We claim that the number of ways to color a $n \times n$ square grid using two colors white and black such that every $2 \times 2$ subsquare contains $2$ white squares and $2$ black squares is $2^{n + 1} - 2$. Proof. First, we establish a coordinate system. Let the bottom left square be $(1, 1)$ and the rest of the squares be labeled in a manner identical to Cartesian Coodinates. If the color of a specific square is known, it will be denoted $(x, y, ?)$, where the question mark will be replaced by either either a $W$, or a $B$. Now, consider the bottom most row of any $n \times n$ square grid, that is, consider squares of the form $(a, 1)$ for $a \in \{1, 2, ..., n\}$. We split the proof into two cases: Case 1 (there are no two consecutive white or black squares) Let the first row be colored in any way that follows the above condition (there are clearly $2$ ways to do this). Now, consider the second row from the bottom, that is, squares of the form $(a, 2)$ for $a \in \{1, 2, ..., n\}$. We claim that the second row can be colored in exactly $2$ ways: (1) Every square above a white square is colored white, and likewise for squares above a black square. More formally, for every $(a, 1, W)$, we have $(a, 2, W)$, and for every $(a, 1, B)$, we have $(a, 2, B)$. (2) Every square above a white square is colored black, and likewise for squares above a black square. More formally, for every $(a, 1, W)$, we have $(a, 2, B)$, and for every $(a, 1, B)$, we have $(a, 2, W)$. To prove this, we let $(1, 2)$ be colored arbitrarily. Now, in the $2 \times 2$ square $(1, 1) (2, 1) (1, 2) (2,2)$ only $(2, 2)$ is uncolored. Since we must have two white squares and two black squares, there is only one possible coloring of $(2, 2)$. If $(1, 2)$ is the same color as $(1, 1)$, we see that $(2, 2)$ will be the same color as $(2, 1)$. Now, if we consider the $2 \times 2$ square $(2, 1) (3, 1) (2, 2) (3, 2)$, we can use the same exact reasoning to deduce that $(3, 2)$ will be the same color as $(3, 1)$. This trend will continue across the entire second row, and gives rise to (1). Similarly, if $(2, 1)$ is the opposite color as $(1, 1)$, we will observe (2). Note that in each of these cases, row two has the same property as row one, in that there are no two consecutive white or black squares. This means that we can apply the same process to rows two and three as we just did to rows one and two, and we will see that row three can be colored in exactly $2$ ways. In general, each of the $n - 1$ rows other than the first row can be colored in exactly $2$ ways. Since there are $2$ ways in which to meet the conditions of Case 1, the total number of associated colorings is $2(2^{n - 1}) = 2^n$. Case 2 (there is at least one pair of consecutive white or black squares) Let the first row be colored in any way that follows the above condition. Since there are $2^n$ ways to color the entire first row, and $2$ that do not satisfy this condition (the ones discussed above), there are exactly $2^n - 2$ ways to do this. Now, consider the second row from the bottom, that is, squares of the form $(a, 2)$ for $a \in \{1, 2, ..., n\}$. We claim that the second row can be colored in exactly $1$ way: Every square above a white square is colored black, and likewise for squares above a black square. More formally, for every $(a, 1, W)$, we have $(a, 2, B)$, and for every $(a, 1, B)$, we have $(a, 2, W)$. Note that this is identical to (2) above. To prove this, we let a set of consecutive squares with the same color in the first row be $(a, 1)$ and $(a + 1, 1)$. Now, consider $(a, 2)$ and $(a + 1, 2)$. Clearly both of these squares will be the opposite color as $(a, 1)$ and $(a + 1, 1)$. We now have the same circumstances as (2) above, in that we have two squares of the form $(a, 1)$ and $(a, 2)$, which are different in color. Using the same exact reasoning as in (2), we see that only one coloring of row two is now possible. Now, note that row two has the same property as row one, in that there is at least one pair of consecutive white or black squares ($(a, 2)$ and $(a + 1, 2)$). This means that we can apply the same process to rows two and three as we just did to rows one and two, and we will see that row three can be colored in exactly $1$ way. In general, each of the $n - 1$ rows other than the first row can be colored in exactly $1$ way. Since there are $n^2 - 2$ ways in which to meet the conditions of Case 2, the total number of associated colorings is $(2^n - 2)(1^{n - 1}) = 2^n - 2$. The total from both cases is then $2^n + (2^n - 2) = 2^{n + 1} - 2$, as desired. Q. E. D.