Tiling With Squares

From RJ's configuration, we see that we can use just a single $1 \times 1$ tile. It remains to show that we cannot use 0 $1 \times 1$ tiles. We will prove this by contradiction.

Suppose there exist a tiling with only $2 \times 2$ and $3 \times 3$ tiles.
Color each alternating row blue and then green.
Then, each $2 \times 2$ tile covers 2 blue and 2 green tiles, and each $3 \times 3$ tile either covers 6 blue and 3 green, or 3 blue and 6 green tiles. Let the number of such tiles be $A, B$ and $C$ respectively.

Since there are $6 \times 11 = 66$ blue tiles and $5 \times 11 = 55$ green tiles, hence we get the system of equations

$2A + 6 B + 3 C = 66 \\ 2A + 3B + 6C = 55$

Taking the difference of the two equations, we get that $3(B-C) = 11 \Rightarrow B-C = \frac{11}{3 }$ . This is not possible since $B$ and $C$ are integers, hence we have a contradiction.

Follow up questions: This solution shows that if $6 \not \mid n$ , then we will need at least 1 tile. Can we figure out the answer for general $n$ ? I suggest looking at cases based on $n \pmod{6}$ .

1. The cases of $6n+ 2$ , $6n+3$ , $6n + 4$ are easy to deal with
2. What about $6n+5$ which is like this? Is 1 tile enough? Can we use the $11 \times 11$ example in our construction?
3. What about $6n + 1$ ? Is 1 tile enough?