# Solution

Is it possible to cover a chessboard with dominoes that each cover exactly two squares? What if the squares in two diagonally opposite corners are removed? Explain.

It is certainly possible to cover a chessboard with dominoes that each cover exactly two squares. Consider the following:

However, if the two diagonally opposite corners are removed, this changes.

We argue indirectly, using the pigeon-hole principle...

Assume it is possible. Note that the diagonally opposite corners of a complete chessboard are of the same color. Without loss of generality, let's suppose the two corners removed were both white. Consequently, there are only 30 white squares left from the original 32 on a complete board, while the number of black squares remains unchanged at 32. These 32 black spaces will play the role of our "pigeons". Each domino, no matter how it is placed on the board to cover exactly two squares will always cover exactly one black square and exactly one white square. With 62 places to be covered, 31 dominoes will be needed. These 31 dominoes will be our "pigeon-holes". By the pigeon-hole principle, two of the black squares must be covered by the same domino. This is clearly impossible, so we have our desired contradiction. The reverse of our assumption must be true:

It is not possible to cover a chessboard with dominoes that each cover exactly two squares if the squares in the two diagonally opposite corners have been removed.

QED.