Can a knight visit every field on a chessboard? I was doing excercises about graphs theory and I came across a quite interesting excercise (which probably has something to do with Hamiltonian Cycle):
"Is it possible to step on every field of a 4x4 or 5x5 chessboard just once and return to the starting point using a knight?"
Does anyone have any idea how to tackle this problem? I am more interested in a outline of how to do it or just some hints.
 A: For the $5 \times 5$ chess board, we colour the squares black and white in a checkerboard fashion.  There are $b:=\lfloor 5^2/2 \rfloor$ squares of one colour (say black) and $w:=\lceil 5^2/2 \rceil$ squares of the other colour (white).
Importantly, the knight moves are always black-to-white or white-to-black.  Thus, for there to be a closed knight's tour, we would need the number $b=w$.  However, $b<w$, so there is no closed knight's tour of this board.
A: Hint: You must enter and exit from each square. Look at the corners. What is special about them?
A: HINT:
$$\begin{array}{|c|c|c|c|} \hline
\cdot&&&\\ \hline
&&\cdot&\\ \hline
&\cdot&&\\ \hline
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\end{array}$$
A: So I was skimming through the questions I've asked here and I remembered that  I found a different solution to my question. Here it is:
If it is possible for a knight to step on every field of a chessboard, that means there is a Hamiltionian Cycle in a graph, in which a vertex is an equivalent to a square on the chessboard and the edges between vertices respond to the possibility of a move of a knight between the squares. There's a theorem that says if we remove a set V of k vertices in a Hamiltonian Graph then there are exactly k components of graph $G-V$. So if we remove four vertices in the middle we would get 4 isolated vertices and the rest of the graph - total of 5 components, which contradicts with the assumption that G was Hamiltonian.
$$\begin{array}{|c|c|c|c|} \hline
\cdot&&&\cdot\\ \hline
&\cdot&\cdot&\\ \hline
&\cdot&\cdot&\\ \hline
\cdot&&&\cdot\\ \hline
\end{array}$$
