<s>[INST] Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$. 
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.

I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
 [/INST] Clearly, $a^2+ab+b^2$ has to be even $\implies a,b$ both must be even(why?)
WLOG $a=2A,b=2B$
Now as $10|(a^2+ab+b^2),5|(a^2+ab+b^2)$
$$a^2+ab+b^2=(A+2B)^2+3A^2\implies(A+2B)^2\equiv-3A^2\pmod5$$
Again, for any integer $r,r\equiv0,\pm1,\pm2\pmod5\implies r^2\equiv0,\pm1$
$\implies(A+2B)^2\equiv0,\pm3\pmod5$ which is only possible if $5|(A+2B)$
$\implies5|B^2\implies5|B$ and $\implies5|(A+2B)\implies5|A$
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