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A circle is centred at $(\pi,e)$. What is the maximum no. of rational points it can have? (A rational point is one with both coordinates rational). 1 rational point is definitely possible, just choose any rational point, and alter the radius to get it through. My book says that only one rational point is possible, as $\pi\neq qe\quad q\in Q$. That's their whole explanation. I don't understand how that's enough. Edit: It has been pointed out that the problem is equivalent to showing $q_1\pi+q_2e=q_3$ has no non trivial solutions. Is this known to be true? Can someone prove it in an elementary way?

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Need a little more, that $1$, $\pi$, $e$ are linearly independent over the rationals. And it is non-obvious even that $\pi\ne qe$. – André Nicolas Feb 27 '13 at 10:37
Here's a better answer. – A.P. Dec 28 '15 at 13:09
up vote 5 down vote accepted

Okay I got it, suppose it passes through (a,b). Then the equation of circle is $x^2-a^2+y^2-b^2-2\pi( x-a)-2e(y-b)=0$ If x and y are both rational then $q_1\pi+q_2e=q_3$ with not everything 0 .I still have to prove this impossible. I don't think it's equivalent to $\pi \neq qe$. Edit: As has been pointed out to me, two rational points are not possible if $\pi $ and $ e$ are linearly independent over the rationals, and this is still an open problem

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The problem is that this is not known to be impossible. – Harald Hanche-Olsen Feb 27 '13 at 10:39
We don't need algebraic independence, just linear independence in this problem, right? Is that proven? – Ishan Banerjee Feb 27 '13 at 10:47
Your book is wrong : there are several rational points on the circle centered on $(\sqrt 2, 1+\sqrt 2)$ and radius $\sqrt2$ (for example, $(0,1+ \sqrt 2)$ and $(\sqrt 2,1)$), even though $\sqrt 2 / (1+ \sqrt 2) \notin \Bbb Q$ – mercio Feb 27 '13 at 11:13
$(\sqrt{2},1)$ isn't a rational point by my definition. Though I admit the book is not to be trusted. – Ishan Banerjee Feb 27 '13 at 11:16
If you followed the link I provided, you would see that it is not even known whether $e+\pi$ is rational. – Harald Hanche-Olsen Feb 27 '13 at 12:02

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