On an $n \times k$ chessboard, a bishop is wandering. The bishop starts from one of the light-colored corners and moves diagonally. When it reaches the edge of the board, it "bounces back". The movement ends when it reaches a corner again. For which pairs of $(n, k)$ is it true that the bishop visits all the light-colored squares?
Let's only consider the case where $k, n > 1$. (The problem is meaningless if $k=1$ or $n=1$.) Number the rows of the table from 0 to $(k-1)$ and the columns from 0 to $(n-1)$, so that the top-left corner is the $(0,0)$ cell. Furthermore, assume that the runner starts from here, which is a light-colored cell. It is not hard to notice that after $x$ steps, the runner's position depends only on the remainder of $x$ when divided by $(2k-2)$ and $(2n-2)$. More precisely, if $x \equiv \pm p \pmod{2n-2}$, where $p=0,1,2, \ldots, n-1$, then the runner is in the $|p|$-th column, and is moving right or left depending on whether a positive or negative sign was given to $p$ (with the agreement that 0 is assigned a positive sign and $(n-1)$ is assigned a negative sign; in these cases, both signs determine the same residue class). Similarly, if $x \equiv \pm q \pmod{2k-2}$, where $q=1,2, \ldots, k-1$, then the runner is in the $|q|$-th row, and is moving downward or upward depending on the sign given to $q$ (0 is assigned a positive sign and $(k-1)$ is assigned a negative sign). The $(p, q)$ cell is light exactly when $p-q$ is even. Thus, the runner visits all light cells if and only if the equation $u(2n-2) \pm p = v(2k-2) \pm q$ has integer solutions $(u, v)$ for every possible $(p, q)$ pair. (By a solution, we mean that the equation holds with the appropriate choice of signs on both sides.) Since $\pm p \pm q$ is even, the equation can be rewritten as $\frac{\pm p \pm q}{2} = v(k-1) - u(n-1)$, still within the integers. If $k-1$ and $n-1$ are coprime, then the equation is always solvable, regardless of the choice of signs. However, if we choose $p$ to be 2 and $q$ to be 0, it is also clear that this condition is necessary for the existence of a solution.

Therefore, the runner visits all light-colored cells on the board if and only if $(n-1, k-1) = 1$.

Based on the work of Ádám Szeidl (Miskolc, Földes F. Gymnasium, 4th grade) and Gábor Braun (Budapest, Szent István Gymnasium, 1st grade)