Find the maximum number of rational points on the circle with center $(0,\sqrt3)$
Let the equation of the circle be $x^2+(y-\sqrt3)^2=r^2$
Let $(a,b)$ be any rational point on the circle $x^2+(y-\sqrt3)^2=r^2$.
Then $a^2+(b-\sqrt3)^2=r^2$.
$a^2+b^2+3-2b\sqrt3=r^2$
How can i find the maximum number of rational points from this equation.I have no idea.Can someone please elaborate?
Best Answer
Let $\sigma$ be the nontrivial automorphism of $\Bbb Q[\sqrt 3]$.
If $P$ is a point with rational coordinates, the circle $C_1$ with center $(0,\sqrt 3)$ passing through $P$ is given by an equation with coefficients in $\Bbb Q[\sqrt 3]$.
But if $Q$ has rational coordinates and is on $C_1$, then $\sigma(Q)$ is on $\sigma(C_1)$, so $Q$ is also on the circle of center $(0,-\sqrt 3)$ passing through $P$.
Since those two circles intersect at at most two points, (well, $P$ and its symmetric opposite the $y$ axis), there can be at most two rational points on $C_1$.
Alternatively, if you have $3$ rational points on a circle then its center (the circumcenter of the triangle) has to have rational coordinates too, and then it can't be $(0,\sqrt 3)$.