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.

So $(a-b)(a^2 + ab + b^2) = 3^3*37$ 
So $(a-b)|3^3*37$.
So $a-b = 3^j*37^k$ where $j = 0,1,2,3$ and $k=0,1$
And $(a^2 + ab +b^2) = \frac {3^3*27}{a-b} = 3^{3-j}*37^{1-k}$
That's 8 possible systems of equation.
But there are some obvious things to note. 
If $37|a-b$ then $a - b \ge 37$ and ... that just seems wrong.
That means $a \ge 37$ and $a^2 + ab + b^2 > 37^2$ but if $37|a+b$ then the very largest that $a^2 + ab + b^2$ can be is $\frac {999}{37} = 27$.
So $37\not \mid a-b$ and $37|a^2 + ab + b^2$.
So $a-b = 3^k; k\le 3; a^2 +ab + b^2 = 3^{3-k}*37$
And there are only 4 systems of equations.
Likewise if $27|a-b$ then $a \ge 27$ and $a^2 + ab + b^2 > 27^2 > 37$ so that's not possible.  
So $a-b = 3^k; k\le 2; a^2 + ab + b^2 = 3^{3-k}*37$.
And there are only 3 systems of equations.
$k = 0,1,2$.
If $k = 0$ we have $a= b+ 1$ and $(b+1)^2 + b(b+1) + b^2 = 999$
or $3b^2 + 3b + 1 = 999$ which isn't possible as LHS and RHS are not equivalent mod $3$.
If $k = 1$ we have $a = b+3$ and $(b + 3)^2 + b(b+3)+b^2 = 333$ and
$3b^2 +  9b + 9 = 333$ or 
$b^2 + 3b - 108 = 0$ so $b = \frac {-3 +\sqrt {9+432}}2 = \frac {-3 + 21}2 = 9$ and $a = 12$
If $k = 2$ we have $a = b+9$ and $(b+9)^2 + b(b+9) + b^2 = 111$
$3b^2 + 27b + 81 = 111$
$b^2 + 9b + 27 = 37$ 
$b^2 + 9 - 10 = 0$
$(b+10)(b - 1) = 0$ so $b = 1$ and $a=10$.
