This is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$. is then $A$ isomorphic to $A^2$? probably no, but how construct a counterexample? you can also ask this in other categories as well, for example rings. if you restrict to Boolean rings, the question becomes a topological one which makes you think about fractals: let $X$ be Stone space such that $X \cong X + X + X$, does it follow that $X \cong X + X$ (here $+$ means disjoint union)?
Edit: In the answers there are already counterexamples. but you may add others in other categories (with products/coproducts), especially if they are easy to understand :).

The answer to the first question is no.  That is, there exists an abelian group $A$ isomorphic to $A^3$ but not $A^2$.  This result is due to A.L.S. (Tony) Corner, and is the case $r = 2$ of the theorem described in the following Mathematical Review.
MR0169905
Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest.
It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m \equiv n \pmod r$.  This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties.  It should be mentioned that the question of the existence of algebraic systems with the property stated above has been considered by several mathematicians.  The author has been too generous in crediting this "conjecture" to the reviewer.
Reviewed by R.S. Pierce
