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:

  1. five $-1$'s and four $1$'s
  2. five $1$'s and four $-1$'s
  3. three $1$'s and six $-1$'s
  4. three $-1$'s and six $1$'s
  5. two $-1$'s and seven $1$'s
  6. two $1$'s and seven $-1$'s
  7. one $1$'s and eight $-1$'s
  8. one $-1$'s and eight $1$'s

regardless, the sum will never be $0$.

but how would i prove that?

  • 1
    $\begingroup$ Look at the 8 sums you calculated for a 3 by 3 board. What do all of these sums have in common? $\endgroup$ – Joey Zou Nov 21 '15 at 5:23
  • $\begingroup$ Actually, I'm not so sure that just specifying how many 1's and -1's is enough to determine the sum $a_1+a_2+...+a_n+b_1+...+b_n$. Work out a few examples of the sums you get by filling in a 3 by 3 in different ways. What do you notice about the sums? $\endgroup$ – Joey Zou Nov 21 '15 at 5:30

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.


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