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Let $k,m\in \mathbb{N}$. Let $a_1,a_2,\cdots,a_k\ >0$ and $b_1,b_2,\cdots,b_m \ >0$ such that $$\sqrt[n]{a_1}+\sqrt[n]{a_2}+\cdots+\sqrt[n]{a_k}= \sqrt[n]{b_1}+\sqrt[n]{b_2}+\cdots+\sqrt[n]{b_m}$$ for all natural number $n,m>1$.

  1. Prove that $k=m.$
  2. Prove that $a_1a_2\cdots a_k=b_1b_2\cdots b_m$
  3. Prove that if each of the two sets of numbers sort of growth, then these sets will be the same.

I've proved that $k=m$.

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    $\begingroup$ If memory serves correctly, this question has been asked here before, and several users have shown simple counterexamples, such as $\sqrt[2]{4}+\sqrt[2]{4}=\sqrt[2]{1}+\sqrt[2]{9}$ which is a counterexample for your 2nd argument, or $\sqrt[2]{9}+\sqrt[2]{16}=\sqrt[2]{49}$ which is a counterexample for your 1st argument. $\endgroup$ – barak manos Oct 14 '14 at 8:44
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    $\begingroup$ @barak manos ,those counterexamples wasn't right because this equality should be right for all natural n>1 (not only for n=2) $\endgroup$ – Andrew Oct 14 '14 at 8:47
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    $\begingroup$ Well, if this equality is wrong for $n=2$ then it sure as hell not right for all natural $n>1$. $\endgroup$ – barak manos Oct 14 '14 at 8:48
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    $\begingroup$ In any case, perhaps I've misinterpreted something in your description, but one thing I do know for sure is that this question has recently been asked here (probably by one of your fellow students), so you might as well just search for it. $\endgroup$ – barak manos Oct 14 '14 at 8:50
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    $\begingroup$ @barakmanos As I understand the question, it goes: Assume that there are there numbers such that the equality holds for all $n$. Then prove 1.-3. To see 1. pass to the limit $n\to\infty$. Indeed, a related question has been asked recently, but on MO: mathoverflow.net/questions/48927/two-equal-series $\endgroup$ – Dirk Oct 14 '14 at 8:53
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From $\lim_{n\to\infty}\sqrt[n]x=1$ for arbitrary $x>0$, we conclude that the left hand side converges to $k$ and the right hand side to $m$ ans $n\to\infty$. This shows the first claim.

Let $x_i=\sqrt[k!]{a_i}$ and $y_i=\sqrt[k!]{b_i}$. Then for $j=1,2,\ldots, k$, we have $x_1^j+\ldots +x_k^j=y_1^j+\ldots +y_k^j$ (just let $n=k!/j$). It follows that the elemtary symmetric polynomials in the $x_i$ have the same values as those in the $y_i$. Since the product is one of the elementary polynomials, we conclude that $a_1\cdots a_k=(x_1\cdots x_k)^{k!}=(y_1\cdots y_k)^{k!}=b_1\cdots b_k$., the second claim.

In fact, as all elementary symmetric polynomials coincide, we have the polynomial identity $\prod_{i=1}^k(X-x_i)=\prod_{i=1}^k(X-y_i)$, that is, the $x_i$ and the $y_i$ are the same as multiset, or: are the same up to reordering. This is the third claim.

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  • $\begingroup$ Seems it's true. $\endgroup$ – Andrew Oct 14 '14 at 9:03
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    $\begingroup$ "It follows that the elemtary symmetric polynomials in the $x_i$ have the same values as those in the $y_i$. Since the product is one of the elementary polynomials, we conclude that...". I understand that it's true but can you show me proof of this fact? $\endgroup$ – Andrew Oct 15 '14 at 7:18

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