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I an wondering if there are non-homeomorphic spaces $X$ and $Y$ such that $X^2$ is homeomorphic to $Y^2$.

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This is not an answer to your question, but relates to a linear version of your question, so I mention it as a comment: There exist Banach spaces $X$ and $Y$ such that there exists a linear homeomorphism of $X\oplus X$ onto $Y\oplus Y$, but there is no linear homeomorphism of $X$ onto $Y$. For an example, take $X$ to be the space of Gowers and Maurey (Banach spaces with small spaces of operators, Math. Ann. 307 (1997), no.4, p.543--568) with the property that $X$ is linearly homeomorphic to its closed subspaces of even codimension, but not to its closed subspaces of odd codimension. –  Philip Brooker May 26 '11 at 7:35
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is Moebius^2 homeomorphic to cylinder^2? –  Yaakov Baruch May 26 '11 at 18:54
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Yaakov -- no: they are both fibered over the 2-torus with fiber a square, so they are topological manifolds with boundary. However, the rational cohomology modulo the boundary is the cohomology of the torus shifted 2 degrees up in one case and zero in the other. –  algori May 26 '11 at 19:44
    
... and in general, there are no counter-examples among 2-polyhedra, see the article by Rosicki mentioned below. –  algori May 26 '11 at 19:47
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Here is an extract from MR0562824 (81d:54005), Trnková, V. Homeomorphisms of products of spaces. (Russian) Uspekhi Mat. Nauk 34 (1979), no. 6(210), 124–138:

S. Ulam raised the following question in 1933: Is there a space $X$ which has nonhomeomorphic square roots, i.e., $X\cong A\times A\cong B\times B$ for some nonhomeomorphic $A,B$? This problem was solved by R. H. Fox in 1947: he constructed two nonhomeomorphic four-dimensional manifolds $A$ and $B$ such that $A\times A\cong B\times B$.

upd: The reference is Fox, R. H. On a problem of S. Ulam concerning Cartesian products. Fund. Math. 34, (1947). 278–287.

The answer to Ulam's question for 3-manifolds is positive as well, see Glimm, James Two Cartesian products which are Euclidean spaces. Bull. Soc. Math. France 88 1960 131–135.

The answer for 2-polyhedra is negative, see W. Rosicki, "On a problem of S. Ulam concerning Cartesian squares of 2-dimensional polyhedra.", Fund. Math. 127 (1987), no. 2, 101–125. This paper also gives the following elementary example:

Take $A$ to be the disjoint union of the Hilbert cube and $\mathbb{N}$ and $B$ to be the disjoint union of two copies of the HIlbert cube and $\mathbb{N}$. Then both $A^2$ and $B^2$ are homeomorphic to the disjoint union of a countable family of Hilbert cubes and $\mathbb{N}$.

Finally, in this example one can replace the Hilbert cube by any space homeomorphic to its square and not homeomorphic to two copies of itself, e.g., by $\left\{1/n\mid n\in\mathbb{Z}_{>0} \right\}\cup\{0\}$.

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Yes. Let $M$ be the Whitehead Manifold, which has the property that $M \not\cong \mathbb{R}^3$, but $M\times\mathbb{R}^3 \cong \mathbb{R}^6$. (In fact $M\times\mathbb{R} \cong \mathbb{R}^4$.) Let $$ X \;=\; \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots $$ and $$ Y \;=\; \mathbb{R}^3 \:\uplus\: \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots\text{,} $$ where $\uplus$ denotes the disjoint union of topological spaces. Then $X$ and $Y$ are not homeomorphic, but $$ X^2 \;\cong\; Y^2 \;\cong\; (\mathbb{R}^6 \:\uplus\: \mathbb{R}^6 \:\uplus\: \cdots) \:\uplus\: (M^2 \:\uplus\: M^2 \:\uplus\: \cdots). $$

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Can you modify this example so that neither space is homeomorphic to $X \times Y$? –  Qiaochu Yuan May 26 '11 at 9:28
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Qiaochu -- in Jim's example neither space is homeomorphic to $X\times Y$: both $X$ and $Y$ are 3-manifolds and so their product is a 6-manifold. –  algori May 26 '11 at 17:12
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