Rational points on a circle with centre $(\pi, e)$
Solution 1:
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