# Does $\sum_{i=1}^n a_i^j=\sum_{i=1}^n b_i^j$ for infinitely many $j$ imply $\{a_i\}=\{b_i\}$?

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.

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

## 1 Answer

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