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"Suppose $a, b, m$, and $n$ are integers satisfying $\gcd(a, b)=\gcd(m, n)=1$ and $b, n \neq 0$. Then $(m/n)^{a/b}$ is rational if and only if $m$ and $n$ are $|b|$th powers of integers." Hint: The proof relies on the Fundamental Theorem of Arithmetic.

The proof of the converse is straightforward, but I’m having difficulty with the proof in the other direction.

I begin my proof as follows: Assume $\left( \frac{m}{n} \right)^{\frac{a}{b}} = \frac{c}{d}$ for some integers $c$ and $d$ with $\gcd(c, d)=1$. We can write:
$$\left( \frac{m}{n} \right)^{\frac{a}{b}} = \frac{m^{a/b}}{n^{a/b}} = \frac{c}{d}.$$
This implies that $dm^{a/b} = cn^{a/b}$, but I am unsure how to apply the Fundamental Theorem of Arithmetic (FTA) to conclude that $m$ and $n$ must be $|b|$-th powers of integers.

Bill Dubuque
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    Raise both sides to the $b$'th power, multiply through by $n^ad^b$, apply the FTA and count carefully to prove $d=1$. Also, assume $(m,n)=(a,b)=1$. – Mastrem Feb 08 '25 at 20:24
  • @Mastrem: But $d$ is not equal to $1$ in general. I think you must mean something different! – TonyK Feb 08 '25 at 20:34
  • Raising your displayed equation to the $b$'th power shows $,m^a = c^b,\ n^a = d^b,$ by unique fractionization in 2nd dupe, thus $,m,n,$ are $,b$'th powers by the first dupe. $\ \ $ – Bill Dubuque Feb 08 '25 at 21:21

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This answer assumes that $a,b,m,n$ are positive integers. If not, we have to deal with various cases, but the procedure is basically the same.

Raising your equation to the $b$th power gives $$\frac{m^a}{n^a}=\frac{c^b}{d^b}$$ And $(m^a,n^a)=(c^b,d^b)=1$, because $(m,n)=(c,d)=1$. So what we have are two rationals in their lowest terms, which are equal to each other. Therefore their numerators and denominators must be equal up to sign. So we have: $$m^a=c^b$$ $$n^a=d^b$$ Now take a prime $p$, and let $p^k$ be the largest power of $p$ that divides $m$. Then $p^{ak}$ is the largest power of $p$ that divides $m^a$, and is therefore the largest power of $p$ that divides $c^b$. But the largest power of $p$ that divides $c^b$ must be a multiple of $b$; i.e. $ak$ is a multiple of $b$. And because $(a,b)=1$, it follows that the $k$ must be a multiple of $b$.

Hence $m$ is a $b$th power! (And similarly $n$ is too.)

TonyK
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