 = 2
\\\beta_0 = 1
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_1 = 10
\\\beta_1 = 7
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_2 = 58
\\\beta_2 = 41
\end{matrix}\right.$$
From here, you can deduce the recoursive relation: 
$$\left\{\begin{matrix}
\alpha_{n} = 3\alpha_{n-1}+4\beta_{n-1}
\\\beta_{n} = 2\alpha_{n-1}+3\beta_{n-1}
\end{matrix}\right.$$
This is the list of the values of $\alpha_n$ and $\beta_n$:

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