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$a_1 + a_22^k + a_33^k + \dots + a_mm^k = 0$

Source: All-Russian Olympiad 1996, Grade 10, First Day, Problem 4

April 18, 2013
algebrapolynomialalgebra proposed

Problem Statement

Show that if the integers a1a_1; …\dots ama_m are nonzero and for each k=0;1;… ;nk =0; 1; \dots ;n (n<m−1n < m - 1), a1+a22k+a33k+⋯+ammk=0a_1 + a_22^k + a_33^k + \dots + a_mm^k = 0; then the sequence a1,…,ama_1, \dots, a_m contains at least n+1n+1 pairs of consecutive terms having opposite signs.
O. Musin
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