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.

You have $(a-b)(a^2+ab+b^2)=999$, so $a-b$ is a factor of $999$, that is
one of $1,3,9,27,37,111,333$ and $999$ and $a^2+ab+b^2$ is the complementary factor. There are now eight cases. If $a-b=1$, then
$a=b+1$ and $999=a^2+ab+b^2=3b^2+3b+1$. This is a quadratic equation;
has it any integer solution? Once this is decided, seven more cases to go!
