Let a and b be positive integers with $gcd(a,bc)$ = $gcd(a,b)gcd(a,c) $ ∀ positive integer c $=>$ $gcd(a,b)=1$
Let $gcd(a,bc)=d$ , $gcd(a,b)=d_{1}$ , $gcd(a,c)=d_{2}$
I try to prove $d_{2}$|$d$ and $d$|$d_{2}$
It is easy to show $d_{2}$|$d$
,but I don't know how to show $d$|$d_{2}$,
or show
$d$ is common divisor of (a,c)
Since if $d$ is a common divisor of (a,c), I can know $d$ ≤ $d_{2}$
and $d_{2}$|$d$ => $d_{2}$ ≤ $d$,
then I have $d_{2}$ = $d$
Hence, I can conclude $d_{1}$ = 1
Best Answer
$$1)\text{ }\text{ }\text{ }\text{ }\text{gcd}(a,b) = \text{gcd}(a,b).$$ $$2)\text{ }\text{ }\forall n \in \mathbb{N} \text{ }\text{ }\text{ }\text{gcd}(a,b^n) = \text{gcd}(a,b)\text{gcd}(a,b^{n-1}).$$ $$3) \text{ }\text{gcd}(a,b^n) = \text{gcd}(a,b)^n.$$
$$4)\text{ } \text{gcd}(a,b^n) \leq a.$$
$$5) \text{ }\text{gcd}(a,b)\geq 2 \Rightarrow \text{gcd}(a,b^n) = \text{gcd}(a,b)^n \geq 2^n.$$
$$6)\text{ }\text{ }\text{ }\text{ } 4), 5) \Rightarrow \text{gcd}(a,b) \leq 1. $$
$$7)\text{ }\text{ }a,b \geq 1 \Rightarrow \text{gcd}(a,b) = 1.$$