Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get 
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get 
$$a\equiv b \text{ mod } 3  \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$,  but I don't know how to go on. 
I would appreciate any hints.

By Fermat Little Theorem 
$$x^3 \equiv x \pmod{3}$$
Therefore
$$0 \equiv a^3-b^3 \equiv a-b \pmod{3}$$
This shows that $a-b=3k$.
Then
$$3^3 \cdot 37 =a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^3+3ab)=3k(9k^2+3ab) \Rightarrow \\
3 \cdot 37=k(3k^2+ab)$$
Since $k <3k^2+ab$ the only posibilities are 
$$k=1 \\
3k^2+ab=3 \cdot 37$$
or
$$k=3 \\
3k^2+ab= 37$$
This leads to
$$a-b=3k=3 \\
ab=108$$
or
$$a-b=9 \\
ab=10$$
which are easy to solve.
