What rectangles does the F hexomino tile? OK, so I stumbled upon a problem. Namely, in what ways can I arrange (using any number of pieces) pieces like this:

in such a way to form a rectangle? We can flip and rotate the pieces but not cut them in any manner. 
I know I can create a $3 \times 4$ rectangles:

using two of these blocks. It implies we can also create any $3a \times 4b$ rectangle (for positive integers $a$ and $b$) But what else? Doesn't seem there's anything... I can't prove there isn't, though. Could you help me?
 A: In fact, there's a simple coloring proof that every rectangle tiled by this hexomino has one side divisible by $4$; consider a $(4m+2)\times(4n+2)$ rectangle and color the cells with both coordinates even black, all other coordinates white.  Each F hexomino covers an even number of black cells no matter its placement (since it consists of a single shape copied with a two-cell displacement), but the $(4m+2)\times(4n+2)$ rectangle has an odd number of black cells, so it can't be tiled.  (And of course, if a $(4m+2)\times\mathrm{odd}$ rectangle were tileable, then it could just be doubled up into a $(4m+2)\times(4n+2)$ rectangle).
On the other hand, the rectangles of size $3a\times 4b$ don't exhaust all the possible tilings, because with the $3\times 4$ rectangle as building block you can craft other shapes - for instance, a $12\times 7$ rectangle can be done by vertically stacking four 'horizontal' boxes next to three 'vertical' boxes also stacked vertically, and this along with the trivial $12\times 6$ and $12\times 8$ tilings means that $12\times n$ rectangles are possible for all $n\geq 6$.  This leaves only the $12\times 5$ rectangle (which obviously can't be tiled by $3\times 4$ rectangles), and unfortunately I don't have an answer for that one; the page at http://www.math.ucf.edu/~reid/Polyomino/f6_rect.html (from which I shamelessly stole the proof above) suggests that the $3\times 4$ is the only 'prime' rectangle tiling, but gives no proof beyond that one side must be divisible by $4$.
Addendum: After some playing around it turns out to be relatively straightforward to show that the $12\times 5$ rectangle is impossible.  Consider the top ($5$-wide) edge, and the pieces that make it up; since the extremal pieces of the F are either $1$, $2$, or $4$ cells wide, then this has to be either $4+1$, $2+2+1$ or $2+1+1+1$.  The latter is easy to rule out, since at least one of the $1$-cell Fs will extend beyond the bounds of the box; similarly, in a $2+2+1$ configuration, one of the $2$-cell Fs will either leave unfillable holes against the outside wall of the box or unfillable holes in the interior.  Finally, for the $4+1$ configuration it's easy to see that there's only one legal configuration of these pieces that covers that edge without going past the bounds of the box: an F covering $(0,0)$, $(0,1)$, $(1,1)$, $(0, 2)$, $(0, 3)$, and $(1,3)$ and an F covering $(1,0)$, $(2,0)$, $(2,1)$, $(3,0)$, $(4,0)$ and $(4,1)$.  Now consider the cell at $(3,1)$; this cell can't be covered by an F laid 'horizontally' (any such would either extend past the bounds of the box or overlaps the cell at $(1,1)$), and filling it with a vertical F leaves an unfillable hole either at $(1,2)$ or $(4,3)$.
