Japanese Mathematical Olympiad - The First Round

We call the squares corners, edges or the center, according to their position. We first consider the case that the center is colored red. Call the edges $A$, $B$, $C$ and $D$ clockwise. We divide the cases by how many edges are colored red. Note that any $2 \times 2$ square must include the center.

* If there are no red edge, or if just one was red, or if two opposing edges were colored red, then however we color the 4 corners, no $2 \times 2$ red square appears. So we can color each of these 4 squares in arbitrary color. On the second case, we have 4 ways each to choose which edge to be colored red. On the third case, we have 2 ways. Hence this case has $(1+4+2) \times 2^4 = 112$ ways.

* If two non-opposing edges were colored red, one corner which is between two red edges must be colored blue, and the other squares can be colored arbitrarily. So this case has $4 \times 2^3 = 32$ ways.

* If there are exactly three red edges, there are exactly 2 corners which are between two red edges and they must be colored blue. Others can be arbitrary color. So this case has $4 \times 2^2 = 16$ ways.

* If all the four edges are colored red, all the corners must be colored blue. There is only 1 way of coloring.

There are the same number of ways of coloring if the center square was colored blue. So we have $(112 + 32 + 16 + 1) \times 2 = 322$ ways of coloring.