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Consider an n x n chessboard with all 4 corner squares removed. Prove that if the board can be covered with L-tetrominoes then n-2 is a multiple of 4. Is the converse true? (an L-tetromino is a plane figure shown below, constructed from four unit squares arranged in the form of L)

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Hint: By counting the number of squares, $n$ is even.

Hint: By using the standard coloring of $i+j \pmod{4}$ for square $(i,j)$, show that $n \neq 4k$.

Hence $n= 4k+2$.


I'm not certain about the converse.

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  • $\begingroup$ I'm not sure how to use your hints to solve the problem $\endgroup$ – Jebediah Nov 18 '13 at 7:14
  • $\begingroup$ @Jebediah How many squares are there in a $n\times n$ chessboard with 4 removed? How many squares are covered by an integer number of $L-$tetrominoes? Hence, show that $n$ is even. $\endgroup$ – Calvin Lin Nov 18 '13 at 15:03
  • $\begingroup$ There are $n^2 - 4$ squares after the removal. An L-tetromino covers a multiple of 4 squares per block. $\endgroup$ – Jebediah Nov 20 '13 at 15:04
  • $\begingroup$ @Jebediah Great. Do you see why that shows that $n^2-4$ must be a multiple of 4, and hence $n$ is even? $\endgroup$ – Calvin Lin Nov 20 '13 at 21:20
  • $\begingroup$ I don't see why $n^2-4$ must be a multiple of 4. Care to explain? $\endgroup$ – Jebediah Nov 21 '13 at 4:34

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