Showing that a tiling of a chessboard is impossible

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Homework Statement


Delete one of the corner squares of a chessboard. Is it possible to cover the remaining area by ##21## dominoes of size ##1 \times 3##?

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The Attempt at a Solution


I claim that it is impossible. Here is my attempt to describe my solution in words. Go to the top left corner of the chessboard, and color it red. Move right, and color the square orange. Move right and color the square yellow. Move to the right and color the square red. Continue this pattern, moving right, and when reaching the end, starting from the very left of the next eight squares. In the end, there will be 22 red squares, 21 orange squares, and 21 yellow squares. Note that in this arrangement the top left square is red, the top right is orange, the bottom left is yellow and the bottom right is red. Now, WLOG delete the top right orange square, then the chessboard could not be tiled, since we must have an equal number of of red, orange, and yellow squares since each domino covers precisely one red, one orange, and one yellow square. If the orange square is deleted, we will have 22 red squares, 20 orange squares, and 21 yellow squares.

This proves that board cannot be tiled, since deleting anyone corner is the same as deleting any other corner, because of symmetry.
 
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I agree. QED
 
.Scott said:
I agree. QED
I have one question. Say that instead of deleting the orange square I deleted the top left red square. Then we would have 21 red squares, 21 orange squares, and 21 yellow squares. I know that, by symmetry, I proved that this cannot be tiled, but is there a direct explanation in this case as to why this arrangement cannot be tiled? Of course we can't, but it "seems" like we should be able to, since we have 21 of each color.
 
Since removing a corner square can result in a 20-21-22 result, it is demonstrated that it is not possible.
Just because you get the 21-21-21 result with a different (rotated) configuration does not mean that it can be tiled. Nothing is excluded by the 21-21-21 result.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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