Continuing from On circles and ellipses drawn on an infinite planar square lattice, let us record two broad questions: In what follows, "contains" means "either contains within or passes thru".

  1. Given an integer n, to find the smallest circle that contains at least n integer lattice points.

For n =3 and 4, the required circles have same radius =2. Is there any parallelogram lattice for which the smallest circle that contains at least n lattice points n has a unique radius for every value of n?

  1. Given an integer n, to find the smallest perimeter ellipse that contains at least n integer points (that there is no closed expression known for perimeter of ellipse need not make this question invalid).

Only for n <=3, the required ellipses seem degenerate. What can one say about sequences of successive values of n for which the least perimeter ellipses containing at least n integer lattice points are the same throughout? Is there any parallelogram lattice for which the least perimeter ellipse that contains at least n lattice points n is unique for every value of n?

Note 1: Instead of at least n lattice points we could also look for containers that contain exactly n lattice points.

Note 2: Analogous questions are conceivable with other shapes of container such as triangle, isosceles triangle etc. I am not sure if for all n, there are equilateral triangles that contain exactly n integer lattice points and if so whether all these triangles have unique sizes.

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    $\begingroup$ In any parallelogram lattice centered at the origin, every origin centered circle contains an odd number of lattice points. The best you can do is that, for all n, the smallest circle for $n$ has smaller radius than that for $n+2.$ This is easy to achieve even with rectangles, let the lattice be generated by $(r,0)$ and $(0,1)$ with $r^2$ irrational. $\endgroup$ Feb 21 at 22:52
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    $\begingroup$ On a square lattice, doesn't radius 1 already get up to n = 5? $\endgroup$ Feb 22 at 14:34
  • $\begingroup$ Thank you. I understand that the circle that contains at least n integer lattice points is necessarily centered on a lattice point and as the circle grows in radius, the number of lattice points it contains is always odd. So, the circle case of the question seems to have a simple answer. Not quite sure if this property of center being at a lattice point applies also to least perimeter ellipses - and other centrally symmetric containers - that contain at least n lattice points. $\endgroup$ Feb 22 at 17:34
  • $\begingroup$ The smallest circle containing two points has center at $(\frac12,0)$ and the smallest containing three also contains four and has center at $(\frac12,\frac12)$ . So what you say isn't quite correct. It is easy to find a point $P$ so that a circle centered at $P$ has at most $1$ lattice point (indeed only one point with both coordinates rational) on the boundary. $P=(\sqrt{2},\sqrt{3})$ should work. $\endgroup$ Feb 24 at 1:17
  • $\begingroup$ Thanks for the correction! And with center chosen at a point with irrational coordinates, it appears that if we gradually expand the circle, the number of contained integer lattice points increases in steps of 1. Not sure about the behavior of the smallest circles that contain exactly n integer lattice points. Hope you could make a consolidated answer of your inputs. $\endgroup$ Feb 24 at 4:28

1 Answer 1


If you want for each $n$ the minimal radius of a disk containing at least $n$ points we have the easily found pairs $(n,r^2)=(1,0),(2,\frac14),(4,\frac12),(5,1),(6,\frac54).$

At this point I would be tempted to say that the center can be taken to be $(\frac{a}2,\frac{b}2)$ for $a,b$ integers. But it turns out I would be wrong. I'm not sure how much is known.

The circles shown below for $(6,\frac54),(9,2)$ and $(12,\frac52)$ do have centers of this form. However a slight enlargement of the first can be shifted to give $(7,\frac{25}{16})$ shown here with center $(\frac34,3).$

And the circle for $n=12$ can be enlarged to give circles for $(n,r^2)=(13,\frac{169}{50})$ and $(14,\frac{65}{18})$ shown here with centers $(\frac{43}{10},\frac{43}{10})$ and $(\frac{25}6,\frac{17}{2}).$

enter image description here

That is enough data to find that sequence $$ {1, 2, 4, 5,6}, 7, {9}, {12}, 13, 14 \cdots$$ and some illustrations.

LATER A slightly new question got added. Let $r_n$ be the minimal radius of any disk $D_n$ containing exactly $n$ (usual) lattice points. Then $r_3$ is undefined and $r_8=\frac32>\sqrt{2}=r_9.$ If $r_{10}$ and $r_{11}$ are both defined, then they are strictly between $r_9$ and $r_{12}.$ I wouldn't guess which is larger.

A disk of type $D_n$ will have a rational center: If it has two antipodal lattice points on the boundary the center has half integer coordinates. Otherwise there will be at least $3$ lattice points on the boundary (forcing a rational center.) Otherwise we could slightly shrink and shift to get the same points contained and a smaller radius.

For an appropriate fixed center $(x,y)$ (say $x,y,\frac{x}{y}$ all irrational) it should be the case that every disk with that center has at most one lattice point one the boundary. A disk

If we eliminate a certain countable set of lines from $\mathbb{R}^2$ "most" points will remain and for any one, no disk centered at it has more than one rational point on the boundary.

  • $\begingroup$ Thanks for the explanation and the link to the integer sequence. The smallest circle that contains exactly 8 integer lattice points (its center seems to lie on the diagonal of a unit square) appears bigger than that containing exactly 9 of them. For some other lattices, the radii of the smallest circles containing exactly n points might strictly increase with n. $\endgroup$ Feb 24 at 17:18

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