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.

  • $\begingroup$ +1, on some level, I expect this to be true even if the original equality holds for $j=1,\ldots,2n$... $\endgroup$ – gt6989b Feb 4 at 15:32
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    $\begingroup$ see this math.stackexchange.com/questions/88424/… $\endgroup$ – daw Feb 4 at 16:00
  • $\begingroup$ Thanks but I don't see how this answers my question. Am I missing something? $\endgroup$ – Landon Carter Feb 4 at 16:10
  • $\begingroup$ 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 en.wikipedia.org/wiki/… $\endgroup$ – daw Feb 5 at 10:50

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 $\sum_{i=1}^{n} a_i^2 = 130 = \sum_{i=1}^{n} b_i^2$.


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