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It is known that P pentominoes cannot tile a 5x5 square board.

Q1: If the east and west edges of the 5x5 square board are "wrapping around" (if you move a piece through one of the edges, the part appears on the opposite edge), can you tile it with P pentominoes?

Q2: In addition to the above, the north and south edges are also wrapping around (thus forming a toroidal surface). Can you tile it with P pentominoes?

Q3: Identify all distinct pentominoes that can tile the 5x5 toroidal board.

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Q1: Here is a perhaps not elegant but at least not super ugly proof that it is

not possible:

Assume to reach a contradiction that we have a tiling. Consider all pentos in the tiling that touch the bottom edge. Observe that any such pento has at least 2 squares in the 4th row (1st is top 5th is bottom). As there are only 5 squares available there can be no more than 2 pentos touching the bottom.

Because these pentos need to cover the bottom row it must be 2 and one must have its 3-long side hugging the bottom. The other must contribute 2 more meaning that these 2 pentos cover at maximum 1 square in the middle row.

As the same argument holds for the top we have with 4 pentos spent no more than 2 of the 5 middle row squares covered. As the last pento can cover no more than 3 it must be 3 from the last pento and 1 from the top pair and 1 from the bottom pair. However, with 3 squares of the last pento aligned to the middle row the remaining 2 can only either both go up or both go down. In either case we have a 9-to-11 imbalance of squares occupied in the top 2 and bottom 2 rows. Contradiction.

Q3 (includes Q2): These are tilings for

all but U,F and T. Color follows wrap-around. enter image description here

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Paul did the hard parts, so here's the answer to Q1:

It's impossible. To cut down the number of variations, we can chop the leg off of all the P-pentominoes, and show that it's impossible to fit 5 square tetrominoes onto the board:

As long as we cannot cross the top boundary, nothing can be gained by not placing two of the squares along the top. Since the sides loop, there's essentially only one way to do that. This leaves a 5x3 strip, which is not tall enough for two squares, and not wide enough for three.
enter image description here

(EDIT: while I was creating the pretty but oh so unnecessary picture, Paul edited his answer to include a solution to Q1 too. Mine seems somewhat more straightforward, so I'm leaving it up for now at least.)

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  • $\begingroup$ Great idea chopping off the nose! Can you explain why your argument doesn't work for torus? (Sorry, always nitpicking.) $\endgroup$ Dec 15 '20 at 11:18
  • $\begingroup$ @PaulPanzer well, surely the part about "As long as we cannot cross the top boundary" won't apply to a torus, so there could, in theory, be something that's gained by not placing (any) two squares side by side :-) $\endgroup$
    – Bass
    Dec 15 '20 at 11:29
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    $\begingroup$ With the 2x2 squares, there will be exactly 5 uncovered little squares. Every row and every column must have exactly one of those. This forces the placement of the 2x2 squares (up to translation/reflection). $\endgroup$ Dec 15 '20 at 12:34
  • $\begingroup$ @JaapScherphuis now there's a clean argument! $\endgroup$ Dec 15 '20 at 14:32

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