Covering a chessboard with L-tetrominoes

Question

We all know the classic puzzle of trying to tile a mutilated chessboard with $2\times1$ dominoes. Let's imagine that, instead of dominoes, we have $n^2$ L-shaped tetrominoes and we want to tile a $2n\times2n$ chessboard using them (in any orientation, i.e. we're allowed to turn and flip tetrominoes if required). For which values of $n$ is this possible?

_{I got the idea for this from a question in the Danish Mathematical Olympiad.}

Answer 1

First of all, notice that

we can put two L-tetrominoes together to make a 4x2 block, and that such blocks will do the job whenever $n$ is even.

Next,

suppose $n$ is odd; then we have an odd number of tetrominoes. Now, colour the rows of our chessboard alternately red and blue; however you place an L-tetromino it will cover either 1 or 3 red squares; therefore the number of red squares covered has the same parity as the number of tetrominoes and is therefore odd. But in fact there are $2n^2$ red squares, an even number. So the covering can't be done for odd $n$.

In conclusion, the covering is possible if and only if

$n$ is even.

Answer 2

The total area will be $4n^2$, so the board is a $2nx2n$ one. If $2n$ is divisible by 4 ($n$ being even), it can be completed by building 4x2 rectangles to combine into such a board with 2 tetrominoes each.

Otherwise...

If $n$ is odd, an odd number of tetrominoes will be used. Paint the square this way:

....................
4...1...2...3...
3...4...1...2...
2...3...4...1...
1...2...3...4...

The different rotations and flips of the tetromino can be written like this:

image

In the shapes in the lower half of the picture, some numbers are repeated, but all the numbers must be distributed equally throughout the square. For every given $a$, there are two kinds of those shapes.

$m_1 = p_{-3} + p_{3}$
$m_2 = r_{-3} + r_{3}$
$m_3 = s_{-3} + s_{3}$
$m_4 = t_{-3} + t_{3}$

$p_{-3}+ p_{3} + 2r_{-3} + s_{-3} + s_{3} + 2t_{3} = s_{-3}+ s_{3} + 2t_{-3} + p_{-3} + p_{3} + 2r_{3}$
$2r_{-3} + 2t_{3} = 2r_{3} + 2t_{-3}$
$r_{-3} + t_{3} = r_{3} + t_{-3} => m_2 + m_4$ is even (by the same token, so is $m_1 + m_3$).

If the latter-half tetrominoes are used an even number of times, the number of the rest must be odd, but the opposite can be said when we flip the square vertically, creating a contradiction.

In short:

The square can be made if and only if $n$ is even.