Let $rS^{d-1}$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside).

More precisely, let $$ N(r,d):=\text{number of lattice points on the sphere of raduis } r=\#\{x\in rS^{d-1}: x\in \mathbb{Z}^d\}. $$

I'm especially interested in the lower bound of $N(r,d)$ for any $d\ge 3$ and large $r$ (with $r^2\in\mathbb{Z}$, of course).

For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor):

$N(r,d)\gtrsim r^{d-2}$ for $d\ge 4$.

So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now.

One can also ask a weaker question: is there a sequence of $r$ tending to $\infty$ such that the above inequality holds with a better lower bound?


2 Answers 2


My answer to this MO question contains the answer to your question, especially if you take into account that $L\left(1,\left(\frac{D}{\cdot}\right)\right)$ can be estimated unconditionally (i.e. without GRH): $$|D|^{-\varepsilon}\ll_\varepsilon L\left(1,\left(\tfrac{D}{\cdot}\right)\right)\ll \log|D|.$$ The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.


It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)

For $d\ge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.

  • $\begingroup$ Thanks. Now I see that the exponent $d-2$ can't be improved for $d\ge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^{1-\epsilon}$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880 $\endgroup$
    – Tony B
    Commented Aug 4, 2019 at 19:42
  • $\begingroup$ @TonyB: For $d\in\{3,4\}$ the quantity is $r^{d-2}$ up to a factor of $\log\log r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details. $\endgroup$
    – GH from MO
    Commented Aug 4, 2019 at 20:30

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