# Covering chessboard with L-tetrominoes

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)

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.

• I'm not sure how to use your hints to solve the problem – Jebediah Nov 18 '13 at 7:14
• @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. – Calvin Lin Nov 18 '13 at 15:03
• There are $n^2 - 4$ squares after the removal. An L-tetromino covers a multiple of 4 squares per block. – Jebediah Nov 20 '13 at 15:04
• @Jebediah Great. Do you see why that shows that $n^2-4$ must be a multiple of 4, and hence $n$ is even? – Calvin Lin Nov 20 '13 at 21:20
• I don't see why $n^2-4$ must be a multiple of 4. Care to explain? – Jebediah Nov 21 '13 at 4:34