Consider the corner squares 2x2. At least 2 cells of 3 2x2 corner square must be covered by L tromino, because if only 1 cell is covered we can place tromino. Note that a tromino can't cover cells from 2 different corner squares 2x2. Therefore we need at least 1 tromino for 3 of the 2x2 corner square.Therefore we need at least 3. We show this is not sufficent. Consider the 2x2 corner square that has 0 cells covered by L tromino (WLOG let it be the upper left). Because of above, we must have covered at least 2 of the cells of 3 2x2 corner squares, so at most 1 cell of every tromino is not in the 2x2 square. Suppose we can't place another tromino in the upper left square 3x3 (or otherwise we are done). Therefore the upper right and the down left trominoes don't have other cells outside their 2x2 squares. Note that the central square can't be covered if every tromino covers at least 2 cells of its 2x2 square. Thus we are left with a down right 3x3 square, in which only 3 squares are covered by tromino. It's easy to see that in the 3x3 square we can place one more tromino. But in the start we placed tromino in the non-occupied 2x2 square, and then we have placed 2 more trominoes: contradiction. Thus we need at least 4 trominoes. Example is easy to make: 01000 01122 03020 33440 00040 Where 0 means non-covered square.