WakeUp wrote: A square board with $8\text{cm}$ sides is divided into $64$ squares square with each side $1\text{cm}$. Each box can be painted white or black. Find the total number of ways to colour the board so that each square of side $2\text{cm}$ formed by four squares with a common vertex contains two white and two black squares. Suppose in a valid coloring, there are two adjacent cells with the same color, wlog in a row. So the two cells above and the two cells below have the opposite color. etc, so the two columns are zebras, meaning either bwbwbwbw or wbwbwbwb. The neighboring columns (and thus all columns) must also be zebras. There are $2^8$ ways of combining 8 zebra columns, and another $2^8$ ways of combining 8 zebra rows. All yield valid colorings. The two chessboard-like colorings (i.e. there are no two adjacent cells with the same color) are already contained and have been counted twice in the process. So the answer is $2^9-2=510$.

I know I am reviving an old thread. But couldn't resist after seeing the similarity, See ISL 1996 C2

bump... Does there exist any other solution?