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

Question

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.

Answer 1

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 $$(x-a)^2 + (y-b)^2 = (x - a + i (y-b))(x - a - i(y-b)) = R^2.$$