As usual in chess, let the bottom left-hand square be black. If we number the rows from the bottom and the columns from the left, then every black square is either in an odd-numbered row or an even-numbered column. That is, all the black squares can be found in rows 1, 3, 5, 7 and columns 2, 4, 6, 8.

Let the black squares in rows 1, 3, 5, 7 be marked A and the black squares in columns 2, 4, 6, 8 be marked B, and let the white squares in rows 1, 3, 5, 7 be marked C. Leave the rest of the white squares unmarked. Finally, let *a*, *b*, *c* denote the total number of pieces that sit on squares marked A, B, C, respectively. We want to show that *a* + *b* is an even number.

Each row contains an odd number of pieces. That means that *a* + *c*, the total number of pieces in rows 1, 3, 5, 7, is the sum of four odd numbers, which means it’s even. Likewise, *b* + *c*, the total number of pieces in columns 2, 4, 6, 8, is also the sum of four odd numbers and thus even. So

(*a* + *c*) + (*b* + *c*) = *a* + *b* + 2*c*

is even, and this implies that *a* + *b* is even.

(From V. Proizvolov, “Problems Teaching Us How to Think,” *Quantum*, January-February 2002, via Ross Honsberger’s *Mathematical Delights*, 2004.)