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Let $gcd(a, b) = d$

Let $a = rd$ and $b = sd$, where $gcd(r, s) = 1$.

$\therefore gcd(r^n, s^n) = 1$

Now, how do I show that $r = 1$?

Bill Dubuque
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Paul
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    Don't prove $r=1$. Prove $d=a$. $\gcd(a^n,b^n) = a^n$ because you were told that. But $a^n =d^nr^n$ and $b^n=s^nd^n$ so $\gcd(d^nr^n,d^ns^n) = d^n\gcd(r^n,s^n) = d^n$ so $d^n = a^n $ and so $d=a$. – fleablood Jan 22 '21 at 18:19
  • Don't forget $a^n| b^n$ so if $a=rd; b=sd$ AND if $b^n = a^nK$ then $s^nd^n = Kr^nd^n$ so $s^n=Kr^n$. So $s = \sqrt[n]Kr$ and $s$ and $r$ are integers $\sqrt[n]K$ is rational and is an integer. So $1=\gcd(r,s) =\gcd(r,r\sqrt[n]K) = r\gcd(1,\sqrt[n]K)=r\cdot 1 = r$. – fleablood Jan 22 '21 at 18:50
  • https://math.stackexchange.com/questions/25716/show-that-an-mid-bn-implies-a-mid-b – Neat Math Jan 22 '21 at 20:25

2 Answers2

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You're on the right track:

$b^n = a^n t \implies s^n = r^n t \implies r \mid s^n \implies r = 1$ because $\gcd(r,s)=1$.

Another way

$a^n\mid b^n$ iff $v_p(a^n) \le v_p(b^n)$ for all primes $p$. Here, $v_p(x)$ is the exponent of $p$ in the prime factorization of $x$.

Now, $v_p(a^n) \le v_p(b^n)$ iff $n v_p(a) \le n v_p(b)$ iff $v_p(a) \le v_p(b)$ iff $a \mid b$.

lhf
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Okay. It shouldn't be hard to prove $\gcd(m^k, n^k) = [\gcd(m,n)]^k$.

You can do it by considering prime factors, or by considering $$\gcd(m^k,n^k)=\\\gcd(m^{k-1}\frac m{\gcd(m,n)}\gcd(m,n), n^{k-1}\frac n{\gcd(m,n)}\gcd(m,n))=\\\gcd(m,n)\gcd(\frac m{\gcd(m,n)}m^{k-1},\frac m{\gcd(m,n)}n^{k-1})=\\\gcd(m,n)\gcd(m^{k-1},n^{k-1})$$ and using induction.

Or we could do it your way.

If $\gcd(m,n) = d$ and $m=rd;n=sd$ with $\gcd(r,s) =1$. then $\gcd(m^k,n^k)=\gcd(d^kr^k,d^ks^k)=d^k\gcd(r^k,s^k) = d^k*1 = d^k$.

.......

So now bear in mind $g|h \iff \gcd(g,h) = g$. (That's too obvious to require any discussion isn't it? If $g$ is a common divisor $g$ and $g$ then $g|h$. ANd if $g|h$ than $g$ is a common divisor and there can't be any greater divisor of $g$ than $g$ itself... right?)

So we are done.

......

$a^n|b^n\implies a^n = \gcd(a^n,b^n) = d^n\gcd(r^n,s^n) =d^n*1 =d^n\implies a=d=\gcd(a,b) \implies a|b$.

That's it.

fleablood
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  • You said $gcd(r^k, s^k)=d^k$ but that's not true. $gcd(r^k, s^k)= 1$. – Paul Jan 22 '21 at 19:42
  • But your comments and answer did help me greatly! So, thanks! – Paul Jan 22 '21 at 19:42
  • "You said $\gcd(r^k,s^k) = d^k$" No I didn't. I never said anything even remotely resembling that. I said $\gcd(d^kr^k, d^ks^k) = d^k \gcd(r^k,s^k)$ which is, of course true. ..$\gcd(\color{red}hw,\color{red}hv) = \color{red}h\gcd(w,v)$... And as $\gcd(r,s)=1$ the $\gcd(\color{red}{d^k}\color{blue}{r^k},\color{red}{d^k}\color{blue}{s^k})=\color{red}{d^k}\color{blue}{\gcd(r^k,s^k)} = \color{red}{d^k}\cdot \color{blue}1 = \color{red}{d^k}$. .... I am, of course, entirely correct. – fleablood Jan 22 '21 at 21:41
  • Sorry, but it continued onto the next line and I thought that. – Paul Jan 23 '21 at 04:04