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Source: IMO LongList 1959-1966 Problem 42

September 2, 2004
SequenceDivisibilitynumber theoryIMO LonglistIMO Shortlist

Problem Statement

Given a finite sequence of integers a1,a_{1}, a2,a_{2}, ...,..., ana_{n} for n2.n\geq 2. Show that there exists a subsequence ak1,a_{k_{1}}, ak2,a_{k_{2}}, ...,..., akm,a_{k_{m}}, where 1k1k2...kmn,1\leq k_{1}\leq k_{2}\leq...\leq k_{m}\leq n, such that the number ak12+ak22+...+akm2a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2} is divisible by n.n. Note by Darij: Of course, the 1k1k2...kmn1\leq k_{1}\leq k_{2}\leq ...\leq k_{m}\leq n should be understood as 1k1<k2<...<kmn;1\leq k_{1}<k_{2}<...<k_{m}\leq n; else, we could take m=nm=n and k1=k2=...=km,k_{1}=k_{2}=...=k_{m}, so that the number ak12+ak22+...+akm2=n2ak12a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2}=n^{2}a_{k_{1}}^{2} will surely be divisible by n.n.