Show that there are infinitely many solutions of distinct natural numbers $m,n$ such that $n^3+m^2\mid m^3+n^2$ 
Show that there are infinitely many solutions of distinct natural numbers $m,n$ such that $n^3+m^2\mid m^3+n^2$. 

This question appeared in Round $2$ of British Math Olympiad $2007-08$. I have been trying this problem since two days. And got no sequence of numbers satisfying it.  
 A: Choose $a\in\mathbb{N}$, and set $m=an$.  Then the condition becomes $n^3+a^2n^2|a^3n^3+n^2$, which is equivalent to $$n+a^2|a^3n+1$$
We now set $n=a^5-a^2-1$ and see that $$(a^3-1)(n+a^2)=(a^3-1)(a^5-1)=a^8-a^5-a^3+1=a^3n+1$$
Update, pulling back the veil: I looked for $k$ such that $k(n+a^2)=a^3n+1$ and I could solve for $n$.  After one or two false attempts I tried $k=a^3-1$.
A: As Vadim showed, with $m=an$ yields
$$ n + a^2 | a^3 n + 1 $$
This implies that 
$$ a^5 - 1 \equiv a^5 + a^3 n - a^3n - 1 \equiv 0 \pmod{a^2 + n}$$
This suggests immediately to use $n = a^5 - a^2 - 1$, as in Vadim's solution. The advantage is that I don't need to do any guesswork, or expand the product.
In fact, since $ a^5 - 1 = (a-1)(a^4+a^3+a^2+a+1)$, we could also use
$$n  = a^4 + a^3 + a + 1 $$

This also shows you how to characterize all solutions. Start with any positive integer $a$, calculate the factors of $a^5-1$ that are greater than $a^2$, and that determines the value of $n$.
Note: Because we want $n$ to be positive, none of the other factors (with integer coefficients) will work to generate a family of solutions.
