Here is a one way to attack the n-dimensional case of regular polygons whose vertices lie in Z^n. Suppose there's a regular p-gon P, p in {3,4,5,...} with vertices in Z^n. If n >= p then R^p \sub R^n, so the standard basis vectors of R^p show that such a p-gon always exists. So we shall assume n < p. Let the vertices of P be V(j) in R^n, j = 1,...,p. Now consider the n x n matrix M whose rows are V(j+1)-V(j), j = 1,...,n. As Rich points out, these row vectors V(j+1)-V(j) all have length sqrt(N) for some N in Z+. Thus we may write: M = sqrt(N) M_1 for an n x n matrix M_1 whose rows have unit length. Hence M and M_1 have the same eigenvalues. Now to find the eigenvalues of M_1, we may use any basis we choose, so let the first row be (1, 0, ..., 0); let the second row be (cos(2pi/p), sin(2pi/p), 0, ..., 0); ...; let the jth row be (cos(2(j-1)pi/p), sin(2(j-1)pi/p), 0, ..., 0). Then it's easy to see that this matrix has sin(2pi/p) as an eigenvalue, and hence so does M. But M is an integer matrix, so the roots of its characteristic polynomial are algebraic of degree <= n. So in cases where sin(2pi/p) is algebraic of degree > n, we have reached a contradiction. E.g., for a pentagon in Z^3, sin(2pi/5) is algebraic of degree 4. Since 4 > 3, the 3 x 3 integer matrix M in this case cannot have sin(2pi/5) as an eigenvalue. E.g., for the case of p prime, the pth cyclotomic polynomial is of degree p-1, which shows that a regular p-gon cannot have its vertices on Z^(p-2). The general case that's easy to state is the following. (Note that p is not necessarily prime.) ----- Let phi denote the Euler phi function. If phi(p) > n then there does not exist any p-gon in R^n with vertices in Z^n. ----- For, phi(p) is the degree of the algebraic number sin(2pi/p). —Dan ----- I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral. -----