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$RTP$ : 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.

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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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