Is there a regular hexagon with integral corners? I'm looking for a regular hexagon in $\mathbb{R}^2$, whose corners are integral, i.e. the coordinates are integers.
The hexagon cannot lie "flat" (with upper and lower line segments horizontal), since then $h = \frac{\sqrt{3}}{2} w$ with $w$ width and $h$ height of the hexagon, making the ratio $\frac{h}{w}$ irrational.
 A: In $\mathbb R^2$, there is no regular hexagon whose vertices are on the lattice points.
Proof : 
Suppose that there exists a regular hexagon $A_1A_2\cdots A_6$ whose vertices are on the lattice points.
Then, the triangle $A_1A_3A_5$ is an equilateral triangle. WLOG, we may set $A_1(0,0),A_3(a,b),A_5(c,d)$ where $(a,b)\not=(0,0)$. Then, considering the area, we get
$$\frac 12|ad-bc|=\frac{\sqrt 3}{4}(a^2+b^2),$$
i.e.
$$\sqrt 3=\frac{2|ad-bc|}{a^2+b^2}$$
The LHS is irrational, the RHS is rational, a contradiction. 

By the way, in $\mathbb R^3$, there are regular hexagons whose vertices are on lattice points.
Here is an example :
$$(0,-1,-1),\quad (1,0,-1),\quad (1,1,0)$$
$$(0,1,1),\quad (-1,0,1),\quad (-1,-1,0)$$
These are the midpoints of the six sides of the cube whose vertices are $(\pm 1,\pm 1,\pm 1)$.
A: The area of a polygon with lattice vertices is half an integer by the shoelace formula.
The area of a regular hexagon of side $L$ is $\dfrac{3 \sqrt 3}{2} L^2$ and so is not half an integer because $ \sqrt 3$ is irrational and $L^2$ is an integer, since $L$ is the distance between two lattice points.
A: Yes we can have a regular hexagon with integral vettex coordinates ... in three dimensions.  Think of a face-centered cubic lattice.  If we set the length unit to half the edge of a cubic unit cell then all lattice points have integer coordinates -- including those forming a hexagonal close-packed plane perpendicular to any body diagonal of the cube.
