Prove that if $10|A$ then $100|A$. For : $A=a^2+ab+b^2$ with $a,b \in \mathbb{N}$. 
We known that $10|A$ prove that $100|A$.

We show that $4 \mid A$ and $25 \mid A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 2$. As $a \equiv 1 \equiv b$ is not a solution, at least one of them is $\equiv 0$, forcing the other to be $\equiv 0$. So $2\mid a,b$, showing $4\mid a^2+ab+b^2 = A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 5$. Then $(a+3b)^2 \equiv a^2+ab+4b^2\equiv 3b^2 \bmod 5$. But $3$ is not a quadratic residue $\bmod 5$, hence $b\equiv 0$ showing $a\equiv 0$. Thus $25 \mid a^2+ab+b^2=A$.
