# On circles and ellipses drawn on an infinite planar square lattice

Consider a plane with a square lattice formed by all points with both coordinates as integers. As can be easily seen, a simple parabola can be found that passes through infinitely many of the square lattice points. However,

1. Given any positive integer n, can we always find a sufficiently large circle drawn on the plane that passes through at least n lattice points? Can such circles be found if the center is not to be a lattice point? What if we require the circle to pass through exactly n lattice points?

2. Question 1 has a natural restatement if instead of circles, we look at ellipses (either all with a given eccentricity e or with e that can be freely chosen). The ellipses need not be axis parallel.

3. And what can one say if the lattice of points has as unit cell not a square but a general parallelogram?

Note 1: Lattice points on the boundary of an ellipse discusses a related question.

Additional Question (added after Prof. Elkies's affirmative answer to questions 1 to 3): What happens in 3D and higher dimensions?

• For your first question in 1), Wikipedia states that for a given integer $n$ there exist $n$ different pythagorean triples with the same hypothenuse. This would mean that we can always find a sufficiently large circle that passes through at least $n$ lattice points if the center is at a lattice point. en.wikipedia.org/wiki/Pythagorean_triple#Special_cases Jan 18, 2021 at 18:41
• If a circle passes through 3 points with rational coordinates then the center must have rational coordinates. (The center can be obtained by intersecting the perpendicular bisectors which are given by rational equations) Jan 18, 2021 at 19:04

(1-2) Yes. For each integer $$n > 0$$ the circle $$x^2 + y^2 = 13^{n-1}$$ passes through exactly $$4n$$ lattice points, namely those with $$z := x+iy = \zeta (3+2i)^a (3-2i)^b$$ with $$a,b$$ nonnegative integers such that $$a+b=n-1$$, and $$\zeta \in \{1, i, -1, -i\}$$. Given $$(a,b)$$, exactly one of the four choices of $$\zeta$$ makes $$x \equiv 2y+1 \bmod 5$$ (because $$\{1, i, -1, -i\}$$ is a complete set of nonzero residues modulo $$2+i$$, and $$z \notin (2+i){\bf Z}[i]$$). So the circle $$(2x-y+1)^2 + (x+2y)^2 = 13^{n-1}$$ passes through exactly $$n$$ lattice points. The left-hand side $$(2x-y+1)^2 + (x+2y)^2$$ is also $$5(x^2+y^2) + 4x - 2y + 1 = 5 \left[ \Bigl(x + \frac25\Bigr)^{\!2} + \Bigl(y - \frac15\Bigr)^{\!2} \right],$$ so we have a circle centered at $$(x,y) = (-2/5, 1/5)$$.

(3) Yes. Use the construction of (2), and (if you don't accept a circle as a special case of an ellipse) apply a linear change of variable such as replacing $$y$$ by $$x+y$$.

• Thanks very much! Will some more general claim on the lines: "for any n and for any planar lattice, conics of any type (ellipse, parabola, hyperbola) may be found that pass thru exactly n lattice points" hold? Jan 19, 2021 at 6:06
• You're welcome. For the "more general" question --  Ellipse, yes: start with the circle that works for the square lattice, and apply a linear transformation to take the square lattice to "any planar lattice". Parabola and hyperbola, I don't think so: once such a conic has a few integral points (probably 5 is enough) it has infinitely many. In the hyperbolic case this will come down to a Fermat-Pell equation. Jan 19, 2021 at 15:43
• Thanks again. That the conics have different behaviors w r to the integer lattice is a surprise! Jan 22, 2021 at 6:23

Such circles are known as Schinzel Circles. See also Kulikowski's Theorem for the sphere. According to zbmath Kulikowski proved his theorem in arbitrary dimension.

• Thanks very much for those pointers! Just one further query: Kulikowski's proof (as given in Honsberger's 'Gems') shows how to construct a sphere such that exactly n points lie on it but those points are all coplanar. Is there a way to show the existence of a sphere in a 3D lattice such that it passes thru exactly n points, say, in general position, for any n? Jan 19, 2021 at 18:35
• Yes, but the easiest way to do that is not very exciting: start with a Schinzel circle (or some variant such as the one I gave) with n-1 points, make it the intersection of R^2 with a large sphere in R^3 whose center has a positive z-coordinate, and use the lattice generated by Z^2 and the north pole . . . Jan 22, 2021 at 4:41
• Thank you. I guess this gets n-1 of the 3D lattice points except the north pole on one plane. How could one possibly get a sphere to pass through n lattice points such that no 4 of them are coplanar? Jan 22, 2021 at 6:35