It is well-known that no two-dimensional point lattice contains a regular pentagon. (See for example <http://mathworld.wolfram.com/LatticePolygon.html>.) The same is true for lattices in $\mathbb{R}^n$, simply because any such polygon would lie in a two-dimensional sublattice. If we relax the requirement that the pentagon lie in a plane, we can easily find a closed path of 5 equal length sides: for example, in $\mathbb{R}^5$ with the integer lattice, we can use the path $$(1,0,0,0,0)\to(0,1,0,0,0)\to(0,0,1,0,0)\to(0,0,0,1,0)\to(0,0,0,0,1).$$ (These are the vertices of the standard 4-simplex.) The same construction, of course, works to find a nonplanar equilateral lattice "$n$-gon" for any $n$, as the vertices of the standard $n+1$-simplex in the integer lattice. We can improve this to a lattice in $\mathbb{R}^{n-1}$ by restricting to the plane $x\_1+\cdots+x\_n=1$. My question is if this is the best we can do, dimension-wise. So is it possible, for example, to find a nonplanar equilateral lattice pentagon in some lattice in $\mathbb{R}^3$? What can be said about the minimal dimension of lattice containing a nonplanar equilateral lattice $n$-gon?


(Note: This was originally posted on math stackexchange [here](https://math.stackexchange.com/questions/604238/nonplanar-equilateral-lattice-pentagons), but nobody answered it.)



Nonplanar equilateral lattice "pentagons"
Here are the vertices of an equilateral pentagon of side $\sqrt{2}$ in $\mathbb{Q}^3$ in order: 


$(0,0,0), (1,1,0), (1,2,1), (0,1,1), (-\frac{1}{3}, -\frac{1}{3}, \frac{4}{3}).$


The first $4$ points form a rhombus. The last point satisfies $z=1-y, x^2+y^2+(1-y)^2=2.$ This has the rational solution $(1,1,0)$ and lines of any rational slope through $(1,1)$ in the $xy$-plane intersect $x^2+y^2+(1-y)^2=2$ in another rational point. I chose $x=y$. Expanding this by a factor of $3$ gives a nonplanar equilateral pentagon in $\mathbb{Z}^3.$


