# Three questions concerning lattice points on sphere surfaces

Pardon my ignorance of this topic.

Q1. In which dimensions $d$ is it the case that, for every natural number $n$, there exists a sphere having exactly $n$ lattice points on it $(d{-}1)$-dimensional surface?

Schinzel's Theorem establishes this in $\mathbb{R}^2$, and Kulikowski's Theorem establishes this in $\mathbb{R}^3$. Is it known in higher dimensions?

Q2. Why is it that the smallest integral radius of a circle (in $\mathbb{R}^2$) centered at $(0,0)$ that has $n$ lattice points on its circumference, is always a multiple of $5$? Or rather: Is there a simple, intuitive explanation for the multiple-of-$5$?

OEIS A046122: $1, 5, 25, 125, 65, 3125, 15625, 325, \ldots$. A046122 link

And this likely follows from answers to the previous two questions:

Q3. Does the multiple-of-$5$ phenomenon occur in higher dimensions as well?

I realize these questions may be answered in the literature I have not encountered, so pointers are welcomed—Thanks!

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For question 2, the number of lattice points on the circle relates to the number of prime factors congruent to 1 modulo 4, and 5 is the smallest such prime. – Gerry Myerson Aug 25 '13 at 23:23
For question 3, the corresponding sequence is tabulated at oeis.org/A071609 and starts $1,2,5,9,14,26,41,74,89,101,146,194,269\dots$. You are welcome to look for patterns. – Gerry Myerson Aug 25 '13 at 23:29
Honsberger (Mathematical Gems, Chapter 11) gives a proof of Kulikowski's Theorem, and then writes, "This proof generalizes immediately to a space of any number of dimensions." Details can be found at cut-the-knot.org/arithmetic/algebra/Kulikowski.shtml – Gerry Myerson Aug 25 '13 at 23:35
Thanks, Gerry & Lev: All three questions are answered! – Joseph O'Rourke Aug 25 '13 at 23:41

It appears that Kulikowski's paper proved the result for all dimensions. I haven't seen the Kulikowski paper, but a short proof is given at this site.

For question 3, the sequence is tabulated at the OEIS and a glance at the first few terms
($1,2,5,9,14,26,41,74,89,101,146,194,269\dots$) does not suggest any divisibility properties (although once you get past $14$ there seem to be a lot of prime factors congruent to 1 modulo 4).

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Great, Gerry! Here is an observation concerning the 3D sequence: Up to the 47th term listed in OEIS, $8126$, every term factors into primes each to the 1st power, e.g., $11^1 \cdot 31^1 = 341$; $2^1 \cdot 397^1 = 794$; $2^1 \cdot 19^1 \cdot 43^1 = 1634$; etc. – Joseph O'Rourke Aug 25 '13 at 23:53
341 is a famous number, the smallest pseudoprime (but I expect that's a coincidence). – Gerry Myerson Aug 25 '13 at 23:58

It is not really my area, but the answer to part 2 is that you are looking for the smallest $N$ such that there are exactly $8n-4$ ways of writing it in the form $a^2+b^2$.

If such $N$ is even but is not $2$ mod $8$ then both $a$ and $b$ need to be even and $N/4$ will give the same number of solutions as $N$. If $N=2$ mod $8$ then any solution would be a pair of odd $(a,b)$ and $(a+b)/2,(a-b)/2$ would provide the same number of solutions for $N/2$. The minimality of $N$ thus assures that it is odd.

Similarly, if $N$ has a prime factor $p$ which is $3$ mod $4$ then $N/p^2$ will have the same number of solutions as $N$.

Thus, $N$ has only prime factors which are $1$ mod $4$. The number of solutions depends only on the powers of such primes, so the minimum $N$ would be achieved when the prime with the highest power is the smallest prime of $1$ mod $4$ type, namely $p=5$.

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