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How can we find $\operatorname{GCD} (a^2, a\cdot b, b^2)$ where $a$ and $b$ are integers? I have only a slight idea of how to solve this, any help would be greatly appreciated. Thanks

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    $\begingroup$ Hint: show that it is at most as large as $\text{GCD}(a^2,b^2)$. Determine this latter quantity, and then show that it is at most as large as $\text{GCD}(a^2,ab,b^2)$ and thus that the two GCDs are equal. $\endgroup$ – Erick Wong Feb 10 '14 at 2:07
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Hint $\ \ (a^2,ab,b^2) = (a,b)^2\,$ by basic gcd laws (distributive, commutative, associative) since

$$ (a,b)^2 = (a,b)(a,b) = (a(a,b),b(a,b)) = ((a^2,ab),(ab,b^2)) = (a^2,ab,b^2)$$

Remark $\ $ Similarly $\ (a,b)^n = (a^n, a^{n-1}b,\ldots, a b^{n-1}, b^n)\ $ is the simple Binomial Theorem for gcds. Further $\ (a,b)^n = (a^n,b^n)\ $ by the GCD Freshman's Dream.

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