Primes of the J hexomino

3 × 4
complete

smallest rectangle: 3 × 4

Proposition. Any rectangle tiled by the J hexomino has one side divisible by 4.
Proof. It suffices to show that it doesn't tile any (4m + 2) × (4n + 2) rectangle. Note that they must mate in pairs

         1 2 2 2           1 1 1
         1 2 1 2    or     1 2 1
         1 1 1 2             2 1 2
                             2 2 2
  

so we need only consider tilings by these two shapes. Now consider the numbering

         (x, y) |---> { 1  if  x  or  y , but not both, are divisible by 4
                      { 0  otherwise.
  

No matter how they are placed, either of these shapes covers an odd total. A (4m + 2) × (4n + 2) rectangle covers an even total. However, it would be tiled by an odd number of the two shapes above, which would cover an odd total, a contradiction. QED.

Remark. The J hexomino has a signed tiling of a 2 × 3 rectangle.

   . . .       * * *       * * *       . . .
   . . *       * . .       * . *       . . .
   * . *   +   * * .   -   * . .   =   * * *
   * * *       . . .       . . .       * * *
  

Problem 3 from the 2004 International Mathematical Olympiad asks which rectangles can be tiled by this hexomino. The proposition above is key to the solution.

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Updated August 23, 2011.