, 2018

Solution:

For the purpose of explaining this solution, let's label the squares as
11121314
21222324
31323334
41424344

Note that since the corner squares $11,14,41,44$ each only have two neighbors, each corner square is the same color as both of its neighbors (for example, $11,12$, and $21$ are the same color, $31,41$, and $42$ are the same color, etc.). This corner square constraint heavily limits the possible colorings. We will now use casework.

Case 1: Suppose two corner squares on the same side (without loss of generality, let them be $11$ and $14$) have the same color (without loss of generality, red). Then $21,11,12,13,14,24$ are all red, and $12$ has two red neighbors ($11$ and $13$) so its third neighbor ($22$) is a color different from red (without loss of generality, magenta). But $22$ has two red neighbors ($12$ and $21$), so its other two neighbors ($23$ and $32$) must be magenta. Applying the same logic symmetrically, we find that all four interior squares $(22,23,32,33)$ have the same color. Furthermore, $21$ has one magenta neighbor $22$, so $31$ must be red. Symmetrically, $34$ is red, and by the corner square constraint we have that all the exterior squares are the same color. Thus in general, this case is equivalent to a window taking the following form (with distinct colors $A$ and $B$):

$$\begin{array}{llll} A & A & A & A \\ A & B & B & A \\ A & B & B & A \\ A & A & A & A \end{array}$$

The number of choices of $A$ and $B$ is $3 \cdot 2=6$.

Case 2: No two corner squares on the same side have the same color.
Then from the corner square constraint $12$ has neighbor $11$ of the same color and neighbor $13$ of a different color, so its neighbor $22$ must be the same color as $12$. Therefore, this case is equivalent to coloring each quadrant entirely in one color such that two quadrants sharing a side have different colors. (A quadrant refers to the four squares on one vertical half and one horizontal half, e.g. $13,14,23,24$). If only two colors are used, the window will take the form (with distinct colors $A$ and $B$):

$$\begin{array}{llll} A & A & B & B \\ A & A & B & B \\ B & B & A & A \\ B & B & A & A \end{array}$$

Again there are $3 \cdot 2=6$ ways to choose $A$ and $B$.

If all three colors are used, the window will take the form (with distinct colors $A, B$ and $C$):

$$\begin{array}{llll} A & A & B & B \\ A & A & B & B \\ C & C & A & A \\ C & C & A & A \end{array}$$

or

Therefore, there are $6$ colorings in Case 1 and $6+6+6$ in Case 2, for a total of $24$ colorings.