Grids don't contain 'L'

You can put together four L shapes to make another L shape that is twice the size: Now, given a $2^n\times2^n$ grid, we can fill it with L shapes of decreasing size, starting with a $2^n$ length L shape, then going down to $2^{n-1}$ , and so on. We can also do this in such a way as to avoid any particular blacked out square on the board, by simply selecting the quadrant which the blacked out square is in and placing the L shape only in the other quadrants. Then we can simply cut each large L shape into four smaller L shapes, and repeat until all the L's are 3 units in area.

Some other examples for different locations of the black square are shown below.

Here, you can subtract the area of the grids with 1 and then check if the differences acquired are divisible by 3 : $S_{1}=16-1=15$ $S_{2}=64-1=63$ $S_{3}=256-1=255$ We can now see that the above given differences are divisible by 3. Thus, the answer being Yes

I will only show $2^{2n}-1$ is divisible by $3$

$2^{2n}-1=(2^n-1)(2^n+1)$

Since $2^{n}$ is not divisible by $3$ , it is either $3k+1$ or $3k-1$

Then, $2^{2n}-1=3k(3k+2)$ or $2^{2n}-1=3k(3k-2)$

The length of the square is 2^n. The area of the square is 2^(2n). Removing the black square, the area is 2^(2n)-1.

We prove by recurrence that 2^(2n)-1 is divisible by 3.

For n=0, 2^0-1=0, divisible by 3 (though not applicable to the problem).

For n=1, 2^2-1=3, divisible by 3 (applicable to a smaller square)

For n=2, 2^4-1=15, divisible by 3 (initial square of the problem)

Let's suppose that 2^(2n)-1 mod 3 == 0

Then 2^(2n) mod 3 == 1

2^(2(n+1)) mod 3 == 4 mod 3 == 1

2^(2(n+1)) - 1 mod 3 == 0

So it's always possible to tile the square of length 2^n with L-shaped tiles of area 3 each, with one remaining blacked square.

Yes bcuz anything is possible

I don't know if this makes sense but I reasoned like this.

If we know that we can solve this problem in a grid with side $2^n$ how can we see the problem as if it were in a grid with side $2^{n+1}$ ?. Well a grid with side $2^n$ has $2^n * 2^n$ 1x1 tiles, a good way to look at a grid with side $2^{n+1}$ is as if it were the same grid but with 2x2 tiles. But how can this help ? Actually it's straightforward, just add an L shape to make to black dot a 2x2 black dot and we do this in a way that the black 2x2 dot is positioned with even coordinates and this is possible because squares with do not touch the edges(in this case it's easy to see why), we get four possibilities of L shapes which accound of (odd, even), (odd, odd), (even, odd) and (even, even). Once this done, deal with the grid as if it were a $2^n$ one and by induction it can be solved.

EDIT 1: Adds picture of recursion

EDIT 2: Adds second picture of a similar approach

We will use induction - 1) prove that n = 1 works. (meaning a grid of size 2^1 = 2 works. 2) prove if n - 1 works, then n works.

1) Obviously n = 1 works, because there is only an L shape left.

2) 2^n = 2 × 2^n - 1 Therefore, we can split a 2^n × 2^n into four 2^n - 1 × 2^n - 1 grids. Since n - 1 works, we can fill the three grids without the blackened square with one large L shape and use n - 1's solution for the remaining grid.

Thus, the answer is yes.

For any grid of length n the area of the all the L shapes combined is (2^n) - 1)(for integer n greater than or equal to 2)

For L shapes to always be with the grid, the ratio of ((2^n) - 1) and 3 must be an integer

Which is the case therefore the answer is yes.

From the initial position we are turning at each step one of the L-shaped pieces and we see how the square blacked out can appear in any position without limitations.

Just make tiles of 3x2 rectangles with 2 Ls and order them up around the 8 by 8, and repeat the process around every third in progression n^2 where n is odd, because in every third there will be 2 tiles on the side empty to fit the last or the first horizontal 3x2, so we have that case solved, next we have the inbetween progressions with even ns where multiple paterns work, among which making vertical 3x2 and following the same pattern.

This is a computed proof for the first 64 grid sizes (any value higher is not possible in C#)

For 2n by 2n square, solving for n=1, 2, 3. By analysis of pattern, it is apparent that L shape can be formed for n=n.

Can a 2nx2n square with a 1x1 black square be tilled by L pieces?

LET'S ASSUME WE CAN TILE AN nxn SQUARE. Now I'll show that we can tile a 2nx2n square.

The 2nx2n is just four nxn squares: square nxn A with a black; and three squares nxn B,C,D black free.

Temporarily whiten the black square. And blacken the four 1x1 squares in the center of ABCD.

Now by the induction assumption we can obviously tile each nxn square. But note that the blacks in BCD can be tiled by a small L piece. So we're done with tiling BCD.

If we go back back to the real situation our tiling of BCD still works! Now all we need is to tile A which we can do since A is nxn.

QED

(To use induction we require that the base case is verified. And of course, we know we can tile a 4x4 square with a black square using L pieces.)

When a shape’s side is raised by scale factor of 2, the area is increased by a factor of 4. This means the new number of L’s will be equal to the previous amount x 4. This will fill all but the black square which will also me multiplied by 4, giving 4 black squares. However since only one is required. 3 of the 4 become a new L while the other remains as the black square.