Cover Chessboard with Domino Tiles

Solution 1:

How about this:

Let's number the rows $1$ through $8$ from top to bottom, and the columns $1$ through $8$ from left to right. Cell $(x,y)$ means the cell in row $x$ and column $y$

You need a tile to cover cell $(1,1)$.This can be done in two ways, but by symmetry, we only have to consider one of these, so let's consider placing it horizontally, i.e. cover cells $(1,1)$ and $(1,2)$:

Now we need to cover cell $(2,1)$. In order to avoid making a $2\times 2$ subsquare made up of two tiles, there is only way way to place a tile under it, so that will cover cells $(2,1)$ and $(3,1)$:

Now we need to cover $(2,2)$. Again, there is only way way to do this in order to avoid making a $2\times 2$ subsquare made up of two tiles: cover cells $(2,2)$ and $(2,3)$.

OK, and now keep placing tiles along this basic 'diagonal' of the chess-board: you'll find all the placements are forced if you want to avoid a $2\times 2$ subsquare made up of two tiles. But, at the end of the diagonal, you end up having to use one tile to cover $(7,6)$ and $(8,6)$, and another one to cover $(7,7)$ and $(7,8)$:

... and now you are forced to also place one on $(8,7)$ and $(8,8)$, and get a $2\times 2$ subsquare made up of two tiles. (also note that the two parts of the board as of yet uncovered each have an odd number of squares left, so they can no longer be completely covered)

So, it is indeed impossible to do a complete tiling without getting a $2\times 2$ subsquare made up of two tiles.