Question and Solution to the process of proving the perfect coverage of the Board

The so-called perfect checkerboard coverage is like this. On the 8-by-8 chess board, cover it with a 1-by-2 rectangle. Shows the checkerboard:

If all the grids on the board are covered, and there is no leakage or repeated coverage, then such coverage is perfect coverage. it is required to prove that there is no perfect coverage in the following chessboard, where the red lattice is not allowed to be covered.

The teacher's or textbook's proof is like this. Edit the two lattice numbers of the rectangle into 1 and 2 respectively, and then overwrite the checker as follows:

Because the Red Square cannot be covered, the two ones in the Red Square need to be removed, so the number of one on the entire board is not equal to the number of two. however, if a chessboard is perfectly covered, the number of Squares used must be an integer, so the number of 1 on the board must be equal to the number of 2. this introduces contradictions. therefore, perfect coverage does not exist. I firmly believe that this conclusion is correct, but I doubt the process of proof. as there is no limit on the placement of blocks in the covering board, we will change the last block in the way shown below. Pay attention to the two numbers at the rightmost of the bottom right corner, instead of 2, 1, and 2.

In such coverage, after removing the numbers in the red lattice, the number of 1 in the entire board is equal to the number of 2, therefore, the above proof method cannot come to the conclusion that there is no perfect coverage. therefore, I think that the proof in the textbook is not very strict (or even wrong). This is where I am questioning. my proof is given below.

Proof 1:
Select one of all existing perfect covers, and then remove the squares that are not allowed to be covered in red. because the two red squares are not connected, you must remove the two squares. in this way, after removing the Red Square (that is, two squares), there will be two more uncovered grids. They are A1, A2, B1, or B2, as shown below.

If we can use a square to cover A1, A2, B1, or B2, the perfect coverage will exist, but it can be seen that this is impossible, it should be because square a and square B are not connected at all, so this is impossible. this proves that perfect coverage does not exist. I believe this is a proof of concept. however, there is room to be refuted. that is, there is no coverage method that can connect non-covered squares. If so, perfect coverage will exist. obviously, it is difficult to determine whether such coverage exists from the perspective of the visual method. It should be said that it cannot be seen clearly. well, the second proof is provided below, which proves that it is unrecoverable.

Before providing the second proof, we should first discuss a problem as a proof of preparation. assume that only one row is required to cover the board, and two grids in this row are not allowed to be overwritten. so when the two grids that cannot be covered meet what conditions are there perfect coverage? As shown in the following figure, there are three scenarios where two uncovered grids split a Single Board:

We can see the three situations from the square number on the left of the first red grid, the square number in the middle of the two red grids, and the square number on the right of the second red grid. obviously, only when these three numbers are even (including zero) will there be perfect coverage. This is the condition to be met. the second proof is given below:

The first row in Figure 2 does not exist because the number of grids that can be covered is an odd number. therefore, at least one square must be placed vertically to overwrite the first line. in this way, the square occupies the first and second rows. therefore, for the coverage of the second row, there is an extra grid that cannot be overwritten (otherwise, it is overwritten again ). we can assume that there is no limit on the number of blocks placed vertically in the first line. then these vertical blocks separate spaces in the first line into several parts. obviously, the number of squares in each part must be an even number (including zero). Otherwise, the first row cannot be overwritten.

Let's look at the number of squares adjacent to the first square in the first line, assuming n. obviously, n is an even number (otherwise the first row cannot be overwritten), so the N in the second row is an odd number, because the lattice under the first row of the red lattice needs to be filled, the number of the second row is n + 1, so it is an odd number. to overwrite the space where N is located in the second row, at least one square in the space where N is located in the second row is placed vertically, so the number of squares where N is located in the third row becomes an even number. now, after recursion, we find that in an odd number of rows, the number of squares where N is located is an even number, while the number of even rows is an odd number, which cannot be perfectly overwritten.

Is there such a possibility that two uncovered grids overwrite 8th rows are exactly in the first and last, so that there are 6 grids in the middle to overwrite them. In this case, you need to place a square in a row along the horizontal direction, as shown in:

Obviously, this is impossible, because the board is symmetric. If you rotate the board, you can find that this placement method is exactly the same as the one discussed currently. Therefore, the original proposition is proved.