Parity of number of pieces on the black squares of a chessboard On a chess board there is an odd number of pieces in each row and in each column. What can we say about the total number of pieces on a black square on the board?


*

*The number must be even.

*The number must be odd.

*The number can be either both even or odd.

 A: Orient the board so that a white square is in the top left hand corner. Label each square with a piece on it with $1$, and $0$ otherwise. Let $R_1,R_2,\ldots, R_8$ be the rows, and $C_1,C_2,\ldots, C_8$ be the columns. 

Add together $R_1+R_3+R_5+R_7+C_1+C_3+C_5+C_7$, which is even. Each white square in these squares gets accounted for $2$ times, so they have an even contribution to the sum. So this leaves the black squares, and an even minus an even is even, and all the black squares are accounted for exactly once, so there are an even number of pieces on the black squares.
The answer is $1$).
A: It depends. If diagonals are all black, $B$ should be always odd. Else if diagonals are all white, $B $ should be even. Assume the former case. Because there are odd number of pieces, the chess board is of size $(2n+1)×(2n+1)$. Let the position of $i$-th piece be $(a_i , b_i)$ $(1\le i \le 2n+1) $. Then it is on the black square if $a_i +b_i $ is even and on the white square otherwise. Then $S:=\sum_{i=1}^{2n+1} {(a_i +b_i )} $ is odd if $B$ is even and $S $ is even if $B$ is odd. But as pieces are on each row and each column, $S=\sum_{i=1}^{2n+1}{(i+i)}$ is even. Hence $B $ must be odd.
