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 $100|A$.
 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 $25 \mid a^2+ab+b^2=A$.
A: Solution 1:  Suppose $10|A$. If either $a$ or $b$ is odd, then $A$ is also odd, so $a$ and $b$ must be even and hence $4|A$.  Now consider $A$ modulo $5$.  If either $a$ or $b$ is divisible by $5$, the other must be as well, and so $25|A$.  Otherwise, $a^2$ and $b^2$ can be either of $\pm 1$ modulo $5$.  If one is $1$ and the other $-1$, then $A\equiv ab \pmod{5}$ is not divisible by $5$.  If both are $1$, then $A\equiv ab+2$, and $a, b \equiv \pm 1 \pmod{5}$ so that $A \equiv 1, 3\pmod{5}$ and is not divisible by $5$.  A similar argument shows that if $a^2\equiv b^2 \equiv -1\pmod{5}$, then $ab \equiv \pm 1\pmod{5}$ and hence $A$ is still not divisible by $5$.  So $a$ and $b$ must both be divisible by $5$ and thus $25|A$.  Therefore $100|A$.
Massively Overkill Solution 2:  The numbers 2 and 5 are Eisenstein primes, so for any Eisenstein integer $a-b\omega$, if $5|N(a-b\omega)$, then $5|(a-b\omega)$ and thus $5|a$ and $5|b$.  The same goes for $2$ in place of $5$ and the result follows since $N(a-b\omega) = a^2+ab+b^2$.
A: $2|A$ hence $a,b$ are both even so $4|A$
It remains to show that $5|A$ implies $5^2|A$
A: It is simply the special case $\, n,k = 10,3\,$ of the following
Theorem $\,\ n\rm\ \color{#c00}{squarefree},\,$ $\,\color{blue}{(n,k)}=1=\color{#0a0}{(\phi(n),k)},\ $ $\ n\mid A = (a^k-b^k)/(a-b)\ \Rightarrow\ n^{k-1}\mid A$
Proof $\ \ {\rm mod}\ n\!:\ a^k\equiv b^k\overset{\color{#0a0}{(\phi(n),k)\,=\,1}}{\color{#c00}\Rightarrow}\! a\equiv b,\ $ so $\,\ 0\equiv \color{orange}A = a^{k-1}\!+a^{k-2}b+\cdots+b^k\equiv kb^{k-1}\,$   
Therefore $ \ n\mid kb^{k-1}\overset{\color{blue}{(n,k)\,=\,1}}\Rightarrow \! n\mid b^{k-1}\overset{\rm n \ \color{#c00}{sqfree}}\Rightarrow\! n\mid b\overset{a\,\equiv\, b}\Rightarrow n\mid a\,\Rightarrow\, n^{k-1}\mid \color{orange}A.\ \ $ QED
