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

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 $1+1+1\equiv 0\pmod 3$.
