Maybe I’m missing something here. Suppose there is an L-reptile of area 3(long side length 2, short side 1). If there is an L-reptile with six pieces, then each piece has an area of 1/2. From that, we can deduce each piece has a long side of sqrt(2)/sqrt(3). That is irrational, therefore there is no additive combination of such pieces that would fit completely into a side of 1 or 2 as they would be irrational also.

I’ll leave as an exercise to show this is also true for an L-irreptile also. (OK, I’m lazy…but the argument should be similar – there would need to be at least one irrationally sided piece.)

]]>More interestingly, does any polyomino irreptile not have a square irreptile?

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