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I am intrigued by the existence of regular hexagons with vertices in $\mathbb{Z}^3$. A simple example is the hexagon with vertices $$(0,−1,−1),(1,0,−1),(1,1,0), (0,1,1),(−1,0,1),(−1,−1,0).$$

Can anyone think of a more exotic (less obvious) example with a larger side length? i.e. one with a greater variety of entries for the $x$, $y$ and $z$ coordinates. Here we only have $3$ different entries: $-1, 1, 0$. I'd be interested to have a look at some more examples.

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  • $\begingroup$ See my Edit with the reference herein. $\endgroup$
    – Jean Marie
    Apr 13 '20 at 7:01
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Just apply an isometry matrix (rotation, symmetry...) (in order to preserve planarity and equality of lengthes) with rational entries for example to the 6 points you have given as an example, then multiply the coordinates of these rational points by the LCM of their denominators to get integer entries, an "homothety" operation preserving as well
planarity and equality of lengthes.

How can such matrices be obtained ?

1) By using Lebesgue identity that can be found here yielding unit norm vectors $U$ with rational entries $(a/d,b/d,c/d)$,

2) Then by building :

$$S_U=I_3-2UU^T\tag{1}$$

which is known to be the matrix of the symmetry with respect to the plane orthogonal to vector $U$ (see Householder form here). All the computations involved in the RHS of (1) are between rational numbers ; therefore $S_U$ has rational entries ; it remains to multiply $S_U$ by the LCM of the denominators of its entries.

If one desires rotations, one can compute the product $S_VS_U$ of 2 symmetries, opening the way to a huge number of solutions...

Here is for example one

$$\begin{array}{rrrrrr}8& 11&3&-8&-11&-3\\ 7&4&-3&-7&-4&3\\ 7&-5&-12&-7&5&12\end{array}$$

obtained by using vector $U=(1/3,2/3,2/3).$

Here is a graphical representation of the original hexagon (in red) together with other hexagons obtained by the process described above :

enter image description here:

Fig 1: The four hexagons generated by using $U=(\pm1/3,\pm2/3,2/3)$ in (1).

Remark : we haven't mentionned translations by integer coordinates vectors that evidently give other solutions...

Important edit : I just discovered the very interesting article by E. Ionescu and A. Markov : I have realized, reading it, that our hexagon issue is naturally reducible (by taking one vertex out of two) to the description of equilateral triangles with coordinates in $\mathbb{Z^3}$.

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    $\begingroup$ Possibly worth noting for OP: the fact that one has (essentially) three 'unique' vertices here and three 'reflected' vertices is going to be a hallmark of any solution to the problem, because a regular hexagon is innately symmetric about its center point. $\endgroup$ Apr 8 '20 at 17:56
  • $\begingroup$ @Steven Stadnicki : good remark. $\endgroup$
    – Jean Marie
    Apr 8 '20 at 17:58
  • $\begingroup$ I very much enjoyed reading this answer. In this example, what is $U$? $\endgroup$
    – wrb98
    Apr 11 '20 at 17:13
  • $\begingroup$ $U=(1/3, 2/3, 2/3)$. Sorry, I called it $V$ in my answer... $\endgroup$
    – Jean Marie
    Apr 11 '20 at 17:32

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