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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?

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  • $\begingroup$ I think your title should read "If $n$ is rational, when is $n^{1/n}$ rational?" $\endgroup$ – Charles Hudgins Sep 11 at 16:56
  • $\begingroup$ $n^{\frac 1k}$ is rational if and only if $n$ is a $k$powered integer and $n^{\frac 1k}$ is an integer. $\endgroup$ – fleablood Sep 11 at 16:59
  • $\begingroup$ Oh, $n$ is rational. Okay, a little more work but same idea. $\endgroup$ – fleablood Sep 11 at 17:07
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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.

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  • $\begingroup$ I'm not sure I understand your last sentence. $\endgroup$ – Timothy James Sep 11 at 18:00
  • $\begingroup$ Could you elaborate more and go through it a bit more thoroughly? $\endgroup$ – Timothy James Sep 11 at 18:00
  • $\begingroup$ I edited, is it clearer now? $\endgroup$ – Mindlack Sep 11 at 20:16
  • $\begingroup$ "But as $p^q$ is a $p$-th power, said valuation is a multiple of $p$, and thus $p|νq$.". How come it implies that the valuation is a multiple of $p$? $\endgroup$ – Timothy James Sep 11 at 20:36
  • $\begingroup$ Let $c=p^q=a^p$. Let $\nu$ be the $\pi$-adic valuation of $p$, $\mu$ that of $a$. Then the $\pi$-adic valuation of $c=p^q$ is $\nu q$, but the $\pi$-adic valuation of $c=a^p$ is $p \mu$. So $p\mu=q\nu$ thus $p|q\nu$. $\endgroup$ – Mindlack Sep 11 at 20:42

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