, 2009

Solution:
Every "corner" covers exactly 3 squares, so a necessary condition for the tiling to exist is $3 \mid m n$.

First, we shall prove that for a tiling with our condition to exist, it is necessary that both $m, n$ for $m, n>3$ to be even. Suppose the contrary, i.e. suppose that $m>3$ is odd (without losing generality). Look at the "corners" that cover squares on the side of length $m$ of table $m \times n$. Because $m$ is odd, there must be a "corner" which covers exactly one square of that side. But any placement of that corner forces existence of a $2 \times 3$ rectangle in the tiling. Thus, $m$ and $n$ for $m, n>3$ must be even and at least one of them is divisible by 3.

Notice that in the corners of table $m \times n$, the "corner" must be placed such that it covers the square in the corner of the rectangle and its two neighboring squares, otherwise, again, a $2 \times 3$ rectangle would form.

If one of $m$ and $n$ is 2 then condition forces that the only convenient tables are $2 \times 3$ and $3 \times 2$. If we try to find the desired tiling when $m=4$, then we are forced to stop at table $4 \times 6$ because of the conditions of problem.

We easily find an example of a desired tiling for the table $6 \times 6$ and, more generally, a tiling for a $6 \times 2 k$ table.

Thus, it will be helpful to prove that the desired tiling exists for tables $6 k \times 4 \ell$, for $k, \ell \geq 2$. Divide that table at rectangle $6 \times 4$ and tile that rectangle as we described. Now, change placement of problematic "corners" as in figure.

Thus, we get desired tiling for this type of table.

Similarly, we prove existence in case $6 k \times (4 \ell+2)$ where $k, \ell \geq 2$. But, we first divide table at two tables $6 k \times 6$ and $6 k \times 4(\ell-1)$. Divide them at rectangles $6 \times 6$ and $6 \times 4$. Tile them as we described earlier, and arrange problematic "corners" as in previous case. So, $2 \times 3, 3 \times 2, 6 \times 2 k, 2 k \times 6, k \geq 2$, and $6 k \times 4 \ell$ for $k, \ell \geq 2$ and $6 k \times (4 \ell+2)$ for $k, \ell \geq 2$ are the convenient pairs.