$RTP$ : Let's examine a $2^n⋅2^n$ square. Show by induction, that if we remove one block from this square, then the remaining square can be covered with...

.. these squares in such a manner that the images do not overlap each other.

$Solution$ : The remaining square can be covered, if the product $2^n⋅2^n-1$ is divisible by 3 for all $n∈N$, i.e our statement to prove by induction is$$3∣2^n⋅2^n-1$$ Lets show the base case is true, $n=1$. That is $3∣3$ which is true. Assume true for $n=k$ and show it's true for $n=k+1$, where $k∈N$. So we assume that$$3∣2^k⋅2^k-1$$ is true. Now let's take a look at $n=k+1$ $$3∣2^{k+1}⋅2^{k+1}-1$$$$3∣(2^{k}⋅2)⋅(2^{k}⋅2)-1$$$$3∣2^{k}⋅2^k-1⋅2⋅2$$

Based on our assumption, we can say that $2^k ⋅2^k -1$ is divisible by 3. Now if we multiple this by $2⋅2$ it will still be divisible by 3. Thus our statement is true by induction. Is this correct? I'm unsure about my analysis of the remaining square.


No, it is not correct. It assumes that if, after removing a square from the bord, the number of remaing squares is a multiple of $3$, then the board (minus the removed square) can be tiled with those pieces that you have mentioned. This is not true in general. If you have a $5\times5$ board from which you remove a square from the second row, then the remaining squares cannot by tiled in that way, in spite of the fact that there are $24(=8\times3)$ such squares.

You will find here a proof of the statement that you want to prove.


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