# L-tromino pair!

Amy is playing with different polyominoes. She suddenly thinks of a problem as follows.

Choose two positive integers $$m,n$$. If we can use only L-trominos to tessellate a $$m\times n$$ rectangle with no gaps, overlaps, or any square hanging out the rectangles, then we call the pair $$(m,n)$$ L-tromino pair.

She calls her brother Ben, and the genie, and try to figure out all L-tromino pairs. The genie is super smart and found all of them with a proof. Can you?

Problem by myself.

Here is a picture of L-tromino, if you want to see it:

Obviously both dimensions of a tileable rectangle must be at least $$2$$. Also, since the area of a tromino is $$3$$, the area of a tileable rectangle is a multiple of $$3$$, and hence at least one of the dimensions is a multiple of 3.

First some easy cases:

$$3k\times2n$$: Two trominoes form a $$3\times2$$ rectangle. Therefore any $$3k\times2n$$ rectangle is trivially tileable.

$$6k\times(2n+3)$$: This rectangle splits into a $$6k\times3$$ and a $$6k\times2n$$ rectangle, both of which are instances of the trivially tileable case above.

The trickest case is this:

The above cases deals with all rectangles where one of the dimensions is even. So there now remains only those with odd dimensions.

$$9\times5$$: This rectangle can be tiled:

$$(6k+9) \times (2n+5)$$: Any rectangle with odd dimensions, one dimension a multiple of 3, and no smaller than $$9\times5$$, can be tiled. You can calve off a rectangle of size $$6k\times(2n+5)$$ which has already been shown to be tileable to reduce it to $$9\times(2n+5)$$. You can then calve off a tileable rectangle of size $$9\times2n$$, leaving the tileable $$9\times5$$.

Now it just remains to be shown that $$3\times(2n+1)$$ is not tileable. This is fairly obvious when you try it. The only ways that you can fill the short edge of the rectangle will create a $$3\times2$$ block. Therefore the rectangle unavoidably gets reduced to the untileable $$3\times1$$ shape.

To recap, the L-tromino pairs are $$(m,n)$$ where $$m,n\ge2$$, at least one of $$m$$ or $$n$$ is divisible by 3, and if they are both odd then $$m,n\ge5$$.

• Yes, this is the correct answer! I used an hour or so to solve this problem, not as fast as the genie. – Culver Kwan Aug 15 '20 at 7:28