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Selection tests for the Gulf Mathematical Olympiad, 2013

Combinatorics Difficulty 7.8 National Olympiad, round 2 Prove it Saudi Arabia

Let F0=0F_{0}=0, F1=1F_{1}=1 and Fn+1=Fn+Fn−1F_{n+1}=F_{n}+F_{n-1}, for all positive integer nn, be the Fibonacci sequence. Prove that for any positive integer mm there exist infinitely many positive integers nn such that
Fn+2≡Fn+1+1≡Fn+2 mod m F_{n}+2 \equiv F_{n+1}+1 \equiv F_{n+2} \quad \bmod m

Solution

Let mm be a positive integer and consider the infinite set of pairs (Fk,Fk+1)(F_{k}, F_{k+1}), for k∈Nk \in \mathbb{N}. By the pigeonhole principle, there exists a pair (a,b)(a, b) of integers 0≤a,b≤m−10 \leq a, b \leq m-1 and an infinite sequence of integers 0<k1<k2<⋯0<k_{1}<k_{2}<\cdots such that
(Fki,Fki+1)≡(a,b) mod m, for all i≥1. (F_{k_{i}}, F_{k_{i}+1}) \equiv (a, b) \quad \bmod m, \text{ for all } i \geq 1.
Therefore
(Fki−1,Fki)=(Fki+1−Fki,Fki)≡(b−a,a) mod m, for all i≥1. (F_{k_{i}-1}, F_{k_{i}}) = (F_{k_{i}+1}-F_{k_{i}}, F_{k_{i}}) \equiv (b-a, a) \bmod m, \text{ for all } i \geq 1.
We keep descending in this way until we get
(Fki−k1+2,Fki−k1+3)≡(F2,F3)≡(1,2) mod m, for all i≥1. (F_{k_{i}-k_{1}+2}, F_{k_{i}-k_{1}+3}) \equiv (F_{2}, F_{3}) \equiv (1,2) \quad \bmod m, \quad \text{ for all } i \geq 1.
Let ni=ki−k1+2n_{i}=k_{i}-k_{1}+2, for all i≥1i \geq 1. Clearly, the infinite sequence 2=n1<n2<⋯2=n_{1}< n_{2}<\cdots is increasing and we have
Fni+2≡F2+2≡3 mod m,Fni+1+1≡F3+1≡3 mod m and Fni+2≡Fni+1+Fni≡F3+F2≡3 mod m \begin{gathered} F_{n_{i}}+2 \equiv F_{2}+2 \equiv 3 \bmod m, \quad F_{n_{i}+1}+1 \equiv F_{3}+1 \equiv 3 \bmod m \\ \text{ and } F_{n_{i}+2} \equiv F_{n_{i}+1}+F_{n_{i}} \equiv F_{3}+F_{2} \equiv 3 \bmod m \end{gathered}

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