Here's another way to think about tilings an Aztec diamond with skew tetrominoes (it's not really a new proof that no such tilings exist, but it's a new way to present the old one). Suppose this region can be tiled by skew tetrominos. Take the 1-skeleton of the tiling, and color its edges with colors A, B, C, and D as shown below: o---B---o---A---o | | | D C D | | | o---B---o---A---o---B---o---A---o | | | | | D C D C D | | | | | o---B---o---A---o---B---o---A---o---B---o---A---o | | | | | | | D C D C D C D | | | | | | | o---A---o---B---o---A---o---B---o---A---o---B---o | | | | | | | C D C D C D C | | | | | | | o---B---o---A---o---B---o---A---o---B---o---A---o | | | | | C D C D C | | | | | o---B---o---A---o---B---o---A---o | | | C D C | | | o---B---o---A---o Now shrink all the segments colored A and D down to length 0. You can check that every tile-boundary is a closed (possibly degenerate) polygon that, if superimposed on a checkerboard, encloses equal numbers of black and white squares. But the boundary of the Aztec diamond shrinks down to o---B---o---B---o---B---o | | C C | | o---B---o o | | C C | | o-------o o | | C C | | o---B---o which does not enclose equal numbers of black and white squares. This proof sweeps a few details under the rug (such as the way the possibly degenerate polygons fit together), and one still is using a hidden argument about winding numbers. But it's still essentially a complete proof.