Regular $n$-gon in the plane with vertices on integers? For which $n \geq 3$ is it possible to draw a regular $n$-gon in the plane ($\mathbb{R}^2$) such that all vertices have integer coordinates? I figured out that $n=3$ is not possible. Is $n=4$ the only possibility?
 A: This is impossible for $n=3$ as you mentioned, and for all $n > 4$, but the result is nontrivial.
See here.
Citation: Klobučar, D. (1998). On nonexistence of an integer regular polygon. Mathematical Communications, 3(1), 143-146.
Edit: In response to the comment, this is possible for polyhedra in $\mathbb{Z}^3$. An example of an octahedron with integer coordinates can be found in the third slide here. If you are curious about general higher dimension characterizations, though, I suggest you post a separate question.
A: We could think of this like placing vertices on lattice and creating regular polygons.
And for $\mathbb{R^2}$ the only regular $n$-gon with all vertices having integer coordinates is the square. And if the polygon isn't regular, then every $2n$-gon will do the job.
The proof is a little bit longer, but it's nicely explained here and I think you'd be interested in this.
A: For n>6 it's easy. Fix n>6, pick the smallest regular n-gon with integer coordinates. Move the edges without rotating so that they form a star in (0,0). The endpoints now are on integer coordinates and form a regular n-gon which is smaller than the smallest. That's impossible, so no such thing exists.  For 5 and 6 you need an additional argument with mirroring etc, not too difficult.
