TITLE: Prove that the sum $a_1+a_2+....+a_n+b_1+b_2+...+b_n$ cannot equal to $0$
QUESTION [7 upvotes]: We are given an $n \times n$ board, where $n$ is an odd number. In each cell
  of the board either $+1$ or $-1$ is written. Let $a_k$ and $b_k$
  denote the products of the numbers in the $k$-th row and the $k$-th
  column, respectively. Prove that the sum
  $a_1+a_2+\cdots+a_n+b_1+b_2+\cdots+b_n$ cannot equal to $0$

I started off with some examples:

If its a $3 \times 3$ board, then:
you can either have:

five $-1$'s and four $1$'s
five $1$'s and four $-1$'s
three $1$'s and six $-1$'s
three $-1$'s and six $1$'s
two $-1$'s and seven $1$'s
two $1$'s and seven $-1$'s
one $1$'s and eight $-1$'s
one $-1$'s and eight $1$'s

regardless, the sum will never be $0$.
but how would i prove that?

REPLY [3 votes]: The product of the $a_i$ is equal to the product of the $b_i$. This is because each of these products is the product of all the elements in the $n\times n$ array.
So the numbers of $-1$'s among the $a_i$, and among the  $b_i$, have the same parity (both numbers are even or both are odd).
It follows that the combined number of $-1$'s among the $a_i$ and $b_i$ is even.
This ensures that the sum of all the $a_i$ and all the $b_i$ cannot be $0$. For if the sum is $0$, the total number of $-1$'s among the $a_i$ and $b_i$ must be $n$.  And $n$ is odd.