, 2013

Let us associate a $1$ to each cell with a black color and a $0$ to each cell with a white color. The condition is equivalent to the sum of numbers in each $2 \times 3$ rectangle and in each $3 \times 2$ rectangle being even. Let $a_{i, j}$ be this number at the cell in the $i^{\text{th}}$ row and $j^{\text{th}}$ column, for $1 \leq i, j \leq 50$.

Consider, for a fixed pair $i, j$, with $1 \leq i \leq 48$ and $1 \leq j \leq 47$, the $3 \times 4$ rectangle:

By applying the condition to the two $2 \times 3$ rectangles which contain cells from the second and the third rows, we get
$$a_{i+1, j}+a_{i+1, j+1}+a_{i+1, j+2}+a_{i+2, j}+a_{i+2, j+1}+a_{i+2, j+2} \equiv 0 \pmod{2},$$
and
$$a_{i+1, j+1}+a_{i+1, j+2}+a_{i+1, j+3}+a_{i+2, j+1}+a_{i+2, j+2}+a_{i+2, j+3} \equiv 0 \pmod{2}$$
By applying the condition to all the $3 \times 2$ rectangles, we get
$$\begin{gathered} a_{i, j}+a_{i, j+1}+a_{i+1, j}+a_{i+1, j+1}+a_{i+2, j}+a_{i+2, j+1} \equiv 0 \pmod{2}, \\ a_{i, j+1}+a_{i, j+2}+a_{i+1, j+1}+a_{i+1, j+2}+a_{i+2, j+1}+a_{i+2, j+2} \equiv 0 \pmod{2}, \end{gathered}$$
and
$$a_{i, j+2}+a_{i, j+3}+a_{i+1, j+2}+a_{i+1, j+3}+a_{i+2, j+2}+a_{i+2, j+3} \equiv 0 \pmod{2} .$$
By adding these 5 relations and cancelling all even numbers we get
$$a_{i, j}+a_{i, j+3} \equiv 0 \quad \pmod{2} .$$
This proves that
$$a_{i, j+3}=a_{i, j}$$
We prove in a similar way that
$$a_{i+3, j}=a_{i, j}$$
Therefore, it is enough to know the numbers in the $3 \times 3$ rectangleto deduce, by periodicity, the numbers in all the other cells.

Applying the condition to the first $2 \times 3$ rectangle, we will get
$$a_{2,3} \equiv a_{1,1}+a_{1,2}+a_{1,3}+a_{2,1}+a_{2,2} \quad \pmod{2}$$
Applying the condition to the second $2 \times 3$ rectangle and to the second $3 \times 2$ rectangle and adding the two relations, we will get
$$a_{3,1} \equiv a_{2,1}+a_{1,2}+a_{1,3} \quad \pmod{2}$$
We obtain in a similar way
$$a_{3,2} \equiv a_{2,2}+a_{1,3}+a_{1,1} \quad \pmod{2}$$
$$a_{3,3} \equiv a_{2,3}+a_{1,1}+a_{1,2} \quad \pmod{2}$$
Therefore, it is enough to know the 5 numbers
$$a_{1,1}, a_{1,2}, a_{1,3}, a_{2,1}, a_{2,2}$$
to deduce all the numbers in the cells of the $50 \times 50$ chessboard.

Conversely, choose a value in $\{0,1\}$ for each number $a_{1,1}, a_{1,2}, a_{1,3}, a_{2,1}, a_{2,2}$, and deduce the values in all the other cells of the chessboard. It is easy to check that the condition on the two $2 \times 3$ rectangles and the two $3 \times 2$ rectangles in the $3 \times 3$ rectangle above is satisfied. We deduce, by periodicity, that it is satisfied in all the chessboard. Hence there are $2^{5}$ ways to color the chessboard.