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When the origin is $(0,0)$, the number of lattice points on the circumference is directly related to the sum of squares function. And sum of square function can be arbitrary large according to its formula. I am wondering if the same holds for a more general case: given a rational origin, then either for arbitrary large $n$, there exist a circle with radius $r$ such that the number of lattice points on its circumference is larger than $n$; or there can only be at most $n$ points regardless or $r$.

My guess is the former, but my attempts on proving it with similar methods as $(0,0)$ case failed. Any help or reference is appreciated.

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  • $\begingroup$ An idea: multiply by the gcd $d$ of the coordinates of the origin, now you have an integral point. You need to show some kind of result that if I have a circle with $n$ points on it then "on average" $n/d^2$ of them have both coordinates divisible by $d$. $\endgroup$ – MCT Feb 15 '18 at 18:54
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    $\begingroup$ More number theoretically: given $d>0$ and a point $(a, b)$ I want to know that the number of ways to represent $n$ as $(dn-a)^2 + (dm-b)^2$ is arbitrarly large for any $d, a, b$. $\endgroup$ – MCT Feb 15 '18 at 19:00
  • $\begingroup$ @MCT That was one of my ideas where I got stuck. $\endgroup$ – happy fish Feb 16 '18 at 7:42

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