# If $n$ is an rational, when is $n^{1/n}$ rational?

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?

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

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.
• "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 $p$? – Timothy James Sep 11 at 20:36
• 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 $p|q\nu$. – Mindlack Sep 11 at 20:42