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If $a$, $b$, $c$ are positive integers, prove that $a^\frac{b}{c}$ can only be an integer if $\frac{b}{c}$ is an integer or $\sqrt[c]{a}$ is an integer. The first case is trivial, it's an integer to an integer power. So assuming $\sqrt[c]{a}$ is non-integer, it's intuitive that $\sqrt[c]{a^b}$ can only be an integer if b is a multiple of c, i.e. $\frac{b}{c}$ is an integer. But I am stuck on how to prove it.

Edit: Also assume $\frac{b}{c}$ is in lowest terms.

Drew Weymouth
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    $4^{\tfrac{6}{4}}$ is a counterexample. $\dfrac{6}{4}$ is not an integer and $\sqrt[4]{4}$ is not an integer, yet $4^{\tfrac{6}{4}} = 8$ is an integer. You need that $\dfrac{b}{c}$ is in lowest terms. – SlipEternal Mar 18 '19 at 18:52
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    Well, your argument shows one direction of the implication. Now you got to show the reverse. – Rushabh Mehta Mar 18 '19 at 18:52
  • Yep, it is not true, unless you have that $b$ and $c$ are relatively prime. – Thomas Andrews Mar 18 '19 at 18:58
  • On ce you assume $b,c$ are relatively prime, consider the prime factorization of $a^{b/c}$ and the prime factorization of $a.$ Then compare two ways of writing the prime factorization of $a^b$ in two different ways. – Thomas Andrews Mar 18 '19 at 19:01
  • Beware that the answer you accepted is incorrect (or incomplete) - it claims without proof that $,r= a^{\large 1/c}\in\Bbb Z$ but the (Bezout) gcd argument only shows that $r\in \Bbb Q$. See my answer for one way to correctly complete the proof (using the Rational Root Test). – Bill Dubuque Mar 19 '19 at 14:06

3 Answers3

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You showed the $\longleftarrow$ part of the implication, now you need to show the $\longrightarrow$ side. However, this isn't too difficult.

Suppose for the sake of contradiction that $a^{ \frac bc}$ is an integer, $\frac bc$ is in lowest terms, and $a^{\frac1c}$ is not an integer.

Let $z=a^{\frac bc}\in\mathbb Z$ Hence, $z^{\frac1b}\notin\mathbb Z$. But, we know that $a=(z^{\frac 1b})^c\in\mathbb Z$. Let $r=z^{\frac1b}\notin\mathbb Z$. So, $r^c,r^b\in\mathbb Z$. But, we know that this implies $r^{\gcd(b,c)}=r\in\mathbb Z$, a contradiction.

Rushabh Mehta
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Let $\,w = a^{\large 1/c}.\,$ Note $\,w^{\large b},w^{\large c}\!\in\Bbb Q\,\Rightarrow\, w^{\large jb+kc}\!\in \Bbb Q\,$ for all $\,j,k\in\Bbb Z,\,$ so for $\, jb\!+\!kc = \gcd(b,c)=1\,$ (Bezout) we infer $\,w\in \Bbb Q,\,$ so $\,w \in \Bbb Z\,$ by applying the Rational Root Test to root $w$ of $\,x^{\large c}-a=0.$

Bill Dubuque
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If "a" is a integer, then "a" can be expressed in terms of primes say

$a = p^xq^y...r^z$

Now $p^{(m/n)}$ is not integer if n does not divide m(if n divides m, then obviously it is integer). Suppose it is integer then

$p^{(m/n)} = Q^Y..R^Z$, Where Q,..,Z are primes and Y,..Zs are integers. Then $p^m = Q^{nY}..R^{nZ}$, now $p^m$ can be divided only by p or some power(less than m) of p . Now $\dfrac{p^m}{Q^{nY}} = S^{nW}..R^{nZ}$, as $(S^{nW}..R^{nZ})$ is integer so $\dfrac{p^m}{Q^{nY}}$ is integer that indicates $Q=p$, and $\dfrac{p^m}{Q^{nY}}$ is some power of p. Proceeding in the same way we see all Q,..,Z are p only. That indicates n divides m.

Now $a = p^xq^y...r^z$, $a^{m/n} = p^{mx/n}q^{my/n}...r^{mz/n} = b$, for some integer b.

So $p^{mx}q^{my}...r^{mz} = b^n$, as left side contains primes p,q,..,r only right side should also contain the same primes. That indicates b should also have p,q,..,r only.

Writing b as $b = p^{i}q^{j}...r^{k}$, we see that $p^{mx}q^{my}...r^{mz} = p^{ni}q^{nj}...r^{nk}$. Now as both side factored into same primes we can equate the powers of the primes. That leads to

$\dfrac{mx}{n}=i ,\dfrac{my}{n}=j,..,\dfrac{mz}{n}=k$,

that indicates if n does not divide m(and m,n in their lowest terms) then n must divide x,n must divide y,..,n must divide z that indicates $p^\dfrac{x}{n}q^\dfrac{y}{n}...r^\dfrac{z}{n} = a^\dfrac{1}{n}$ is integer.

novice_2
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