Covering a rectangular board with Tetrominoes

combinatorial-geometrycombinatoricsdiscrete mathematicspuzzletiling

I am reading about a puzzle question that is about Tetrominoes and proving that if a rectangular board can be covered with T-Tetrominoes the board's number of squares has to be a multiple of 8.
The solution first proves that if a rectangular board can be covered by Tetrominoes one of the sides must be even. This part is clear.
Now the solution is using a coloring "trick" i.e. it colors both the board and the Tetrominoes. The coloring of the tetrominoes is either e.g.

-----------------------
|-BLACK-|-WHITE-|BLACK|  
 -------|-BLACK-|-----
         -------

or

------------------------- 
|-WHITE-|-BLACK-|-WHITE-|  
 -------|-WHITE-|-------
         -------

Also the rectangular board is considered as colored like a chessboard.
The part that I don't understand is the following.
Based on the fact that was proven earlier i.e. that one of the rectangular boards side must be even if the rectangular board can be covered with Tetrominoes at all, the solution consider this implies that the number of black equals the number of white.
This part (black = white) is not obvious to me. Could someone please help?

Best Answer

Suppose that one side has $2n$ cells. Then every row parallel to that side has $n$ black and $n$ white cells. Thus, if there are $m$ rows, there are $mn$ black and $mn$ white cells.

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