The eight points
$$
(\pm 1, \pm 1, \pm 1)
$$
are the vertices of a cube. The six points
$$
(\pm1, 0,0)\; , \; (0, \pm1, 0)\; , \; (0,0,\pm1)\; , \;
$$
are the vertices of an octahedron. The four points
$$
(-1,-1,-1) \; , \; (-1, 1, 1) \; , \; (1, -1, 1) \; , \; (1, 1, -1)
$$
are the vertices of a regular tetrahedron. But there is no collection of points in $ \mathbb{Z}^3 $ that form the vertices of a regular icosahedron.
Does there exist any larger dimension $ n $ for which some set of twelve integer points form the vertices of a regular icosahedron?
Best Answer
No, nor for a dodecahedron. A core problem in both these cases is that regular pentagons can't be embedded in any $\mathbb{Z}^n$.
Proof: Suppose we had such a pentagon $ABCDE$. Then the points $0$, $B-A$, $(B-A)+(E-D)$, $(B-A)+(E-D)+(C-B)$, $(B-A)+(E-D)+(C-B)+(A-E)$ also form a pentagon with integer coordinates and a shorter side length. (We drew a five-pointed star whose diagonals were the side lengths of our first pentagon.) So by infinite descent, there's no smallest pentagon and thus no pentagon at all.
The same argument generalizes to all regular polygons with a number of sides other than $3,4,$ or $6$ and to all other non-dense lattices.