In a certain chess position, each row and each column contains an

**odd** number of pieces.

Prove that the total number of pieces on black squares is an **even** number.

Source: Proizvolov, “Problems Teaching Us How to Think".

There must be an even number of pieces (the sum of eight odd numbers is even). Furthermore, each of the four stripe parities must each be even (sum of four odd numbers is even): odd rows, even rows, odd columns, and even columns.

Let X be the number of pieces on odd row white tiles. Let Y be the number of pieces on the odd row black tiles. X+Y is even, which means that X and Y are the same parity.

Let V be the number of pieces on odd column white tiles. Let W be the number of pieces on the odd column black tiles. V+W is even, which means that V and W are the same parity.

The odd row white tiles and the odd column white tiles are the same tiles, so X=V. Then all four X, Y, V, and W are the same parity.

All the black tiles are the union of the odd row black tiles and the odd column black tiles. Then the number of pieces on black tiles is Y+W. Since Y and W are the same parity their sum must be even!

*Edited on ***March 24, 2016, 11:40 am**