Jutaro wrote: If $p$, $q$ and $r$ are nonzero rational numbers such that $\sqrt[3]{pq^2}+\sqrt[3]{qr^2}+\sqrt[3]{rp^2}$ is a nonzero rational number, prove that $\frac{1}{\sqrt[3]{pq^2}}+\frac{1}{\sqrt[3]{qr^2}}+\frac{1}{\sqrt[3]{rp^2}}$ is also a rational number. $a=\sqrt[3]{pq^2},b=\sqrt[3]{qr^2},c=\sqrt[3]{rp^2}$. We have: $a+b+c,abc,a^3+b^3+c^3\in Q$. $\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\in Q$. $\Rightarrow a^2+b^2+c^2-ab-bc-ca\in Q$. $\Rightarrow (a+b+c)^2-3(ab+bc+ca)\in Q$. $\Rightarrow ab+bc+ca\in Q$. $\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}\in Q$.

During the olympiad many people tought that the condition $\sqrt[3]{pq^{2}}+\sqrt[3]{qr^{2}}+\sqrt[3]{rp^{2}}\in\mathbb{Q}$ implies $\sqrt[3]{pq^{2}}, \sqrt[3]{qr^{2}}, \sqrt[3]{rp^{2}}\in\mathbb{Q}$ but I can't find an elementary proof for that. May be someone has another idea...

ElChapin wrote: During the olympiad many people tought that the condition $\sqrt[3]{pq^{2}}+\sqrt[3]{qr^{2}}+\sqrt[3]{rp^{2}}\in\mathbb{Q}$ implies $\sqrt[3]{pq^{2}}, \sqrt[3]{qr^{2}}, \sqrt[3]{rp^{2}}\in\mathbb{Q}$ but I can't find an elementary proof for that. May be someone has another idea... We have $(2+\sqrt{2})+(3+\sqrt{3})+(10-\sqrt{2}-\sqrt{3})\in\mathbb{Q}$ but $(2+\sqrt{2}), (3+\sqrt{3}), (10-\sqrt{2}-\sqrt{3})\not\in\mathbb{Q}$

... mmm of course gouthamphilomath but the numbers you provided are not of the form $pq^2, qr^2, ... $ I didn't mean that if any three numbers have rational sum then they are rational. I was talking specifically of the problem.

Actually, the following result holds, I'm told: Let $k,n_1,\dots,n_k$ be positive integers and $a_1,\ldots,a_k$ be positive rational numbers. If \[\sqrt[n_1]{a_1}+\sqrt[n_2]{a_2}+\ldots+\sqrt[n_k]{a_k}\] is rational, then $\sqrt[n_i]{a_i}$ is rational, for all $i$.

enescu wrote: Actually, the following result holds, I'm told: Let $k,n_1,\dots,n_k$ be positive integers and $a_1,\ldots,a_k$ be positive rational numbers. If \[\sqrt[n_1]{a_1}+\sqrt[n_2]{a_2}+\ldots+\sqrt[n_k]{a_k}\] is rational, then $\sqrt[n_i]{a_i}$ is rational, for all $i$. Can this be proved by considering the minimal polynomial of \[\sqrt[n_1]{a_1}+\sqrt[n_2]{a_2}+\ldots+\sqrt[n_k]{a_k}\] :

Maybe someone speaking German can translate problem 835 from http://gdz.sub.uni-goettingen.de/en/dms/load/img/

enescu, can you upload the image/file or whatever that is in that link. I cannot see anything from the link (only got a blank page).

Here are two files. Sorry, they are in reversed order.

It has been posted before by vess, but I can't find it now..

I have found this one http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=6174&p=21087&hilit=rational+sqrt#p21087 but it only has square roots. The idea of pbornzstein's proof quoted there is about the same as I had suggested, using the conjugates of $s$. I think I have understood the German proof. The assumption is $s\in\mathbb Q$ and $a:=\sqrt[m]{a_n}\not\in\mathbb N$, i.e. $k>1$ where $k$ is smallest such that $b:=a^k\in\mathbb N$. The idea is building an integer polynomial (2) which has all the conjugates of $s-a=\sum_{j=1}^{n-1}\sqrt[m]{a_j}$ as roots. These conjugates use all possible combinations of $m$th roots of unity. Then $x^k-b$, being the minimal polynomial of $a$, can be factorized as a product of linear factors using certain of these conjugates. This is (3).

(4) is obtained by comparing coefficients of $x^{k-1}$ in (3). On the other hand, by the triangle inequality $\left|ks-\sum_{t=1}^{n-1} \Bigl(\sum_{j=1}^{k}\omega_{\nu_t^j}\Bigr)a_t^{1/m}\right|$ $\ge ks-\sum_{t=1}^{n-1} \Bigl|\sum_{j=1}^{k}\omega_{\nu_t^j}\Bigr|a_t^{1/m}\ge ks-\sum_{t=1}^{n-1} ka_t^{1/m}=ka>0.$ Contradiction

enescu wrote: Actually, the following result holds, I'm told: Let $k,n_1,\dots,n_k$ be positive integers and $a_1,\ldots,a_k$ be positive rational numbers. If \[\sqrt[n_1]{a_1}+\sqrt[n_2]{a_2}+\ldots+\sqrt[n_k]{a_k}\] is rational, then $\sqrt[n_i]{a_i}$ is rational, for all $i$. yes.actually,it's Xie's lemma. but could anyone give a complete proof of this?

Thjch Ph4 Trjnh wrote: Jutaro wrote: If $p$, $q$ and $r$ are nonzero rational numbers such that $\sqrt[3]{pq^2}+\sqrt[3]{qr^2}+\sqrt[3]{rp^2}$ is a nonzero rational number, prove that $\frac{1}{\sqrt[3]{pq^2}}+\frac{1}{\sqrt[3]{qr^2}}+\frac{1}{\sqrt[3]{rp^2}}$ is also a rational number. $a=\sqrt[3]{pq^2},b=\sqrt[3]{qr^2},c=\sqrt[3]{rp^2}$. We have: $a+b+c,abc,a^3+b^3+c^3\in Q$. $\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\in Q$. $\Rightarrow a^2+b^2+c^2-ab-bc-ca\in Q$. $\Rightarrow (a+b+c)^2-3(ab+bc+ca)\in Q$. $\Rightarrow ab+bc+ca\in Q$. $\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}\in Q$. To prove that $ab+bc+ca$ is rational we just inspect the following identity: $$(a+b+c)^3-(a^3+b^3+c^3)+3abc=3(a+b+c)(ab+bc+ac)$$Because $a+b+c$, $a^3+b^3+c^3$, and $abc$ are all rational it is clear the $ab+bc+ac$ must also be rational.