Since $ \frac {n}{m}$ is an integer, the first thing we think of is to let $ n = mk$. Then the problem reduces to showing that $ k = \lfloor\sqrt [3]{m^2k^2}\rfloor + \lfloor\sqrt {mk}\rfloor + 1$ has a solution for some integer k. Now, we consider the function $ f(k) = k - \lfloor\sqrt [3]{m^2k^2}\rfloor - \lfloor\sqrt {mk}\rfloor - 1$ taking the natural numbers to the integers. When $ k = 1$, $ f(k) < 0$. Note that every time we increase k by 1, $ f(k)$ can increase by at most 1. Also note that $ k$ grows faster than $ \lfloor\sqrt [3]{m^2k^2}\rfloor + \lfloor\sqrt {mk}\rfloor + 1$, since it is an exponential of greater power than either of the terms in $ \lfloor\sqrt [3]{m^2k^2}\rfloor + \lfloor\sqrt {mk}\rfloor + 1$, so there must be an integer m s.t. $ f(m) > 0$. Lastly, consider the function $ g(k) = max(f(1),f(2),f(3),...,f(k))$. Then $ g(1) < 0$, $ g(m) \ge f(m) > 0$, and since $ g(x + 1)$ is always either 0 or 1 greater than $ g(x)$, $ g(x)$ must hit every integer between $ f(1)$ and $ f(m)$ at some integer k between 1 and m. This implies there exists an $ f(k)$ s.t. $ f(k)=0$, so we're done.