# Number of lattice points on spheres with center not at the origin

Let $$k\ge1$$. It is known that the number of lattice points on the $$k$$-sphere $$S^k(0)$$ (center at the origin, radius $$R$$), namely the size of $$\mathbb{Z}^{k+1}\cap S^k(0)$$, is bounded by $$R^{k-1+\epsilon}$$, $$\forall \epsilon>0$$.

Now I have a question about this result. We consider a $$k$$-sphere $$S^k(a)$$ embedded in $$\mathbb{R}^d$$ with arbitrary center $$a\in\mathbb{R}^d$$ and radius $$R$$, where $$d\ge k+1$$. Do we have the same estimates about the size of $$\mathbb{Z}^{d}\cap S^k(a)$$? Can we have uniform estimates independent of the center $$a$$ and dimension $$d$$? I don't know if there are any references about this generalization. Any help and comments will be greatly appreciated.

• The earlier MO question, "What is the smallest sphere whose surface includes 100 integer points?", even though phrased for origin-centered spheres, contains some info on off-origin spheres. For example, "The sphere centred on $(1,1,1)/2$ and radius $\sqrt{131}/2\approx 5.723$ contains $120$ points with integer coordinates." – Joseph O'Rourke Dec 16 '18 at 0:22
• Looks like your question is related to mine mathoverflow.net/questions/319130/… (I didn't see it before posting mine), though I think there is a larger hope a "number theory" argument will help with yours since you want the points to be exactly on the sphere. – Rodrigo Dec 20 '18 at 19:23

Yes, the estimate is uniform in $$a$$ and $$d$$, I think. It suffices to handle the two-dimensional case, by considering two-dimensional slices. So consider the circle $$(x - a)^2 + (y-b)^2 = R^2$$ with $$a,b$$ arbitrary. We may assume $$0 \leq a,b < 1$$. Then for any solution $$(x,y)$$, $$x$$ and $$y$$ must be $$O(R)$$. If there are more than just a few solutions, then we can solve for $$a$$ and $$b$$, proving that $$a, b$$ are algebraic of bounded degree and height $$R^{O(1)}$$ (this is an "effective nullstellensatz"). Now you use the divisor bound in the extension $$K = \mathbf{Q}(i, a,b)$$ to bound the number of solutions to $$(x-a)^2 + (y-b)^2 = (x - a + i (y-b))(x - a - i(y-b)) = R^2.$$
• In fact, $a$ and $b$ must be rational, with denominators $O(R)$. – Sean Eberhard Jan 16 '19 at 7:58