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Are there solutions in $\mathbb{Q}-\mathbb{Z}$ to $X^2+2y^2=1$?

I'm pretty sure that there isnt but Im not sure how to show this. I don't have much experience (yet) with $p$-adics.

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@Jason: please accept answers to your previous questions. It is the polite response and shows that you appreciate the help you are being given. – Qiaochu Yuan Dec 7 '10 at 22:06
How do I do this? Id love to do so. – Jason Smith Dec 7 '10 at 22:38
@Jason Smith: There should be a "transparent" check mark on the left, below the up and down arrows for voting. You can accept one answer per question. – Arturo Magidin Dec 7 '10 at 22:42
Ok, Ive accepted the answers. Thanks. – Jason Smith Dec 7 '10 at 22:46
@Jason Smith: It means it is directed at you; it also lets the system know. Although you'll get notified of any comments made to your questions or answers, you will not in other cases. Had you preceded the previous comment with @Arturo Magidin, I would have been told about it. As it is, I noticed it only because I came back to the page. – Arturo Magidin Dec 7 '10 at 22:50
up vote 12 down vote accepted

Yes, there are solutions. For example $({7 \over 11}, {6 \over 11})$ solves it. In general one can find such solutions to $x^2 + ry^2 = 1$ whenever $r$ is a rational number. One can do this by intersecting a line $ax + by = c$ containing $(1,0)$ ($a,b,$ and $c$ rational) with your ellipse $x^2 + ry^2 = 1$; the second intersection point will also have rational coordinates. Since only finitely many of these will have an integer coordinate the rest of such points will have both their entries in $Q - Z$.

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Yes. E.g. $X = 1/3$, $Y = 2/3$.

[Added: Carrying out the procedure suggested in Zaricuses's answer, i.e. intersecting the line with slope $-p/q$ through $(1,0)$ with the conic $X^2 + 2 Y^2 =1$, we find that the general solution is $X = (2p^2 - q^2)/(2 p^2 + q^2), Y = 2pq/(2 p^2 + q^2).$ The solution above comes form taking $p = 1, q = 1$. The solution in Zaricuse's answer comes from taking $p = 3, q = 2$.]

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