By Fermat's last theorem, the equation $u^3+v^3=w^3$ has no solutions in positive integers $u,v,w$. Now consider the following variant : call $\rho(x)$ the distance between $x$ and the nearest integer, for any real number $x$ (thus $\rho(3)=0,\rho(3.2)=0.2,\rho(3.5)=0.5, \rho(3.9)=0.1$ etc).

An "approximate" version of the Fermat equation is to ask for $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ to be arbitrarily small. A trivial way to achieve this is to make one of the variables very small compared to the other, say $v$ very small compared to $u$, so that ${( u^3+v^3 )}^{\frac{1}{3}}$ is very near to $u$.

It is therefore natural to ask if there is an absolute constant $C>0$ such that $\rho({( u^3+v^3 )}^{\frac{1}{3}})$ can be made arbitrarily small with $u,v$ positive and $u \leq C v, v \leq C u$ (so that neither of $u$ or $v$ dominates).

Can a (reasonably) explicit $(u_n,v_n)$ sequence be found, such that $\rho(u_n^3+v_n^3)$ tends to $0$ as $n$ goes to infinity and $u_n \leq C v_n, v_n \leq C u_n$ ? "Closed-form" formula would be the best, of course, but even a simple recursion would be nice.