Induction: Each square can be covered by L-stones

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mathmari
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Hey! :o

A square with the side length $2^n$ length units (LU) is divided in sub-squares with the side length $1$. One of the sub-squares in the corners has been removed. All other sub-squares should now be covered completely and without overlapping with L-stones. An L-stone consists of three sub-squares that together form an L.

I want to draw the problem for the first three cases described above ($1 \leq n \leq 3$). Then I want to show the following using induction:

For all $n \in N$ the square with side length $2^n$ LU can be covered completely and without overlapping with L-stones, after one of the sub-squares in the corners has been removed.
For the first part:

View attachment 9354

Is the drawing correct? (Wondering)
 

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Can we use the sketch of the case $n=2$ to get the one of the case $n=3$ ? (Wondering)

Is it maybe as follows?

The upper right sub-square is the one of case $n=2$. For the other sub-squares we have to fill them completely.

(Wondering)
 
mathmari said:
Can we use the sketch of the case $n=2$ to get the one of the case $n=3$ ? (Wondering)

Is it maybe as follows?

The upper right sub-square is the one of case $n=2$. For the other sub-squares we have to fill them completely.

Hey mathmari!

I think so yes.
Suppose we use the same case $n=2$ square to fill each of the 4 sub squares of the case $n=3$.
Then we have 3 cells left that we still have to fill don't we?
Can we align them so that we can put another L-square into it? (Wondering)
 
Klaas van Aarsen said:
I think so yes.
Suppose we use the same case $n=2$ square to fill each of the 4 sub squares of the case $n=3$.
Then we have 3 cells left that we still have to fill don't we?
Can we align them so that we can put another L-square into it? (Wondering)

To do that we have to make the empty cell in that corner so that the three empty cells make a L, or not? (Wondering)
 
mathmari said:
To do that we have to make the empty cell in that corner so that the three empty cells make a L, or not?

Yes. So the sub squares at left-top, left-bottom, and right-bottom would have their empty cell at the center.
Those empty cells have the shape of an L then, allowing for another piece. (Thinking)
 
Klaas van Aarsen said:
Yes. So the sub squares at left-top, left-bottom, and right-bottom would have their empty cell at the center.
Those empty cells have the shape of an L then, allowing for another piece. (Thinking)

I see! Thanks a lot! (Mmm)