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?

  • 4
    $\begingroup$ 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 $\endgroup$
    – sebastian
    Jan 18, 2021 at 18:41
  • 5
    $\begingroup$ 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) $\endgroup$ Jan 18, 2021 at 19:04

2 Answers 2


(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 $x+y$.

  • $\begingroup$ 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? $\endgroup$ Jan 19, 2021 at 6:06
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    $\begingroup$ 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. $\endgroup$ Jan 19, 2021 at 15:43
  • $\begingroup$ Thanks again. That the conics have different behaviors w r to the integer lattice is a surprise! $\endgroup$ 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.

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    $\begingroup$ 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? $\endgroup$ Jan 19, 2021 at 18:35
  • $\begingroup$ 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 . . . $\endgroup$ Jan 22, 2021 at 4:41
  • $\begingroup$ 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? $\endgroup$ Jan 22, 2021 at 6:35

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