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So my idea is as such: first I note that if $\gcd(a,b)=1$ and $\gcd(a,c)=1$ so $\gcd(a,bc)=1$.

Now It enough to prove $\gcd\left(a^{k},b\right)=1$ for every $k\geq1$: indeed, given such result, for arbitrary positive integer $m$ we can apply the observation above $m$ times to get $\gcd\left(a^{m},b^{m}\right)=1$. We will now prove this by induction on $k$. the case $k=1$ is clear. Now assume $\gcd\left(a^{k},b\right)=1$. So we have $x,y$ positive integers such that $a^{k}x+by=1$ which implies

$$\begin{align} a^{k+1}x+b\left(ay\right)=a &\implies\gcd\left(a^{k+1},b\right)\lvert \;a\\ &\implies\gcd\left(a^{k+1},b\right)\le\gcd\left(a,b\right)=1\\ &\implies\gcd\left(a^{k+1},b\right)=1 \end{align}$$

and we are done. Is it correct? It feels a bit off to me.

Other approches and thoughts about ring thoery

As people commented, another option is to raise both sides of $ax+by=1$ to the $2k$ power and grouping terms to find integers such that $a^kz+b^kw=1$. In my abstract algebra course we were asked to show that in a commutative ring $R$ the radical of an ideal $I$ defined as

$\sqrt{I}=\left\{ r\in R\;\lvert\;r^{n}\in I\;\text{for some }n\in\mathbb{N}\right\} $

is an ideal itself. to show that $\sqrt{I}$ is closed under addition, I raised $a+b$ to the power of $n+m$ where $a^n,b^m \in I$ and grouped them using the same way. I wonder if there is a connection.any insight would be great. I never learned elementary number theory and I self study from Burton's book, so I still don't have a deep understanding of these algebraic structures in the context of $\mathbb{Z}$.

Oria
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    If $\gcd(a,b)=1$, then $a,b$ have no common prime factor, and this carries to $a^m$ and $b^n$. –  Oct 13 '21 at 16:08
  • of course, I should have mentioned I try to prove this relying only on elementary properties of the $\gcd$ – Oria Oct 13 '21 at 16:10
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    See my answer in this post, I did it with $\gcd$ properties. https://math.stackexchange.com/a/4034535/399263 – zwim Oct 13 '21 at 16:16
  • @zwim I see NovaDenizen did something similar to me, but I wonder if my adaptation is fine. – Oria Oct 13 '21 at 16:22
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    Appropriate answers will depend on what you're allowed to use. Here's one possibility. Since gcd$(a,b)=1$, there are integers $x$ and $y$ with $ax+by=1$. Take the $2k$-th power of both sides, $(ax+by)^{2k}=1$. When you multiply out the left side, you get a lot of monomials in $a,b,x,y$, and each of those monomials has at least $k$ factors $a$ or at least $k$ factors $b$ (or both). Collecting terms, we get an equation of the form $a^ku+b^kv=1$, which implies gcd$(a^k,b^k)=1$. – Andreas Blass Oct 13 '21 at 16:26
  • @andreasblass I'm familiar with this solution, just wondering If mine is fine – Oria Oct 13 '21 at 16:29
  • Why don't you use your lemma with $c=b^k$, giving $\gcd(a,b)=1\land\gcd(a,b^k)=1\implies\gcd(a,b^{k+1})=1$ ?? –  Oct 13 '21 at 16:39
  • @YvesDaoust definitely an option, just thought of applying a different reasoning. – Oria Oct 13 '21 at 16:51
  • You say "my idea is as such", and then don't use it !? Why making it simple when you can complexify... –  Oct 13 '21 at 16:58
  • @YvesDaoust I use it to get from $\gcd(a^m,b)=1$ to $\gcd(a^m,b^m)=1$ – Oria Oct 13 '21 at 17:00
  • Yep, why making it simple when you can complexify... –  Oct 13 '21 at 17:24

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Try the contrapositive.

If $p\mid \gcd(a^k, b^k)$ for some prime $p$ and some $k\in\Bbb N$, then $p\mid a^k$ and $p\mid b^k$. Now Euclid's Lemma gives $p\mid a$ and $p\mid b$. But then $p\mid \gcd(a,b)$.

Shaun
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