Given that $n$ is rational, when is $\sqrt[n]{n}$ rational?
We can make a polynomial $x^{n}-n$ whose root is $\sqrt[n]{n}$ and using RRT we can show that there are no rational roots, but in the process we are assuming that $n$ is an integer (coefficients must be integers for RRT to work).
How can we solve this problem?
If $1/n$ is an integer, it works... conversely, if $(p/q)^{q/p}$ is a rational $a/b$, where $p,q$ are coprime positive (and $a,b$ as well), then $p^q/q^q=a^p/b^p$, both as quotients of coprime positive integers, so $p^q=a^p$.
So if $\pi$ is a prime divisor of $p$ with multiplicity $\nu$ (ie $\pi^{\nu}|p$, $\pi^{\nu+1}$ doesn't divide $p$), then the $\pi$-adic valuation of $p^q$ is $\nu q$. But as $p^q$ is a $p$-th power, said valuation is a multiple of $p$, and thus $p|\nu q$. But as $p$ and $q$ are coprime, $p|\nu$, thus $\nu \geq p$, and hence $p \geq \pi^{\nu} > \nu$, a contradiction. Thus $p=1$ (because $p$ has no prime divisors), whence the conclusion.