A logical task on placing the dices on a chessboard

You are given an 8×8 chessboard. Two opposite catty-corner angles and 31 dominoes were removed. Each domino can close two chessboard squares. Is it possible to fill the whole board with dominoes? Prove your answer.

At first sight, it seems that it is possible. We have the 8×8 board, therefore, 64 cells, then we exclude two cells and 62 ones remain. It looks like 31 dices should fit there, right?

When you try to spread out the dices in the first line, we have only 7 squares at our disposal, one dice goes to the second line. Then we place the dice in the second line, and again one dice goes to the third row.

In each line there will always be one dice that needs to be transferred to the next line, no matter how many layouts we try, we will never be able to decompose all the dices.

The chessboard is divided into 32 black and 32 white cells. Once removing the opposite corners’ cells (the cells have the same colour), we leave 30 cells of one and 32 cells of a different colour. Thus, now we have 30 black and 32 white squares.

Each dice that we put on the board will occupy one black and one white cell. Therefore, 31 dices will occupy 31 white and 31 black cells. However, we have only 30 black and 32 white cells. Therefore, the decomposition of dices is impossible.