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For the Pell's equation where $d=2$:
$$x^2-2y^2=1$$ What are all the integer solutions to the equation. Apparantly there are infinitely many solutions, but how would I represent them in an expression?

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  • $\begingroup$ I don't believe there are infinitely many. I believe there are $2$. What have you attempted? $\endgroup$
    – emka
    Commented Jan 13, 2017 at 3:18
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    $\begingroup$ Maybe you can try using continued fractions. :) $\endgroup$
    – awllower
    Commented Jan 13, 2017 at 3:20
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    $\begingroup$ @emka there are definitely more than $2$ (OP is correct in the statement that there are infinitely many). Here are a few: $(x,y) = (\pm 1,0)$, $(\pm 3, \pm 2)$, $(\pm 17, \pm 12)$. See here for an explanation of awllower's comment. $\endgroup$
    – Stahl
    Commented Jan 13, 2017 at 3:22
  • $\begingroup$ I'm caught up now. I remembered this as an example from a number theory course (a few years ago). The (x,y) pairs will be good approximations for $\sqrt{2}$. $\endgroup$
    – emka
    Commented Jan 13, 2017 at 3:25
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    $\begingroup$ Please do not do that again. I will suspend you immediately if you do. $\endgroup$ Commented Jan 20, 2017 at 1:57

2 Answers 2

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Just note that for any pair $(a,b)$ giving a solution to $x^2-2y^2=1$, you have $$(a+b\sqrt{2})(a-b\sqrt{2})=1.$$ That is, $a+b\sqrt{2}$ is a unit in $\mathbb{Z}[\sqrt{2}]$. The units in $\mathbb{Z}[\sqrt{2}]$ are well known, they are powers $\pm (1+\sqrt{2})^n$, with $n\in \mathbb{Z}$. For instance, $$(1+\sqrt{2})^4=17+12\sqrt{2}$$ gives you a solution $(a,b)=(17,12)$. For more details, see e.g. here : The units of $\mathbb Z[\sqrt{2}]$.

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You can use continued fractions to find natural solutions. $\frac{x}{y} = [1;2,2,2,2,2,..... ] = \frac{p}{q}$ gives pairs (p,q) satisfying the Pell's equation.

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