Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n
Hi everyone, for the proof to the above question,
Can I assume that since $(a, b) = 1$, then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the $nth$ power, they will still have no prime factors in common, and so $(a^n, b^n) = 1$ for all positive integers n.
I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :)
And also, I do not know how to approach the reverse problem where if $(a^n, b^n) = 1 then (a, b) = 1$, any guidance will be much appreciated!
 A: Here's an approach that doesn't use prime factorization.
Lemma: For all $m,r,k\in \Bbb N$, $\gcd(m,k)=1\implies \gcd(m,k^r)=1$.
Proof: Let $m,r,k\in \Bbb N$ be such that $\gcd(m,k)=1$. 
Bézout yields $um+vk=1$, for some $u,v\in \Bbb Z$. 
Thus $$1=1^r=(um+vk)^r=\sum \limits_{j=0}^r\left({r\choose j}(um)^{r-j}(vk)^j\right)=\sum \limits_{j=0}^{r-1}\left({r\choose j}(um)^{r-j}(vk)^j\right)+(vk)^r,$$
which implies $$m\underbrace{\color{blue}{\sum \limits_{j=0}^r\left({r\choose j}u^{r-j}m^{r-j-1}(vk)^j\right)}}_{\huge \in \Bbb Z}+k^r\color{blue}{v^r}=1.$$
Again Bézout says that $\gcd(m,k^r)=1$ (due to the blue scalars).$\,\square$
You can now use the lemma twice: $\gcd(a,b)=1\implies \gcd(a,b^n)=1\implies \gcd(a^n,b^n)=1$.
A: For those type of proofs, assuming $ gcd(a^n,b^n) \neq 1 $ and showing a contradiction is a good way. 
Assume $ gcd(a^n,b^n) \neq 1 $. Then $\exists d_1$ such that $d_1|a^n \land  d_1|b^n \land d_1>1$. This implies:
Let's check out what $\exists d_1$ such that $d_1|a^n  \land d_1>1$ implies to.
$ d_1|a^n \implies (d_1|a \,\,\vee d_1=a^m)$ where $ m|n$.
Samely, $ d_1|b^n \implies (d_1|b \,\,\vee d_1=b^k)$ where $ k|n$.
Then, we have:
$(d_1|a \,\,\vee d_1=a^m)  \land (d_1|b \,\,\vee d_1=b^k) \land m|n \land k|n \land d_1>1$.
We know that there is no $d_1>1$ satisfying $d_1|a \land d_1|b$, because we know that $gcd(a,b)=1$.
On the other hand, as the proposition should be satisfied for all $n$, if we selected $n$ as a prime number, we wouldn't be able to find such $m\neq1$ and $k\neq1$. Then $m=n=1$, that is $(d_1=a \land d_1|b) \vee (d_1=b \land d_1|a)$. Then $gcd(a,b)=a \vee gcd(a,b)=b$ which is a contradiction.
Then our assumption is false.
