We call a tiling of an $m \times n$ rectangle with corners (see figure below) "regular" if there is no sub-rectangle which is tiled with corners. Prove that if for some $m$ and $n$ there exists a "regular" tiling of the $m \times n$ rectangular then there exists a "regular" tiling also for the $2m \times 2n $ rectangle.
For every square 1x1 make it 2x2. It's easy to see that the corner could be made from corners and we can save the covering with corners. Therefore if mxn is regular, 2mx2n is too.
Find an exemple for $m\cdot n$. Then make a "big" corner $(2 \cdot 2)$ Then put the "big corners" on the table $2m \cdot 2n$ the same you put the small corners on the $m \cdot n$ table. It is easy to make a "big corner" with small corner.
The problem is finished.
PS: Now I noticed that VicKmath7's solution is the same.
