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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!

  • 0
    By the way, this is very similar to [this post](https://math.stackexchange.com/questions/2095694/find-all-integer-solutions-to-x2-2y2-1?rq=1).2018-05-20

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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.

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    That would also give you solutions for $x^2-2y^2=2$. To solve $x^2-2y^2=-2$, you need a particular solution of $x^2-2y^2=-2$ with a solution of $x^2-2y^2=1$, or a particular solution of $x^2-2y^2=2$ with a solution of $x^2-2y^2=-1$. Think about how these solutions work, about the norms you get!2012-06-24
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Pell's Theorem is only valid for integer coefficients - so no.

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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 $Y_n^2 -2 X_n^2 = -1$