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By invoking the Künneth formula for cohomology, one can argue that among the Euclidean spaces only the even-dimensional ones can be the square of some space. Clearly these are indeed squares, as $\mathbb{R}^n \times \mathbb{R}^n \cong \mathbb{R}^{2n}$.

This does not guarantee that there are no other decompositions. Fox showed in 1947 that non-homeomorphic spaces can have homeomorphic squares, and subsequently quite reasonable examples of such spaces have been found.

Question. If $\mathbb{R}^{2n} = X \times X$, must $X$ be homeomorphic to $\mathbb{R}^n$?

As a follow-up question, one might consider any decomposition of $\mathbb{R}^N$ as a product $X_1 \times \cdots \times X_k$ and ask whether the $X_i$ must be Euclidean.

Sofie Verbeek
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After some searching I found that the question was asked long ago on MathOverflow. A counterexample is given by the Whitehead manifold, whose square is homeomorphic to $\mathbb{R}^6$.

See https://mathoverflow.net/questions/60395/square-roots-of-mathbb-r2n

Sofie Verbeek
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