Is there a simple proof that any square besides a 3x3 square with area divisible by 3 is tileable with L-trominos?
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RavenclawPrefect
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user1524759
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All such squares have an edge length which is a aum of 6es and 9s, so it suffices to exhibit tilings of the $6\times6 $ and the $9\times 9$ squares. – MJD Aug 28 '14 at 23:30
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2... and of the $9 \times 6$ rectangle. – Robert Israel Aug 28 '14 at 23:38
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Here's a proof of a stronger statement: the $L$-tromino tiles any rectangle with dimensions $3a\times b$ if $a\ge 2, b\ne 1,3$.
Note that the L tromino certainly tiles a $3a\times 2$ rectangle (stack $3\times 2$ rectangles on each other), so if we can show this for $b=5$ we'll be done (since all positive integers other than $1$ and $3$ are a sum of $2$s and $5$s).
To do this when $b=5$, we combine two rectangles together as needed: the $6\times 5$ and the $9\times 5$. They can both be tiled, as seen:
Stacking as necessary, we obtain $3a\times 5$ rectangles for all $a\ge 2$, which completes the proof. The corollary $b=3a$ is immediate.

RavenclawPrefect
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1why do you need it to show for $b=4$ please? Since you do show it for $b=2,5$, it should be enough for your statement to be true unless I missed something... – Marco Bellocchi May 26 '21 at 18:04
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1@MarcoBellocchi: You're right, that also suffices. Thanks for the catch! I'll edit the post accordingly. – RavenclawPrefect May 27 '21 at 02:35
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Thank you for letting me know, I saw the post updated, nice one. – Marco Bellocchi May 27 '21 at 12:19