Dominoes on a board missing opposite corners

Two diagonally opposite corner squares are removed from an $8\times8$ chessboard, leaving $62$ squares. Can the remainder be covered exactly by $31$ dominoes, each covering two adjacent squares?

No. Opposite corners have the same colour, so the board is left with $32$ squares of one colour and $30$ of the other. Every domino covers one light and one dark square, so $31$ dominoes would need $31$ of each.