Start by focusing on the black squares.
• • • A
• • C B
• • E D
• G F •
• I H •
K J • •
M L • •
N • • •
A and B must be different (one has a pawn, the other does not), because the upper right corner touches an odd number of pawns. Looking at the square surrounded by A, B, C and D, this means that C and D must be the same (both have pawns, or both do not). Continuing in this fashion, we get the following pattern:
• • • A
• • B a <-- Lower case letters must the opposite of
• • C B corresponding upper case letter.
• D c L
• E D K
F e J • Two same capital letters must be the same.
G F I •
g H • •
From now on, we use mod 2 arithmetic. Give each square a number so squares without pawns are 0 and squares with pawns are 1. Then the neighbors of any square must add to 1 (mod 2). This means that
H = g + F + 1
I = e + F + H + 1 = e + F + (g + F + 1) + 1 = e + g
J = D + e + I + 1 = D + e + (e + g) + 1 = D + g + 1
K = c + D + J + 1 = c + D + (D + g + 1) + 1 = c + g
L = B + c + K + 1 = B + c + (c + g + 1) + 1 = B + g + 1
Finally, considering the white square surrounded by a, B and L, we must have
a + B + L = 1
∴ a + B + (B + g + 1) = 1
∴ a = g
In other words, we have learned that the H, I, J, K, L diagonal is entirely determined by the numbers above it, and we have learned that a and g are constrained to be equal.
By doing similar work on the other diagonals, you can deduce that the grid must look like this:
• • • A
• • B a
• • C B
• D c •
• C D •
B c • •
A B • •
a • • •
where all of the •'s are forced by the choices of A, B, C and D. However, these four variables can be chosen freely, so there are sixteen ways to place the pawns on the black squares. The white pawns can be chosen independently of the black pawns in the same number of ways, for
16 x 16 = 256
ways total.