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This is homework but I would appreciate a hint :

If $(a,b)=1$ and $(\frac{a}{b})^m=n$ then show $b=1$

From $(\frac{a}{b})^m=n$ we can conclude that $a^m=n b^m$ so $b^m | a^m$ But what next? It’s obvious that because of $(a,b)=1$ b should be 1 to make that happen but how should I write it down ?

Bill Dubuque
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Negar Rezaei Nejad
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    Just add a few steps : you have $b^m|a^m$, so $b|a^m$, so because $(a,b)=1$, you have $b|a$ and again because $(a,b)=1$, you have $b=1$. – TheSilverDoe Oct 20 '20 at 17:08
  • @TheSilverDoe can you explain to me briefly how can we conclude $b|a$ from $b|a^m$ using the fact (a,b)=1? – Negar Rezaei Nejad Oct 20 '20 at 17:36
  • This is a classic result : if $a|bc$ and $(a,b)=1$, then $a|c$ (this is sometimes called Gauss theorem). This is a direct consequence of Bezout theorem. – TheSilverDoe Oct 20 '20 at 17:40
  • @TheSilverDoe i got it thank you . But isn’t it also true that b can be both 1 and -1? – Negar Rezaei Nejad Oct 20 '20 at 17:57
  • I presume that by $(a/b)^m=n$, you mean $(a/b)^m\in\Bbb Z$ for some $m\in\Bbb Z$. If $a,b$ are just integers, not necessarily natural numbers, $b$ can be $-1$ as well. – Prasun Biswas Oct 20 '20 at 17:58
  • @Negar Yes indeed. I guess you should add $b>0$ in your question to make it true :) Otherwise as you pointed, $b=-1$ is also a possibility. – TheSilverDoe Oct 20 '20 at 17:59
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    @TheSilverDoe thank you so much for your time <3 – Negar Rezaei Nejad Oct 20 '20 at 18:00
  • You are welcome ;) – TheSilverDoe Oct 20 '20 at 18:00
  • @TheSilverDoe sorry to bother you this much =,| but I have a related question: there is another question asking “ for n not being mth power of any natural number ,prove $n^\frac{1}{m}$ is irrational. By using contradiction we must show there is no a and b , (a,b)=1 that $n=\frac{a^m}{b^m} $but from previous question we know b can be 1 and -1 ; then (a,b)=1 would be true (???) – Negar Rezaei Nejad Oct 20 '20 at 18:54
  • @Negar: For that, you can wlog assume $b\gt 0$ for $n^{1/m}=a/b$ with $\gcd(a,b)=1$ since if $b\lt 0$, you can just take the negative sign in the numerator. The previous result then implies $b^m=1$ and thus $n=a^m$, contradicting that $n$ is not a perfect $m$th power. – Prasun Biswas Oct 20 '20 at 20:42
  • @PrasunBiswas thank you very much – Negar Rezaei Nejad Oct 21 '20 at 06:33

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Because $\gcd(a,b)=1$, we have that $a$ and $b$ share no prime factors (we also assume both are positive). Then, $a^{m}$ and $b^{m}$ also share no prime factors because exponentiation introduces no new prime factors. Thus, $\frac{a^{m}}{b^{m}}$ is fully reduced, so $b^{m}$ must equal $1$, and thus $\boxed{b = 1.}$

Joshua Wang
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