Is the following constructed number always irrational? Given $\ $ $ a, b, c \in \mathbb{N} $, $\ $ $ a,b,c > 1  $, $\ $ $ b > c $, $\ $ $b$ and $c$ coprime, $a$ and $c$ coprime, is $ a^{\frac{b}{c}} $ irrational?
Examples: $2^{\frac{4}{3}}$, $13^{\frac{25}{12}}$.
 A: For $x\in \Bbb N$ and for prime $p$ let $F_p(x)$ be the largest $n\in \{0\}\cup \Bbb N$ such that $p^n$ is a divisor of $x.$ 
$(\bullet)$ . If $a,c\in \Bbb N$ and if $c$ is a divisor of $F_p(a)$ for $every$ prime $p$ that divides $a$ then $a^{1/c}\in \Bbb N....\;$ I.e. then $a=(a')^c$ for some $a'\in \Bbb N.$
If $a,b,c,d,e\in \Bbb N$ with $\gcd (b,c)=1$ and if $a^{b/c}=d/e$ then $a^be^c=d^c.$
So for any prime $p$ that divides $a$ we have $bF_p(a)+cF_p(e)=F_p(a^be^c)=F_p(d^c)=cF_p(d).$
So $c$ is a divisor of $bF_p(a)$ because $bF_p(a)=c(F_p(d)-F_p(e)\,).$
Now $c|bF_p(a)\implies c|F_p(a)$ because $\gcd (b,c)=1.$ So by $(\bullet)$ we have $a=(a')^c$ for some $a'\in \Bbb N.$ Hence, $d/e=a^{b/c}=((a')^c)^{b/c}=(a')^c\in \Bbb N.$
So if $a,b,c\in \Bbb N$ and $\gcd(b,c)=1$ then either $a^{b/c}$ is an integer or $a^{b/c}$ is irrational.
A: No, it is not true always.
For example, $a=4, b=5, c=2$, then $a^\frac{b}{c} = 32$ which is rational.
However, if you cannot find any $x\in\mathbb{N}$ such that $x^c=a$, then $a^\frac{b}{c}$ is always irrational. In my example, $x$ was $2$.
