Cris Moore>What is the smallest number of dimensions in which we can embed a pentagon with integer vertices? More generally, for an n-gon, the argument below suggests (I think) that we need at least p-1 dimensions where p is the largest prime factor of n. We can do a hexagon in three dimensions, but not, I think, in two. Cris On May 25, 2013, at 11:09 PM, Bill Gosper wrote: Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center. Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix. No, That was DanA. http://www.tweedledum.com/rwg/rectarith12.pdf (end) illustrates hexagons in R^3, followed immediately by "How many dimensions before we see (regular) octagons, pentagons, dodecagons, ...? We won't! Even in an infinite dimensional grid, the only regular polygons of finite size are triangles, hexagons, and squares. We could probably prescribe the nth coordinate of the kth vertex of a regular pentagon, say, but infinitely many coordinates would be nonzero, resulting in infinite size. And it would only approach regularity as we consider higher and higher dimensions." --rwg