Looks correct to me.Suppose we have an 8 by 8 chessboard where the squares always alternate color from black to white. Suppose two squares that are directly next to each other are removed from each of the two opposite corners (so in total, two black squares and two white squares are removed). Can the remaining 60 squares, 30 black and 30 white, can be covered by 15 T-shaped figures (of three whites and one blue one or three blue squares and one white one)?
So here is my attempt of proof:
Suppose we can. Without loss of generality, let us consider the black tiles. So there exists some combination of the 3 black, 1 white T-shapes and 1 black, 3 white T-shapes that cover 30 black squares. Obviously, this combination must add up to 15 (perhaps 15 3-black t-shapes and 0 3-white t-shapes, etc). So the following system of equations must have integer solutions:
, where X is the number of 3-black t-shapes and Y is the number of 1-black t-shapes). However, this is a contradiction. So there is no combination of the 3-black t-shapes and 1-black t-shapes that cover exactly 30 black squares and cover the total 60 squares, since we must use 15 t-shapes to cover the 60 squares. So therefore it cannot be done. Q.E.D.
Is this correct? Thanks.
By the way, you wrote blue in a couple places you meant black.