TITLE: Square roots of $\mathbb R^{2n}$
QUESTION [34 upvotes]: Recently, Richard Dore asked us if $\mathbb R^3$ is the cartesian square of some space, and Tyler Lawson answered beautifully in the negative.
The even powers of $\mathbb R$ were left out in that question because, well, it is obvious that they are squares. Now:

Are they squares in a unique way? In other words, if a space $X$ is such that $X\times X\cong\mathbb R^{2n}$, must $X$ be homeomorphic to $\mathbb R^n$?

One can consider other values of $2$ in Richard's question or here, as well as look for factors instead of only square roots (but if I recall correctly $\mathbb R^5$ has all exotic $\mathbb R^4$s as factors, so the last variant might be «trivial»...)

REPLY [28 votes]: I'm pretty certain the answer is no provided $n \geq 3$.  Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold
It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$.  
$X^2$ I claim is homeomorphic to $\mathbb R^6$.  I don't have a slick proof of this.  The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay.   But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 Y * \pi_1 Y) / <\pi_1 Y^2>$, where $*$ denotes free product and angle brackets "normal closure", and $\pi_1 Y$ is "the fundamental group at infinity for $X$".  So $\pi_1 X^2$ is the trivial group. 
Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is $D^6$.