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.
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.
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 $Q - Z$.
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 $p = 3, q = 2$.]