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There's a hint in the question: [ Put $d = \gcd(a, b)$ and write $a = rd, b = sd$, where $\gcd(r,s) = 1$. If $\gcd(r,s) = 1$ then $\gcd(r^n, s^n)=1$. Show that $r=1$, whence $a = d$.]

My proof is closer to this answer here.

Proof: Let, $d = \gcd(a, b)$ and $a = rd, b = sd, \gcd(r, s) = 1$.
$a^n|b^n$, so let $b^n = pa^n$
$(sd)^n = p(rd)^n$, $s^n = pr^n$.
Since $\gcd(r^n, s^n)=1$, $r^n=1$ (I'm confused if this can be written from the previous line: $s^n = pr^n$) or $r=1$.
This means $a=rd=d$, and therefore, $a|b$.

Kindly tell me if my proof is correct. Thank you.

alu
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Concerning your confusion : Since $\gcd(r^n,s^n)=1$ and $s^n=pr^n$, $s^n$ divides $p$ and $1=\frac{pr^n}{s^n}$. What do you conclude from there ?

Otherwise, the proof is correct.

Bernard
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Igor Haladjian
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  • $s^n$ divides $p$ means $p = p_1 s^n$, so $1 = \frac{pr^n}{s^n} = \frac{s^np_1r^n}{s^n}=p_1r^n$. Therefore $p_1=r^n=1$?

    My reasoning while writing the proof was, since p and $r^n$ both divides $s^n$, but $r^n$ cannot divide $s^n$, that means $r^n=1$. Is this reasoning okay?

     – alu Sep 24 '21 at 21:08
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    Yes, everything you say here is right. A small subtelty here : If $r^n$ is not equal to $1$, then it cannot divide $s^n$, therefore $r^n=1$. This is just a matter of saying things and doesn't change the mathematical content. – Igor Haladjian Sep 24 '21 at 21:58
  • Alright, thanks. – alu Sep 24 '21 at 22:01
  • Why did you write $\gcd(r^n,s^n)=1$ and $s^n=pr^n$ implies $s^n$ divides $p$? Is this from the Euclid's Lemma? If a = bc and gcd(a, c)= 1, a|bc, so we can apply Euclid's Lemma here? – alu Oct 01 '21 at 14:26
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    Yes, precisely. – Igor Haladjian Oct 02 '21 at 19:56
  • Alright, I understand now – alu Oct 02 '21 at 21:02