[pozorvlak]

Suppose the square can be dissected into n pieces. Then any symmetry of the square induces a permutation of those n pieces. Hence, the symmetry group of the square (D_8) must be embeddable as a subgroup of S_n. Hence there is no dissection into three pieces, since |D_8| = 8, which does not divide |S_3| = 6.

I suspect this line of argument can be taken further, but wanted to post it before someone else did :-)

Edit: no, that doesn't work, because we already have a dissection into two pieces, and |S_2| = 2. Dammit. Not all of the symmetries of the square must map pieces to pieces.