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Suppose $a_1,...,a_n,b_1,...,b_n$ are real numbers with $\sum_{i=1}^na_i=\sum_{i=1}^n b_i,\sum_{i=1}^n a_i^2=\sum_{i=1}^n b_i^2$ and for infinitely many $j\geq3$, $\sum_{i=1}^n a_i^j=\sum_{i=1}^n b_i^j$. Does this imply $\{a_1,...,a_n\}=\{b_1,...,b_n\}$?

The answer is positive if I can find infinitely many even integers $j$ such that $\sum_{i=1}^n a_i^j=\sum_{i=1}^n b_i^j$. But if I don't have this, still is the result true? Intuitively it seems to be.

José Carlos Santos
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Landon Carter
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  • +1, on some level, I expect this to be true even if the original equality holds for $j=1,\ldots,2n$... – gt6989b Feb 04 '19 at 15:32
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    see this https://math.stackexchange.com/questions/88424/if-mathrmtrmk-mathrmtrnk-for-all-1-leq-k-leq-n-then-how-do-w?noredirect=1&lq=1 – daw Feb 04 '19 at 16:00
  • Thanks but I don't see how this answers my question. Am I missing something? – Landon Carter Feb 04 '19 at 16:10
  • It shows that it suffices to have the equality of power sums for powers $j=1\dots n$. So finitely many even and odd powers suffice if enough of them are present. See https://en.wikipedia.org/wiki/Newton's_identities#Application_to_the_roots_of_a_polynomial – daw Feb 05 '19 at 10:50

1 Answers1

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Consider the case $n = 4$ with

$$ (a_1, a_2, a_3, a_4) = (-8, -1, 1, 8), \qquad (b_1, b_2, b_3, b_4) = (-7, -4, 4, 7). $$

Then $ \sum_{i=1}^{n} a_i^j = 0 = \sum_{i=1}^{n} b_i^j$ holds for all positive odd integers $j$, and $\sum_{i=1}^{n} a_i^2 = 130 = \sum_{i=1}^{n} b_i^2$.

Sangchul Lee
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