Primes of the F hexomino

3 × 4
complete

smallest rectangle: 3 × 4

Proposition. Any rectangle tiled by the F hexomino has one side divisible by 4.
Proof. It suffices to show that it doesn't tile any (4m + 2) × (4n + 2) rectangle. Number the squares by

        (x, y) |--> { 1  if  x  and  y  are both even
                    { 0  otherwise.
  

No matter how it is placed, each F hexomino covers an even total. However, a (4m + 2) × (4n + 2) rectangle covers an odd total, so it can't be tiled. QED.

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Updated May 25, 2005.