parmenides51 wrote: Let $n$ be a natural number. Given a chessboard sized $m \times n$. The sides of the small squares of chessboard are not on the perimeter of the chessboard will be colored so that each small square has exactly two sides colored. Prove that a coloring like that is possible if and only if $m \cdot n$ is even. Let assume that we have such a colouring. We are going to construct a graph in the following manner. The center of each unit square is a vertice, and we link two vertex if and only if they have a coloured edge in common. According to the given condition, each vertice has degree two. This let us construct different cycles (on the constructed graph). All the vertex are part of a cycle. If we colour the chess board in the chess manner (what a surprise) a cycle goes from a black case to a white one, so each cycle has an even number of vertex. But there is $mn$ unit square so $mn$ must be even. If $mn$ is even, wlog we assume that $m$ is even. We can do a $2\times n$ like that : __ |__|__|__|__|__|__| __ ----|---|---|---|----|---|---| We can add $m/2$ of those two do colour the edges of a $m\times n$ chess board.