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,

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*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?


*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.


*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?
One more thought (added on 11th July, 2022): In the above questions, one can replace ellipse/circle with "multifocal ellipses (n-ellipses) for any n" or closed curves of constant width.
 A: 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.
A: (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 $x+y$.
