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t_{n+1} = 2(t_n)^2 - 1, gcd (t_n,t_m) = 1

Source: Netherlands - Dutch NMO 1966 p2

January 31, 2023
number theorygreatest common divisorGCDrecurrence relation

Problem Statement

For all nn, tn+1=2(tn)2−1t_{n+1} = 2(t_n)^2 - 1. Prove that gcd (tn,tm)=1(t_n,t_m) = 1 if n≠mn \ne m.
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