Ok, so I've found a proof for the problem that I set yesterday.
Consider the following colouring:
If you try to place an L tetramino on the grid then it will either stand vertically and so cover a column of 3 of the same colours, or it will lie horizontally in which case it will still cover 3 of the same colour. Either way you will end up with either 3 greens and a red, or 3 reds and a green being covered.
Given that the grid has the same number of red and green squares that means that there must be an even number of L tetraminoes. Therefore the most we can fit in is 8 and we have successfully proven that the puzzle from yesterday was impossible without resorting to actually trying it.
This same argument can be applied to any rectangular grid and the necessary (and sufficient) condition to be able to fill an AxB grid with L tetraminoes is that AB is a multiple of 8 and that both A and B are greater than 1.