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ok, i have a proof! it's recursive, i don't know if i can explain it without diagrams.
trivial case: 2x2 should be obvious.
4x4: a 4x4 is made up of 4 2x2 squares. Let's divide it into NW, NE, SW, SE quadrants. The defective square will be in one of those quadrants, let's say in the NW, and since we've already proved that a 2x2 defective square can be tiled, the NW quadrant can be completely tiled. For the remaining 3 quadrants, pretend that the 3 tiles closest to the center of the big 4x4 square are defective. Then you can tile the each of those quadrants, since they're all defective 2x2 squares. Finally, stick the final tile right in the center, in the three tiles you pretended were defective, and you're done!
8x8: Same logic. Divide into four 4x4 quadrants. Whichever quadrant has the defective square can be tiled (as proved above). Pretent the remaining center squares are defective, tile the other three quadrants. Put the final tile in the center.
Etc etc etc. Is there a better proof?
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oh baby oh baby, i like gravy.
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