Prove that on a chessboard that has dimension $4 \times m$ there doesn't exist a knights tour in which we return to square we started at.
I know that we need to turn each square into a vertice and then add edges between vertices where a valid move exists. Now we must show that this graph is not Hamiltonian, which I am unsure how to do. Any help is appreciated, thanks!
HINT
Show that since you can't move between rows 1 and 4, in a (supposed!) Tour you can't move between rows 2 and 3 either.
So, what are all the colors of the squares visited in rows 2 and 3 like in such a Tour?