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F_{n (k+1)} = b_n F_{n k} + c_n F_{n (k-1)} , Fibonacci numbers

Source: 2018 SMT - Stanford Math Tournament , Team Round, Proof Question 4

January 26, 2022
FibonacciFibonacci sequenceFibonacci Numbersnumber theory

Problem Statement

Let FkF_k denote the series of Fibonacci numbers shifted back by one index, so that F0=1F_0 = 1, F1=1,F_1 = 1, and Fk+1=Fk+Fk−1F_{k+1} = F_k +F_{k-1}. It is known that for any fixed n≥1n \ge 1 there exist real constants bnb_n, cnc_n such that the following recurrence holds for all k≥1k \ge 1: Fn⋅(k+1)=bn⋅Fn⋅k+cn⋅Fn⋅(k−1).F_{n\cdot (k+1)} = b_n \cdot F_{n \cdot k} + c_n \cdot F_{n\cdot (k-1)}. Prove that ∣cn∣=1|c_n| = 1 for all n≥1n \ge 1.
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