<s>[INST] 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 $100|A$.
 [/INST] 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 $N(a-b\omega) = a^2+ab+b^2$.
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