Does there exist $X$, $Y \in\mathbb{Q}-\mathbb{Z}$ such that $X^2+2Y^2=1$?

Exist remedies in $\mathbb{Q}-\mathbb{Z}$ to $X^2+2y^2=1$?

I'm rather certain that there isn't yet Im not exactly sure just how to show this. I do not have much experience (yet) with $p$ - adics.

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2019-12-02 03:12:28
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Answers: 2

Yes, there are remedies. As an example $({7 \over 11}, {6 \over 11})$ addresses it. As a whole one can locate such remedies to $x^2 + ry^2 = 1$ whenever $r$ is a sensible number. One can do this by converging a line $ax + by = c$ having $(1,0)$ ($a,b,$ and also $c$ sensible) with your ellipse $x^2 + ry^2 = 1$ ; the 2nd junction factor will certainly additionally have sensible works with. Given that just finitely most of these will certainly have an integer work with the remainder of such factors will certainly have both their access in $Q - Z$.

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2019-12-03 05:08:56
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Yes. E.g. $X = 1/3$, $Y = 2/3$.

[Included : Carrying out the procedure recommended in Zaricuses is solution, i.e. converging the line with incline $-p/q$ via $(1,0)$ with the conic $X^2 + 2 Y^2 =1$, we locate that the basic remedy is $X = (2p^2 - q^2)/(2 p^2 + q^2), Y = 2pq/(2 p^2 + q^2).$ The remedy over comes kind taking $p = 1, q = 1$. The remedy in Zaricuse is solution originates from taking $p = 3, q = 2$. ]

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2019-12-03 05:07:48
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