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".
The simpliest disposition I can think is with 8 pieces in one of the diagonals (i.e white). This meets the asked conditions.
a) The minimal change to the disposition of the 8 pieces in the chessboard without breaking the odd-rows-columns rule involves two pieces, in a reciprocally inversal row (or column) movement. This is the minimal elementary change in the chessboard that conserves the odd-number rule, and it is always "parity-color" (i.e: if the initial disposition was 8 in white diagonal, after this elementary change it will be either: or 8 white and 0 black or 6 white and 2 black). As every possible disposition rule-compliant of the 8 pieces in the chessboard can be adquired as the sum of a number of elementary changes parity-color conservative, we conclude that for any 8 pieces disposition rule-compliant always there will be a even number of pieces placed in black squares.
b) Now there is a similar way of reasoning for more than 8 pieces. If we want now to add more pieces to our 8 pieces chessboard without breaking the rule, we need to add at least 4 more pieces (otherwise the rule is over) and to place them forming a rectangle in the chessboard. But this minimal elementary addition do not change parity-color because it forms a rectangle in the chessboard. As every other rule compliant disposition in the chessboard can be acquired as the sum of a number of elementary additions which are parity-color conservative [this I see now is not true, so the argument is flawed], we conclude that for any disposition rule compliant there will be necessarily an even number of pieces placed in black squares.
I leave this as an unsuccessfull try
Edited on March 19, 2016, 11:47 am
Edited on March 19, 2016, 11:49 am
Posted by armando
on 2016-03-19 05:27:50