Playing with Coordinates
What is the maximum number of points with rational coordinates on the circumference of a circle with centre at ( 3 , 0 ) ?
The answer is 2.
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3 solutions
A potential follow-up question: What is the maximum if the center is ( π , e ) ?
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Following the strategy of my other solution: x 2 + y 2 = − ( π 2 + e 2 ) + 2 x π + 2 y e + r 2 . To make both sides rational, r 2 must be of the form r 2 = ( π 2 + e 2 ) + a π + b e + c , with rational coefficients a , b , c ; so that ( x 2 + y 2 − c ) − ( 2 x + a ) π − ( 2 y + b ) e = 0 . Because 1 , π , and e are linearly independent over Q , all three terms must be zero. The system { 2 x + a = 0 ; 2 y + b = 0 } has at most one solution.
The existence of such a maximum solution is easily proved. Choose a = b = c = 0 , so that r = π 2 + e 2 ; then ( 0 , 0 ) lies on the circle.
1
The equation of the circle is ( x − 3 ) 2 + y 2 = r 2 . ∴ x 2 + y 2 + 3 = 2 3 x + r 2 . The left side of the equation is rational, and therefore the right side of the equation must be as well. This limits the possible values of the radius r to r 2 = a 3 + b , with a and b rational numbers. The equation of the circle becomes x 2 + y 2 − b + 3 = ( 2 x + a ) 3 ; since both x 2 + y 2 − b + 3 and 2 x + a are rational, this equation can only be satisfied if both sides of the equation are zero. The equation 2 x + a = 0 cannot have more than one solution for x , and for every x coordinate there are no more than two y values.
(Of course we ought to prove that a solution with two such points actually exists. The circle centered at 3 , 0 and radius 2 contains the points 0 , ± 1 , so that we made our case.)
Great strategy, particularly when applied to the ( π , e ) case, (mine doesn't narrow things down to a maximum of one for that case). I just thought I should point out that your second and fourth equations should be, respectively, x 2 + y 2 + 3 = 2 3 x + r 2 , and x 2 + y 2 − b + 3 = ( 2 x + a ) 3 .
Here is a brief sketch for a different proof..
1) Lemma: For any three distinct non-collinear points with rational coordinates, the circumcentre is also a rational number.
2)Using this lemma, we conclude that the circle centred at ( 3 , 0 ) has at most 2 " rational points " on it.
3) Finally, a circle of radius 2 does the trick; it cuts the y-axis in ( 0 , 1 ) , ( 0 , − 1 ) .
For any two rational points A , B , the line segment A B has rational slope, and the midpoint is a rational point as well. In addition, the perpendicular bisector of A B will have rational slope since it will be the negative reciprocal of the slope of A B .
Now suppose a circle centered at ( 3 , 0 ) has three or more rational points on its circumference. Choose three of them and label them P , Q , R . The perpendicular bisectors of chords P Q and Q R will intersect at the center of the circle. From the discussion above, we see that the point of intersection of these two lines will be the solution of two linear equations in x , y with rational coefficients, which implies that the point of intersection, i.e., the center, will be a rational point as well. But the center ( 3 , 0 ) is not rational, and so by contradiction there cannot be three (or more) rational points on the circumference of any circle centered at ( 3 , 0 ) . In other words, there can be a maximum of 2 rational points on the circumference. This maximum can be achieved, as an example, for a circle of radius 2 , the two rational points on the circumference being ( 0 , 1 ) and ( 0 , − 1 ) , as both satisfy the circle equation ( x − 3 ) 2 + y 2 = 2 2 .