This is part of an archive of questions and answers that may be of interest to puzzle enthusiasts. Part 1 contains the index to the archive. Read the rec.puzzles FAQ for more information. If one side of the rectangle is not divisible by a, and the other is not divisible by b, then the tiling is impossible. Thus the number of bxb starting at row 2 must BE an a-multiple.
Every square tile covers an a-multiple of black squares.
Every square tile covers an a-multiple of black squares. This statement of how to position the colouring isn't good enough, I'm afraid. Suppose N columns are black with this positioning. The difference between these two numbers of black cells is H, which is not a multiple of a. The original rectangle is the union of four such rectangles.