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I asked a question over at Code Golf Stack Exchange which essentially asked folks to write a program to determine if a collection of $2^n$ points in $\mathbb{Z}^m$ is the vertex set of some $n$-dimensional hypercube.

One of the solutions looked at the squared distance between each pair of points and concluded that the collection of points was the vertex set of a hypercube if and only if there exists some $a$ such that for $j \in \{0, 1, ..., n\}$, the squared distance $ja^2$ occurs $2^n\binom{n}{j}$ times.

I came up with a counterexample to this claim for the case of $n=2$, but only in $\mathbb Q^3$. I can't figure out how (or if it's possible) to rotate/scale it into $\mathbb Z^3$ (or $\mathbb Z^4$, etc):

  • $p_1 = (0,0,0)$
  • $p_2 = (1,0,0)$
  • $p_3 = (\frac 12, \frac{\sqrt{3}}2, 0)$
  • $p_4 = (\frac 12, \frac{\sqrt{3}}2, 1)$

Is there a way to rotate and scale this polytope to give it all integer coordinates? Or does the criteria given in the solution indeed work for sets of points in an integer lattice?

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Code Golf Stack Exchange user l4m2 gave me the answer I was looking for. The tetraheron with vertex set $$ \{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,1,1,1)\}. $$ is similar to the tetrahedron with vertex set $\{ (0,0,0), (1,0,0), (\frac 12 \frac{\sqrt{3}}2, 0), (\frac 12 \frac{\sqrt{3}}2, 1) \}$, as in the question.

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