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a_i = -a_{n+1-i} if sum a_i^{2k+1} = 0 when a_1 <= a_2 <= ... <= a_n

Source: (2021 -) 2022 Dürer Math Competition Regional E+5 https://artofproblemsolving.com/community/c1621671_

November 30, 2022
algebraSequence

Problem Statement

Let a1≤a2≤...≤ana_1 \le a_2 \le ... \le a_n be real numbers for which ∑i=1nai2k+1=0\sum_{i=1}^{n} a_i^{2k+1} = 0 holds for all integers 0≤k<n0 \le k < n. Show that in this case, ai=−an+1−ia_i = -a_{n+1-i} holds for all 1≤i≤n1 \le i \le n.
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