[Guest]

On every square of a 1997x1997 checkerboard is written +1 or -1.

Let, R_i = Product of all the numbers in the ith row, and:

C_i = Product of all the numbers in the ith column.

Prove that Sum (i = 1 to 1997) (R_i + C_i) is never equal to zero.

Edited March 2, 2010 by K Sengupta

[Guest]

Let S be the total sum, that is, S = Sum(i=1 to 1997)(Ri + Ci).

What happens to S if we replace the number of just one square?

Changing the number of square at row i and column j makes Ri and Cj change either from -1 to 1 or from 1 to -1. Hence Ri and Cj either decreases by 2, remains the same or increases by 2. Therefore, Ri + Cj changes by either -4, 0 or 4. Since all others rows and columns do not change, S changes by either -4, 0 or 4. In particular, S mod 4 is invariant.

Now, if we change, one-by-one, all squares with -1, then we end up with a 1997x1997 chessboard filled up with 1s. For this configuration we have Ri = Ci = 1 for all i from 1 to 1997 and, thus, S = 2*1997 = 2 mod 4. Therefore, S for the initial configuration must be 2 mod 4. In particular, S cannot be 0.