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Show that one cannot make a 8×8 square using 15 T-tetrominoes and 1 square tetromino.

Its a coloring problem. Unable to solve. please help.

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  • $\begingroup$ Typically these problems either have a simple solution using congruence/parity arguments (as the Answers show in this case) or they turn out to be very difficult. $\endgroup$ – hardmath Jun 18 '15 at 22:16
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Hint: Color the 8x8 square as a chessboard. The T-tetromino covers three black and one white or three white and one black. In any case the number of black squares covered is congruent to 1 mod 2.

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Color the $8 \times 8$ square like a checkerboard with black and white squares. Now assume you can place 15 T-tetrominoes and one square on the checkerboard. Then there is a certain number (say $n$) of T-tetrominoes that cover 1 black and 3 white squares, there are 15-n T's that cover 3 black and 1 white square, and the $2 \times 2$ square covers two black and two white squares.

Then you have covered $n + 3(15-n) + 2 = 47 - 2n$ black squares. That's an odd number, but there are 32 black squares to be covered. So that doesn't work.

Conclusion: It is impossible.

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HINT: Each $T$-tetromino has $3$ small squares of one color and one of the other, while a square tetromino has $2$ of each. An $8\times 8$ square has $32$ of each. Say the colors are black and white. Call a $T$-tetromino white if it has $3$ white cells and black if it has $3$ black cells. Let $w$ be the number of white $T$-tetrominoes. Show that no matter what $w$ is, your $16$ tetrominoes do not have $32$ white and $32$ black cells.

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