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I suddenly thought of this conjecture. I'm not sure if it's true:

" Let m be a positive integer such that m= $p_{\it 1}^{i_1}$ $p_{\it 2}^{i_2}$... $p_{\it k}^{i_k}$ where $p_1$,...,$p_k$ are distinct primes.

I) If k >1, then $m^ \frac{1}{n} $ is irrational, for n= q+1, q+2,...where q = gcd($ i_1$,...,$i_k$)

II) If k =1, then $m^ \frac{1}{n} $ is irrational, for n= $i_1$+ 1, $i_1$+ 2,... "

Appreciate your advice, thank you.

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    $\begingroup$ How about $m^{1/n}$ is rational, if and only if $n\mid i_1$, $n\mid i_2$, $\ldots$, $n\mid i_k$? $\endgroup$ – Jyrki Lahtonen Jun 29 '13 at 12:49
  • $\begingroup$ @JyrkiLahtonen: Your statement is intuitively true:) $\endgroup$ – Alexy Vincenzo Jun 29 '13 at 12:55
  • $\begingroup$ And follows from the uniqueness of the prime factorization of integers: if $q=r/s$ were the saide rational root, then write factorizations of $r$ and $s$ and work from there... $\endgroup$ – Jyrki Lahtonen Jun 29 '13 at 13:06
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Your conjecture is right. To proof this, we state a more general result:

Thm: $m^{\frac{1}{n}}$ is rational, iff $n\mid \operatorname{ord}_p(m)$ for all $p\in\mathbb P$.

(where $\operatorname{ord}_p(a)$ denotes the exponent of a prime $p$ in the prime factorization of an integer $a$.)

Before giving a proof of this theorem, let's explain, why it implies your conjecture. First, II is just a special case of I. So $m=\prod\limits_{i=1}^{k} p_i^{i_i}>1$ for any $k\in\mathbb N$.

Assume $m$ is rational, then $n\mid i_j$ for all $j$ by the above theorem. But then, $n\mid (i_1,\dots,i_k)$ and $n\leq (i_1,\dots,i_k)$, which is a contradiction to your choice of $n$. Hence, your conjecture is proven.


Proof: $m^{\frac{1}{n}}$ is rational, if and only if there are some $a,b\in\mathbb Z, b\neq 0$, such that $\frac{a}{b}=m^{\frac{1}{n}}$, or in other words $$a^n=m\cdot b^n$$ This again holds, if and only if for every $p\in\mathbb P$: $$\operatorname{ord}_p(a^n)=\operatorname{ord}_p(mb^n)$$ $\operatorname{ord}$ is multiplicative, so this is the same as $n\cdot \operatorname{ord}_p(a)=\operatorname{ord}_p(m) + n\cdot \operatorname{ord}_p(b)$, which is the same as: $$\operatorname{ord}_p(m) = n \left( \operatorname{ord}_p(a) - \operatorname{ord}_p(b) \right)$$ It follows immediately, that $n\mid \operatorname{ord}_{p}(m)$. Conversely, if this holds, set $b=1$ and $$a=\prod\limits_{p \text{ prime}}p^{\frac{\operatorname{ord}_p(m)}{n}}$$ then for every $p\in\mathbb P$: $$n \left( \operatorname{ord}_p(a) - \operatorname{ord}_p(b) \right)=n\left(\frac{\operatorname{ord}_p(m)}{n}-0\right)=\operatorname{ord}_p{m}$$

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