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