13is it only one??...i m not sure[3]
what is the max no of rational pts on a circle's circumference which is centred at (Ï€,e)
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13(x-Ï€)2+(y-e)2=a2
x2-2Ï€x+Ï€2+y2-2ey+e2=a2
when a2=Ï€2+e2
then it passes through (0,0)
this not a proof i think[3] may b just a hint...
341abhirup has shown that there exist circles that have (\pi,e) as centre and have at least one rational point [both x and y coordinates are rational numbers]
We will now prove that there exists at most one such point.
Suppose having found a point (x_1,y_1) that is a rational point on this circle. Now, suppose there exists another point (x_2,y_2) that is also rational.
The equation of the circle is (x-\pi)^2+(y-e)^2 = r^2, so that
(x_1-\pi)^2+(y_1-e)^2 = (x_2-\pi)^2+(y_2-e)^2
from which we obtainx_1^2-x_2^2 + y_1^2-y_2^2 = 2[\pi(x_1-x_2)+e(y_1-y_2)]
Since LHS is rational we must have \pi(x_1-x_2)+e(y_1-y_2) = 0
Again, if x_1 \ne x_2; y_1 \ne y_2, we obtain \frac{\pi}{e} to be rational [which according to the authors of the problem must be a contradiction. I say this because i am not sure how to conclude that \frac{\pi}{e} is irrational, though a rational human would assume so :D]
341(...contd). So that means we must have x_1 = x_2 and y_1 = y_2 and that means at most one rational point exists.