# $\operatorname{GCD} (a^2, a*b, b^2)$ where $a$ and $b$ are integers

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

• 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. – Erick Wong Feb 10 '14 at 2:07

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.