Every distance squared between two points in $\mathbb{Z}^n$ is clearly all $\in \mathbb{N}$. Is the converse true (upto isomorphism)

geometryinteger-latticesnumber theory

Every distance squared between two points in $\mathbb{Z}^n$ is clearly all $\in \mathbb{N}$. Is the converse true (upto isomorphism)?

The title was rather informal, so clarifying the question:

Is every countably infinite set $S \subset \mathbb{R}^n$ such that

$$\forall v_1, v_2 \in S, v_1 \neq v_2 \rightarrow d(v_1, v_2)^2 \in \mathbb{N}$$

subset of $L + \delta$, where $L$ is some lattice in $\mathbb{R}^n$ and $\delta \in \mathbb{R}^n$
?

I suspect it would be true, here is some approach I've took.

  1. Fix a point v. translate $S$, so that $v = 0$. Also transform so that $\dim S = n$. Define $S^*$ as maximal superset of $S$ that also satisfies such property. It would be enough to show that $S^*$ is a lattice.
  2. Fix some set of $k$ points, and denote them as $T$. Denote set of points $p$ in $\mathbb{R}^n$ such that $d^2(p, t) \in \mathbb{N}$ for every $t \in T$ as $P_T$. By definition, $S^* \subset P_T$.
  3. (choosing nice $k$ points). Since $S^*$ is countable, set of every size $k$ subset of $S^*$ is also countable. Therefore we select $k$ points $T$ that has maximal rank with respect to $S$. Formally this would mean selecting $k$ points that $\dim (Span(T) \cap Span(S^*))$ is maximum. Here dim is used in lattice sense.
  4. It would be enough to prove that for $k = n$, such $T$ satisfies $P_T = S^*$.

There are close problems, but non exact search results…
https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Ulam_problem

Seems to be lots of researches with "integral distances" but my google search skills limits me with finding non about distance squared being integral.

https://link.springer.com/article/10.1007/s00454-003-0014-7

Best Answer

Following your approach, we translate so that $0\in S$, and define $S^*$ as a maximal superset of $S$ satisfying the requested property.

To prove that $S^*$ is a lattice, it suffices to show that it is a discrete subgroup of $\mathbb{R}^n$, i.e. that for any points $a,b \in S^*$, their sum and difference $a\pm b$ are also in $S^*$ (discreteness is obvious since each point is at least $1$ unit of distance away from the others).

First, we note that for any $x,y \in S^*$, we have

$$|x|^2 = d(x,0)^2 \in \mathbb{N}\cup\{0\}$$

and

$$2\,x\cdot y = d(x,0)^2+d(y,0)^2-d(x,y)^2 \in \mathbb{Z},$$

where $\cdot$ and $|\:\:|$ denote the standard Euclidean inner product and norm respectively.

Next, given an arbitrary $c \in S^*$ we compute

$$d(a\pm b,c)^2=|a\pm b-c|^2 = |a|^2+|b|^2+|c|^2\pm 2\,a\cdot b-2\,a\cdot c \mp 2\,b \cdot c \in \mathbb{N}\cup\{0\}.$$

We conclude that $a\pm b \in S^*$ by maximality.

Finally, we prove that the rank of the lattice $S^*$ must be $n$: if it were lower, say $m<n$, then any unit-length vector perpendicular to the hyperplane $S^*\otimes \mathbb{R} \simeq \mathbb{R}^m$ would have integral squared distance to any vector in $S^*$ by the Pythagorean theorem, contradicting maximality.

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