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Let $a,b$ be natural numbers so that $a \ge 2, b \ge 2$ and $a^{2}b^{2}$ is divisible by $a^{2}-ab+b^{2}$ Prove that $a$, $b$ are not relatively prime numbers. Any ideas?

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  • $\begingroup$ At least one can try to prove by contradiction so that some work could have been done I suppose? $\endgroup$ – Jack Dec 19 '16 at 19:24
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    $\begingroup$ Step 0: show that $a^2 - ab + b^2 > 1$. Step 1: hence there is a prime $p$ dividing $a^2 - ab + b^2$. Step 2: hence $p$ divides $a^2b^2$ … $\endgroup$ – Daniel Fischer Dec 19 '16 at 19:44
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${\bf Hint}\quad \begin{align} &(\color{#c00}a,\!\overbrace{a^2\!-\!ab\!+\!b^2}^{\Large\ \equiv\ \color{#0a0}{ b^2}\bmod{\color{#c00} a}}\!\!)=(a,\color{#0a0}{b^2})\\ &(b,\ a^2\!-\!ab\!+\!b^2)=(b,a^2) \end{align}\ $ $ \begin{align}\\ \\ {\rm so}\ \ \ (a^2b^2,\,a^2\!-\!ab\!+\!b^2)=1 \iff (a,b)=1\\ \\ \end{align}$

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