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

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 fourdimensional 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 3manifolds 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 2polyhedra is negative, see W. Rosicki, "On a problem of S. Ulam concerning Cartesian squares of 2dimensional 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 $\left\{1/n\mid n\in\mathbb{Z}_{>0} \right\}\cup\{0\}$. 


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 $$ X^2 \;\cong\; Y^2 \;\cong\; (\mathbb{R}^6 \:\uplus\: \mathbb{R}^6 \:\uplus\: \cdots) \:\uplus\: (M^2 \:\uplus\: M^2 \:\uplus\: \cdots). $$ 

