Can points on an integer lattice form the vertices of a regular hexagon Is it possible for six points in the integer lattice to form the vertices of a regular hexagon
 A: No. By Pick's Theorem, the area of such a hexagon is rational (in fact, it is an integer multiple of $\frac 12$).
But by an alternative geometric method, you can show that the area of a regular hexagon is $A = \frac{3\sqrt 3}{2}a^2$, where $a$ is the side length. The square of the side length of a lattice polygon is integral by the Pythagorean theorem, so $A$ is irrational, and you have a contradiction.
A: No. Otherwise you would have two segments with endpoints in the lattice whose distances are in ratio $1$ to $\sqrt 3$ (one side of the hexagon and the diagonal perpendicular to it), but this can not happen in the lattice: it would mean that there are integers $a,b,c,d,$ such that
$$a^2+b^2=3(c^2+d^2),$$
and by infinite descent this is impossible. Indeed, the LHS would be a multiple of $3$. By examining squares modulo $3$, the only way is that both $a$ and $b$ are multiples of $3$, but then you can keep going doing the same for the RHS, etc.
Note that this fails in $3D$ because you can also have $1+1+1\equiv 0\pmod 3$.
