Show a knight can never tour an odd squared chess board by starting and ending at the same place
I can already see that solving this requires Euler's theorem and showing that since there aren't an even degree for each vertex then there can't be a Euler's cycle, but I don't know how to prove each vertex the knight only has an odd number of moves in an odd squared chess board.
An $n\times n$ chessboard has, without loss of generality, $\left\lceil\dfrac{n^2}2\right\rceil$ light squares and $\left\lfloor\dfrac{n^2}2\right\rfloor$ dark squares. While there are equal numbers of light and dark squares if $n$ is even, there is $1$ more light square than there are dark squares if $n$ is odd.
Suppose the knight starts on a light square and makes $m$ moves.
If $m$ is even, then the knight is on the same colour square as the one on which it started.
If $m$ is odd, then the knight is on the opposite colour square.
Since the knight needs to make an even number of moves to return to its starting position, and the chessboard has an odd number of squares for odd $n$, this proves the statement.