Jtdylktuy

I am taking a course on graph theory right now and we were posed the following question:

Show that if $n$ is odd, a knight on an $n times n$ chessboard can not make a closed tour of the chessboard (making only jumps that are allowed by chess rules) in such a way that every square of the board is traversed exactly once.

My answer would move into this direction:

any odd number squared also produces an odd square $n^2=(2k+1)^2= 4k^2 +2k +1= 2(2k^2+2k)+1$ which is the amount of squares on the board. The sum of the white and black squares is odd, therefore the amount of black squares is unequal to the amount of white squares. When a knight makes a jump, he jumps to a square with a different color. Suppose there are $l$ black squares on this chess board and $l+1$ white squares, such that $l+(l+1)=2l+1=n^2$ is the total number of squares again.

My classmates came up with the following argument:

Suppose we start on a black square, we make $2l-1$ jumps, this is an odd number so we end up at a white square, we have already exhausted all of the black squares $1+ 2l-1=2l$ (so just one white left)so now we cannot jump to the last white square because we are missing an intermediate black square.

Now suppose we start on a white square, there are $l$ white squares and $l$ black ones left. We then make $2l$ jumps, divided into $l$ white squares and $l$ black squares, everything is fine and we don't reach a contradiction, what is going on here? what is my reasoning error? My classmates suggested that we would be missing a square to make the final jump, but I don't see it.

Edit:

The contradiction in the white case arises that after we make $2l$ jumps, we have reached all the white squares that we did not start out with, but we have also exhausted all the black squares, we cannot possibly go back, because we would need an additional black square to get back - we don't have that square.

There is no reasoning error in your argument. On some odd-by-odd chessboards, it is possible to have a knight's tour, as the following $5 times 5$ example shows:

The argument by considering black squares and white squares does show that there is no closed knight's tour that visits every square then returns to the start. After all, if there are $l$ black squares and $l+1$ white squares, then any knight's tour must begin and end on a white square - and so there can be no knight's move from the last square of the tour to the first.

If the knight's tour is not required to be closed, then that is a mistake in the problem: the diagram above is a counterexample.

And the $5times 5$ board is not an isolated exception. As Hew Wolff points out in the comments, Wikipedia cites some results showing that a knight's tour exists on any $n times n$ chessboard where $nge 5$ (and in fact on any $n times m$ chessboard where $nge 5$ and $mge 5$).