Tiling for $m\times n$ chessboard problem for both $m, n$ are odd

Question

Consider an $m\times n$ chessboard in which both $m$ and $n$ are odd. The board has one more square of one color, say, black, than of white. Show that, if exactly one black square is forbidden on the board, the resulting board has a tiling with dominoes.

So if at least one of the $m$ or $n$ are even, you can fully tile the chessboard. But what about odd?

Answer 1

Here is a crude drawing of a 7 by 11 chessboard which I will use to illustrate the construciton. The black squares are labeled A and B, because these two categories of black square use different constructions.

A   A   A   A   A   A   A 
  B   B   B   B   B   B   
A   A   A   A   A   A   A 
  B   B   B   B   B   B   
A   A   A   A   A   A   A 
  B   B   B   B   B   B   
A   A   A   A   A   A   A 

If the selected black square, "*," is an "A," then tile the remainder of its row and column with dominoes, then the remaining even by even rectangles with dominoes:

          1  
          1
  3 3 2 2 * 4 4 5 5 6 6
          7 
          7 
          8 
          8 

If the selected black square is a "B," then first surround that square with four dominoes as shown, then break the remaining grid 8 rectangles, each with at least one even (possibly zero) dimension:

       . . . 
       . . .
   . . 3 4 4 . . . . . . 
   . . 3 * 1 . . . . . . 
   . . 2 2 1 . . . . . . 
       . . . 
       . . .