Prove that if $a, b, n\in \mathbb{N}, n\geq2\longrightarrow \sqrt[\leftroot{-2}\uproot{2}n]{a}\in \mathbb{Q} \iff a=b^n$.

Question

I'm at a complete loss here, I tried using the order of a prime function but didn't get anywhere. Any tips to get going?

Thanks in advance!

This can be done as a generalization of the proof that $\sqrt 2$ is irrational. — gary, Apr 21 '15 at 19:20
The statement is false. Take $n=2, a=b=4$. $\sqrt{a}\in \mathbb{Q}$, but $a\neq b^n$. To make it true, you need to fix the quantifiers. — vadim123, Apr 21 '15 at 19:21
@vadim123 this is what the problem says, I'm thinking it should be $\exists b\in \mathbb{N} / \sqrt[\leftroot{-2}\uproot{2}n]{a}\in \mathbb{Q} \iff a=b^n$ right? — YoTengoUnLCD, Apr 21 '15 at 19:28

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

This is an immediate consequence of the Rational Root Test. Apply it to the polynomial $\,x^n\!-a.$

edited Apr 13 '17 at 12:20

Community

answered Apr 21 '15 at 19:21

Bill Dubuque

Could you explain a bit? I think I have to use things like FTA to solve this. – YoTengoUnLCD Apr 21 '15 at 19:29
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@YoTengoUnLCD The linked answer gives a proof of RRT which uses well-known consequences of FTA (gcds and Euclid's Lemma). – Bill Dubuque Apr 21 '15 at 19:33

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