Regular icosahedron with integer vertices 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?
 A: In six dimensions we can specify a set of integer-coordinate points that produce the symmetries of the regular icosahedron/regular dodecahedron
$(1,1,1,-1,-1,-1)$
$(1,1,-1,-1,-1,1)$
$(1,-1,-1,-1,1,1)$
$(1,-1,-1,1,1,-1)$
$(1,-1,1,1,-1,-1)$
$(1,1,-1,1,-1,-1)$
$(1,-1,1,-1,-1,1)$
$(1,1,-1,-1,1,-1)$
$(1,-1,-1,1,-1,1)$
$(1,-1,1,-1,1,-1)$
And their additive inverses.
Note the grouping of the above points: the first five form a fivefold rotation about $(1,0,0,0,0,0)$ and so do the second five. When we include the additive inverse points we similarly find fivefold symmetry about each of the other five orthogonal axes. Thus there are six fivefold axes matching a requirement for icosahedral symmetry. Various other groupings similarly show ten threefold and fifteen twofold rotational axes.
Gaze into the (quasi)crystal ball
Living in six-dimensional space, the complete set of twenty points above do not lie at the vertices of a regular dodecahedron. But, the six-dimensional space may be folded into our ordinary three-dimensional space so that the images do lie on a regular dodecahedron (or the face centers of a regular icosahedron). This folding process will, of course, map different points from the six-dimensional apace into the same point of three-dimensional space -- for real coordinates. If, however, we consider only rational coordinates, those points remain distinct even though they become densely packed. Therefore planes passing through rational points of the six-dimensional space appear as the distinct lattice planes of an icosahedral quasicristal. Scientists therefore can distinguish these planes using sets of six integer Miller indices instead of having to resort to irrational numbers. As explained in http://www.jcrystal.com/steffenweber/qc.html:

Since quasicrystals lost periodicity in at least one dimension it is not possible to describe them in 3D-space as easily as normal crystal structures. Thus it becomes more difficult to find mathematical formalisms for the interpretation and analysis of diffraction data. For normal crystals we can assign three integer values (Miller indices) to label the observable reflections. This is due to the three-dimensional translational periodicity of the structure. In order to assign integer indices to the diffraction intensities of quasicrystals, however, at least 5 linearly independent vectors are necessary. So we need 5 indices for polygonal quasicrystals and 6 indices for icosahedral quasicrystals. We can call them generalized Miller indices. The necessary n vectors span a nD-reciprocal space. Therefore there is also a nD-direct space in which a structure can be built that gives rise to a diffraction pattern as it is observed for quasicrystals. To put it simply we can say that in the higher-dimensional space we can describe a quasiperiodic structure as a periodic one.

So, with a little ingenuity, we can fit integer coordinates to a regular dodecahedron or icosahedron -- and to icosahedral quasicrystals!
Putting two and four together
Why do we need six dimensions Ford a lattice with iceosahedral symmetry, anyway?
Imagine taking a trip around a regular pentagon. You may interpret each side as a unit vector, the resultant of all five of these vectors being of course zero as you return to the starting point.
With that zero resultant all five unit vectors cannot be independent. But you can easily see that any four of them are, and so you need whichever four you choose to build a lattice containing this pentagon. A quasilattice in the plane must incorporate a four-dimensional integer domain to achieve pentagonal symmetry. More generally, the symmetry if a regular $n$-gon requires projecting from a lattice with $\phi(n)$ dimensions, where $\phi$ is the Euler totient function.
When we go to more complex symmetries we need to combine multiple rotational symmetries, and the dimensional requirements add together. Fir example, we can generate cubic symmetry from a threefold axis and a two-fold axis if they are wt the proper angle relative to each other. The two-fold axis we start with must be rotatable to make more such axes, the threefold axis must be rotated around each twofold axis to make another threefold axis, and before we know it we have generated the full cubic symmetry out of just a threefold axis and a twofold axis. These then require a space of $\phi(3)+\phi(2)=2+1=3$ dimensions to generateca cubic lattice, just as we knew all along.
For a pentagonal, octahedral, decahedral or dodecahedral prism as indicated in the above quotation, we need the $n$-fold symmetry from the polygon plus another twofold axis or mirtor plane. The Euler totient function of $5,8,10,12$ is $4$ for all these arguments, so $4+1=5$ dimensions overall would be enough to embed the prsmatic symmetries rendered above. With the regular icosahedron, however, the minimal set of symmetry elements is instead a fivefold axis plus a threefold axis. That means $\phi(5)+\phi(3)$ dimensions -- $4+2=6$, just as we reckoned at the beginning of this discussion.
A: 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.
