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?
 A: Here's a more general result:

Let $\Psi_n(x,y)$ denote the $n$th homogenized cyclotomic polynomial. If $x,y\in\mathbb Z$, every prime divisor of $\Psi_n(x,y)$ is either:
  
  
*
  
*$1\pmod n$
  
*a prime divisor of $n$
  
*a prime divisor of both $x$ and $y$
  

This is a simple corollary of the same result for the usual cyclotomic polynomial $\Phi_n(x)=\Psi_n(x,1)$ (where the third case doesn't occur).
In our case $2\mid\Psi_3(a,b)=a^2+ab+b^2$. As $3\nmid2$ and $2\not\equiv1\pmod3$, we have $2\mid a,b$. Same argument for $5$.

Entirely self-contained proof of that result in the case of prime $n$ (here $n=3$):
We then have $\Psi_n(x,y)=x^{n-1}+xy^{n-2}+\cdots+y^{n-1}$. Let $p\mid\Psi_n(x,y)\mid x^n-y^n$. If $p\mid x$ or $y$, we are done. If not, $y$ is invertible mod $p$, and $(x/y)^n\equiv1$. Let $k$ be the order of $x/y$ mod $p$. $k\mid n$.
If $k=1$, $x\equiv y$ and $p\mid x^{n-1}+xy^{n-2}+\cdots+y^{n-1}\equiv nx^{n-1}$, so $p\mid n$, so $p=n$.
If $k=n$, $n\mid p-1$ by Fermat's Little Theorem.
A: 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 $\implies5|(A+2B)\implies5|A$
A: $10 \mid (a^2+ab+b^2)$ implies $10 \mid a^3-b^3 = (a-b)(a^2+ab+b^2)$, that is, $a^3 \equiv b^3 \bmod 10$.
Now, $x \mapsto x^3$ is a bijection mod $10$. Therefore, $a^3 \equiv b^3 \bmod 10$ implies $a \equiv b \bmod 10$.
Then $a^2+ab+b^2 \equiv 3a^2$ implies $10 \mid a$ and so $10 \mid b$.
Therefore, $1000 \mid a^3$ and $1000 \mid b^3$, hence $1000 \mid (a^3-b^3)$.
