The two pentominos pictured below cannot tile any rectangle.

+--+  +--+   +--+--+
|  |  |  |   |  |  |
+--+--+--+   +--+--+--+--+
|  |  |  |      |  |  |  | 
+--+--+--+      +--+--+--+

Take eight copies of one pentomino and place them on a 4x5 rectangle such that each 1x1 subsquare of the rectangle has exactly two pentominos covering it.

(In reply to considering second pentomino by Charlie)

As far as I can tell your program does not allow for reflections of the polyomino.

Posted by Brian Smith on 2009-09-30 00:45:40