# covering a square with unit squares

Can some square of side length greater than $n$ be covered by $n^2+1$ unit squares? (The unit squares may be rotated. The large square and its interior must be covered.)

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I meant to replace the original "any" with "some", which is done now. (Others may read the questioner's intent differently.) – Steve Huntsman Jun 25 '10 at 16:46
Perhaps it should be stated for the record that this problem was originated by Alexander Soifer. – Joseph O'Rourke Jun 27 '10 at 11:25

This reference is certainly pertinent, being the second Google hit for "covering a square with squares" (after your question). Just reading it now...

http://www.uccs.edu/~faculty/asoifer/docs/untitled.pdf

UPDATE: so far as I can tell from looking at this article, the author regards your question as an unsolved problem. There is a further article by him and Karabash, apparently in (his) journal Geombinatorics, vol. 18, which I cannot access online and which has not been reviewed on MathSciNet.

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To supplement Ben Green's key reference (to "Covering a square of side n + ε with unit squares") , there is some follow-on work: Karabash & Soifer, "A sharp upper bound for cover-up squares," Geombinatorics, v16, 219-226, 2006; "Note on covering square with equal squares," Geombinatorics, v18, 13-17, 2008; Chung & Graham, "Note: Packing equal squares into a large square," Journal of Combinatorial Theory Series A, Volume 116, Issue 6 (August 2009), 1167-1175.

Addendum in response to Ben Green's remark: I do have the 2008 Geombinatorics paper (but not the 2006 one). They define $\Pi(n)$ as the number of unit squares that can cover a square of side length $n+\epsilon$. It appears that the status as of this 2008 paper was that $\Pi(n)=n^2+O(n^{2/3})$ has been established, and they conjecture that $\Pi(n)=n^2+ \Omega(n^{1/2})$.

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Is there a proof that $\Pi(n) > n^2 + 1$? – Ben Green Jun 25 '10 at 19:22
@Ben G.: Not that I can see, but then I have not scoured all this literature. $\Pi(1)=3$, but already for $n=2$ and $n=3$, the known lower bound is $n^2+1$. – Joseph O'Rourke Jun 25 '10 at 20:09