4
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I could clearly fit in 4 squares together to make a square. Or, I could fit in 6 squares into a square as well:

But this is not possible with 3 squares (or so I think!)

How does this generalize? How can we decide for which N such a covering is possible?

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5
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You can tile with 1, 4, and 6 squares already.

You cannot tile with 2, 3, or 5 squares. For 2 and 3, you can see this because no tile can cover more than one of the corners of the big square. For 5 tiles, there must be one tile in each of the corners of the big square and there must remain at least two edges with uncovered gaps which cannot be filled by a single square.

For any even number $N=2m>2$, you can tile using $2m-1$ squares of size $1\times 1$ along two sides of an $(m-1)\times(m-1)$ square. This gives solutions for $N=4, 6, 8, 10,\ldots$.

Finally, if you have a tiling with $n$ squares, you can make one with $n+3$ squares by cutting one of the tiles into 4 smaller ones. Just apply this once to the above solution and you have $N=7, 9, 11, \ldots$ solved.

Thus, there are tilings for all $N$ except 2, 3, and 5.

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  • $\begingroup$ are all squares disjoint ? the OP is similar to the mondrian art puzzle. $\endgroup$ – user451844 Jul 20 '17 at 17:50
  • $\begingroup$ @RoddyMacPhee: Yes, they do not overlap (other than the border if that's included in the squares). $\endgroup$ – Einar Rødland Jul 20 '17 at 19:41

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