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.

  • $\begingroup$ 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." $\endgroup$ – Joseph O'Rourke Dec 16 '18 at 0:22
  • $\begingroup$ 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. $\endgroup$ – 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 $$(x-a)^2 + (y-b)^2 = (x - a + i (y-b))(x - a - i(y-b)) = R^2.$$

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  • $\begingroup$ In fact, $a$ and $b$ must be rational, with denominators $O(R)$. $\endgroup$ – Sean Eberhard Jan 16 '19 at 7:58

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