2 × 4 (smallest rectangle)

3 × 8

complete

smallest rectangle: 2 × 4

**Proposition.** The L tetromino is even.

**Proof.** Number the squares by

(x, y) |---> { 1 if x is even { 0 if x is odd

No matter how it is placed, an L tetromino covers an odd total.
However, any rectangle with area divisible by 4 covers an even total.
Therefore, if it is tiled by L tetrominoes, there must
be an even number of them. QED.

Golomb [1] asks for all rectangles which can be tiled by the L tetromino.
This proposition is the key to Klarner's solution.
He gives a more general criterion using the same numbering in
[2, Theorem 4].

**References**

[1a] S.W. Golomb,
Covering a
Rectangle with L-tetrominoes, Problem E 1543,
*American Mathematical Monthly* **69** (1962) p. 920.

[1b] Solution to
Problem E 1543 by D.A. Klarner,
*American Mathematical Monthly* **70** (1963) pp. 760-761.

[2] David A. Klarner,
Packing a
Rectangle with Congruent *N*-ominoes,
*Journal of Combinatorial Theory* **7** (1969) pp. 107-115.

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Updated August 23, 2011.