Why Can’t a Chessboard with Two Opposite Corners Missing Be Tiled?

A standard chessboard has 8 × 8 = 64 squares. One domino covers two adjacent squares, so 32 dominoes can tile the complete board. Remove two opposite corners and 62 squares remain, apparently just right for 31 dominoes. Yet no arrangement can cover them all.

The key is the alternating colouring. The original board has 32 black and 32 white squares, and opposite corners always have the same colour. If two white corners are removed, the board has 30 white and 32 black squares left. Every domino covers adjacent squares, which are necessarily one black and one white. No matter how the 31 pieces are rotated, they must therefore cover 31 squares of each colour. The damaged board does not contain those numbers, so a tiling cannot exist.

This is not a conjecture or a failure to find a clever arrangement; it is a proof. Colouring exposes an invariant: each placed domino always increases the covered totals by one black and one white square. It also shows why having the right area is necessary but not sufficient. Local shape and structural constraints can still prevent the whole arrangement.

https://pi.math.cornell.edu/~mec/Summer2009/Leung/puzzles_p4.htm