I have got this problem at high-school math-contest seminar on Graph Theory
Let us have a chessboard, where one black and one white lame rooks stand. Lame rook can move to edge-adjacent field only. You can move the same rook as much as you want before moving another. Is it possible to obtain every position exactly once moving the rooks? I.e. Is there a hamiltonian path in the graph of rooks' positions.
Is it true that the answer doesn't depend on the size of the board?
I am completely stuck, because I don't know any property of non-Hamiltonian graph. You see that the number of edges is less that $n^2(n^2-1)/2 $ that should be compared with $\leq 8$. So the only Hamiltonian criterion I know doesn't say much.
UPD Two other, I think harder, questions arise if we decide to place two white rooks instead, and if we treat rotation-coinciding cases as equal. The other version of the question is about ordinary rooks.
As long as there are more questions than answers in graph theory, I leave them be.
UPD: I have written the question such that the number of rooks could be assumed more than two. Now the question is edited.
There are $32^2$ positions when rooks stand on squares of different colour and $2 \cdot 32 \cdot 31$ positions when rooks stand on the same colour. Those two cases are alternating in the sequence of moves, so we got a contradiction.
From @MarkBennet 's comment.