# On the equality $\sqrt[n]{a_1}+\sqrt[n]{a_2}+\cdots+\sqrt[n]{a_k}= \sqrt[n]{b_1}+\sqrt[n]{b_2}+\cdots+\sqrt[n]{b_m}$

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 $k=m$.

• 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. Oct 14 '14 at 8:44
• @barak manos ,those counterexamples wasn't right because this equality should be right for all natural n>1 (not only for n=2) Oct 14 '14 at 8:47
• Well, if this equality is wrong for $n=2$ then it sure as hell not right for all natural $n>1$. Oct 14 '14 at 8:48
• 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. Oct 14 '14 at 8:50
• @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
– Dirk
Oct 14 '14 at 8:53

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.
• "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? Oct 15 '14 at 7:18