I have a question about a mathematical riddle which I already solved but still looking for a shorter/simpler solution:
Following problem: We consider a standard $8 \times 8$ chessboard and we cover it (completely!) with dominos of size $2 \times 1$ (therefore every domino tile cover exactly $2$ fields).
The question is if we can find a cover such that there doesn't exist a $2 \times 2$ subsquare which is covered exactly by two domino tiles or in other words in the cover there don't occure two "direct" neighbour domino tiles from following shape:
I have it already solved in following way: I claim that such covering isn't possible.
Argue via contradiction: Assume that it's possible. Consider the $2 \times 2$ squares of the chess board and consider the partial cover of directly neighboured $2 \times 2$ squares. If a cover as above really exist then up to symmetry on the common edge of the two neighboured squares there could only occure two following cases (here only the vertical pairs; horizontally: analogous):
The two neighboured squares share a common domino tile (the orange one)
they don't share any domino tile on the common edge
Now there are exactly $24$ such pairings between neighboured $2 \times 2$ squares (note we don't consider the diagonal neighbour pairs).
Now we count all domino tiles in following way:
-each pair of neighbour squares which share a unique common domino tile contribute a $1$ (the orange one)
-each pair of neighbour squares don't share a common domino tile contribute a $1$ with the unique tile beeing fully contained in only one square and intersecting the common edge (the grey one).
That's all. But then we get only $24$ tiles althought there are $32$. Contradiction.
I guess this argument works but I think that it's too cumbersome. Does anybody have an easier / not too circumstaneous way to show it?