Let $k,l$ be natural numbers and $\{ a_i, b_i \}$ be real positive numbers such that $a_1\leq a_2 \ldots \leq a_k$, $b_1\leq b_2 \ldots \leq b_l$ and $$ \sqrt[n]{a_1}+\sqrt[n]{a_2}+\cdots+\sqrt[n]{a_k}=\sqrt[n]{b_1}+\sqrt[n]{b_2}+\cdots+\sqrt[n]{b_l} $$

for all $n>1.$

How to prove that $k=l$ and $a_i=b_i$?

I can prove it with brute forse for $k=2$ only.

  • $\begingroup$ Can you post your proof for the case $k=2$? $\endgroup$
    – user170039
    Sep 28, 2014 at 14:19
  • $\begingroup$ I think you mean that $a$ and $b$ are whole numbers. $\endgroup$ Sep 28, 2014 at 14:20
  • 1
    $\begingroup$ k=l by taking limit since the limit of (a)^(1/n) is 1. $\endgroup$
    – Chen Jiang
    Sep 28, 2014 at 15:00

1 Answer 1


I assume you're dealing with positive numbers $a_i$ and $b_i$, for otherwise the $n$th roots won't always make sense. Also, the result is clearly false if some of the numbers are permitted to be zero.

My answer is unfortunately not at the precalculus level.

Let $x_n = \sqrt[n]{a_1}+\sqrt[n]{a_2}+\cdots+\sqrt[n]{a_k}$. Now write $c_i = \log a_i$ for $i=1, \dots, k$.

Using the Taylor expansion of degree $p$ for $e^t$ near $t = 0$, we find $$x_n = \sum_{i=1}^k e^{c_i/n} = k + \left(\sum_{i=1}^k c_i \right)\frac{1}{n} + \frac{1}{2!} \left(\sum_{i=1}^k c_i^2 \right)\frac{1}{n^2} + \dots + \frac{1}{p!}\left(\sum_{i=1}^k c_i^p \right)\frac{1}{n^p} + o\left( \frac{1}{n^p}\right).$$

Since the asymptotic expansion of the sequence $\{ x_n \}$ in powers of $1/n$ is well determined, this proves that the number $k$, and the sums of powers $S_1 = \sum_i c_i$, $S_2 = \sum_i c_i^2$,... , $S_p = \sum_i c_i^p$, are well determined by $\{x_n\}$. This is valid for an arbitrary choice of $p$. In turn, Newton's identities (http://en.wikipedia.org/wiki/Newton%27s_identities ), show that the elementary symmetric polynomials $\sigma_1$, ..., $\sigma_k$ of the $c_i$'s are well determined by $\{ x_n \}$. The numbers $c_1$, ..., $c_k$ are the roots of the polynomial $$x^k - \sigma_1 x^{k-1} + \dots \pm \sigma_k = 0.$$

Hence the numbers $c_i$ are determined by $\{ x_n \}$, up to permutation. Considering that the sequence $c_1$, ..., $c_k$, is arranged in increasing order, this shows that the $c_i$'s, and therefore the $a_i$'s, are well determined by the sequence $\{ x_n \}$.

  • $\begingroup$ Great solution! Very nice way to prove that two set of numbers $a_i$ and $b_i$ are equals up to permutation: just one need to prove that any family of symmetric functions are identical on this sets. Thank you. $\endgroup$
    – Leox
    Sep 28, 2014 at 17:46

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