An Algebra Problem By Mohammed Imran

The answer is Yes. We will prove by induction. If n=1, then we have a $2*2$ square. When one tile is removed from the square, the remaining figure is an "L" tile. Now, assume that the proposition is true for n=k. Where k is a natural number. Then for n=k+1, there will be 4 $2^k*2^k$ grids. Now, from 3 of those $2^k*2^k$ grids, if we take 3 squares in the innermost corner, then, they shall form an "L" tile. Now if we take the final innermost corner of the remaining $2^k*2^k$ grid, we have taken one square. Now if we remove the "L" tile which we formed we have that all the 4 $2^k*2^k$ grids have lost a square tile. This implies that all of the 4 $2^k*2^k$ grids without a square can be filled with "L" shaped tiles which is what we assumed. And when the "L" tile is added back, the $2^k+1*2^k+1$ will not have a square tile and the remaining portion can be filled with "L" shaped tiles. And hence our proposition is true for all n who are natural numbers.