# Plane tesselation, using stairs $n\times n$, is it possible?

The other day I was constructing new mathematical problems for my pupils and thought of something like this:

Given the infinite sequence of "stairs" $$n\times n$$, constructed from $$1\times1$$ squares:

$$\hspace{130pt}$$ $$\cdots$$

tesselate whole plane with it, using $$\underline{\text{every}}$$ "stair" and $$\underline{\text{only once}}$$.

Unfortunately, neither I could prove it, nor my pupils. If you know a solution, please post it here, or any reference link will be much applied. But actually, if we weaken either of two initial conditions it becomes very easy. Let me demonstrate:

1. If we are allowed to use every stair, but not only once, then the solution comes from the idea of merging every $$(2k-1)\times(2k-1)$$ stair with $$2k\times2k$$ stair into square $$2k\times2k$$ like this:

$$\hspace{110pt}$$ $$\cdots$$

Our plane tesselation is constructed like this: we stack $$2\times2$$ squares in an infinte vertical strip (roughly speaking $$2\times\infty$$), then to the right of it we put $$4\times\infty$$ infinte strip of stacked $$4\times4$$ squares, to the left we put $$6\times\infty$$, then to the right we put $$8\times\infty$$, then again to the left we put $$10\times\infty$$ and so on. Plane tesselation complete!

2. If instead we are allowed to use only once, but not every stair, then the solution comes from idea of squaring the square: just like in previous solution, merging every $$(2k-1)\times(2k-1)$$ stair with $$2k\times2k$$ stair give us a sequence of squares $$2k\times2k$$.

Plane tesselation constructed like this: we join some squares to make a bigger square $$S_1$$ with the help of squaring the square method, then we make new huge square $$S_2$$, consisting of obtained square $$S_1$$ as a smallest one, and others to complete construction and so on.

To point it out, we can solve the weaken problem, but is there a solution for a stronger one with both conditions kept? If you have an answer, post it here please, or at least give me a link.

• Interesting problem. It is enough to show that the plane can be tiled by using every square of even order just once, or that the plane can be tiled by using every square just once. I suspect that Beatty sequences (en.wikipedia.org/wiki/Beatty_sequence) are deeply involved. For instance, it is easy to show that a quarter of plane minus the square in the corner can be tiled by using every square $F_n\times F_n$ (where $F_n$ is a Fibonacci number and $n\geq 2$) just once. Apr 21, 2016 at 17:21
• Stairs tile more easily that squares, so a tiling is likely. Will stairs of order 1 to 20 fit into stair order 55? Apr 21, 2016 at 21:26