# Find all integer solutions to $x^2-2y^2=1$ [closed]

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?

• I don't believe there are infinitely many. I believe there are $2$. What have you attempted?
– emka
Commented Jan 13, 2017 at 3:18
• Maybe you can try using continued fractions. :) Commented Jan 13, 2017 at 3:20
• @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. Commented Jan 13, 2017 at 3:22
• 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 $\sqrt{2}$.
– emka
Commented Jan 13, 2017 at 3:25
• Please do not do that again. I will suspend you immediately if you do. Commented Jan 20, 2017 at 1:57

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 $\mathbb Z[\sqrt{2}]$.
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.