On a System of Difference Equations

Abstract

We have investigated the periodical solutions of the system of rational difference equations 𝑥_𝑛+1=𝑦_𝑛−2/(−1±𝑦_𝑛−2𝑥_𝑛−1𝑦_𝑛),𝑦_𝑛+1=𝑥_𝑛−2/(−1±𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛), and 𝑧_𝑛+1=(𝑥_𝑛−2+𝑦_𝑛−2)/(−1±𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛), where 𝑦₀,𝑦₋₁,𝑦₋₂,𝑥₀,𝑥₋₁,𝑥₋₂,𝑧₀,𝑧₋₁,𝑧₋₂∈ℝ.

1. Introduction

Recently, a great interest has arisen on studying difference equation systems. One of the reasons for that is the necessity for some techniques which can be used in investigating equations which originate in mathematical models to describe real-life situations such as population biology, economics, probability theory, genetics, and psychology. There are many papers related to the difference equations system.

In [1], Kurbanli et al. studied the periodicity of solutions of the system of rational difference equations 𝑥_𝑛+1=𝑥_𝑛−1+𝑦_𝑛𝑦_𝑛𝑥_𝑛−1−1,𝑦_𝑛+1=𝑦_𝑛−1+𝑥_𝑛𝑥_𝑛𝑦_𝑛−1.−1 (1)

In [2], Çinar studied the solutions of the systems of difference equations 𝑥_𝑛+1=1𝑦_𝑛,𝑦_𝑛+1=𝑦_𝑛𝑥_𝑛−1𝑦_𝑛−1. (2)

In [3, 4], Özban studied the positive solutions of the system of rational difference equations 𝑥_𝑛=𝑎𝑦_𝑛−3,𝑦_𝑛=𝑏𝑦_𝑛−3𝑥_𝑛−𝑞𝑦_𝑛−𝑞,𝑥_𝑛+1=1𝑦_𝑛−𝑘,𝑦_𝑛+1=𝑦_𝑛𝑥_𝑛−𝑚𝑦_{𝑛−𝑚−𝑘}. (3)

In [5–16], Elsayed studied a variety of systems of rational difference equations; for more, see references.

In this paper, we have investigated the periodical solutions of the system of difference equations 𝑥_𝑛+1=𝑦_𝑛−2−1±𝑦_𝑛−2𝑥_𝑛−1𝑦_𝑛,𝑦_𝑛+1=𝑥_𝑛−2−1±𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑧_𝑛+1=𝑥_𝑛−2+𝑦_𝑛−2−1±𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑛∈ℕ₀, (4) where the initial conditions are arbitrary real numbers.

2. Main Results

Theorem 1. Let 𝑦₀=𝑎, 𝑦₋₁=𝑏, 𝑦₋₂=𝑐, 𝑥₀=𝑑, 𝑥₋₁=𝑒, 𝑥₋₂=𝑓, 𝑧₀=𝑘, 𝑧₋₁=𝑝, and 𝑧₋₂=𝑞 be arbitrary real numbers, and let {𝑥_𝑛,𝑦_𝑛,𝑧_𝑛} be a solution of the system 𝑥_𝑛+1=𝑦_𝑛−2−1+𝑦_𝑛−2𝑥_𝑛−1𝑦_𝑛,𝑦_𝑛+1=𝑥_𝑛−2−1+𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑧_𝑛+1=𝑥_𝑛−2+𝑦_𝑛−2−1+𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑛∈ℕ₀. (5) Also, assume that 𝑏≠0, 𝑒≠0, 𝑓𝑏𝑑≠1, and 𝑐𝑒𝑎≠1. Then, all six-period solutions of (5) are as follows: 𝑥_6𝑛+1=𝑐1−𝑐𝑒𝑎,𝑦_6𝑛+1=𝑓,𝑧𝑓𝑏𝑑−1_6𝑛+1=−𝑓+𝑐,𝑥𝑓𝑏𝑑−1_6𝑛+1=𝑏𝑓𝑏𝑑−1,𝑦_6𝑛+2(),𝑧=𝑒𝑐𝑒𝑎−1_6𝑛+2(),𝑥=−𝑒+𝑏)(𝑐𝑒𝑎+1_6𝑛+1=𝑎1−𝑐𝑒𝑎,𝑦_6𝑛+3=𝑑,𝑧𝑓𝑏𝑑−1_6𝑛+3=−𝑑+𝑎,𝑥𝑓𝑏𝑑+1_6𝑛+1=𝑓,𝑦_6𝑛+4𝑧=𝑐,_6𝑛+4=𝑐𝑓𝑏𝑑+1+𝑓(𝑐𝑒𝑎+1),𝑥𝑓𝑏𝑑+1_6𝑛+1=𝑒,𝑦_6𝑛+5𝑧=𝑏,_6𝑛+5=𝑏𝑓𝑏𝑑+1+𝑒(𝑐𝑒𝑎+1),𝑥𝑓𝑏𝑑+1_6𝑛+1=𝑑,𝑦_6𝑛+6𝑧=𝑎,_6𝑛+6=𝑎𝑓𝑏𝑑+1+𝑑(𝑐𝑒𝑎+1),𝑓𝑏𝑑+1𝑛∈ℕ₀. (6)

Proof. For 𝑛=0,1,2,3,4,5, we have 𝑥₁=𝑦₋₂−1+𝑦₋₂𝑥₋₁𝑦₀=𝑐,𝑦−1+𝑐𝑒𝑎₁=𝑥₋₂−1+𝑥₋₂𝑦₋₁𝑥₀=𝑓,𝑧−1+𝑓𝑏𝑑₁=𝑥₋₂+𝑦₋₂−1+𝑦₋₂𝑥₋₁𝑦₀=𝑓+𝑐,𝑥−1+𝑓𝑏𝑑₂=𝑦₋₁−1+𝑦₋₁𝑥₀𝑦₁=𝑏=𝑏−1+𝑏𝑑𝑓/𝑓𝑏𝑑−1,𝑦1/𝑓𝑏𝑑−1=𝑏𝑓𝑏𝑑−1₂=𝑥₋₁−1+𝑥₋₁𝑦₀𝑥₁=𝑒((=𝑒−1+𝑒𝑎𝑐/𝑐𝑒𝑎−1))()(),𝑧1/𝑐𝑒𝑎−1=𝑒𝑐𝑒𝑎−1₂=𝑥₋₁+𝑦₋₁−1+𝑥₋₁𝑦₀𝑥₁=𝑒+𝑏((=−1+𝑒𝑎𝑐/𝑐𝑒𝑎−1))𝑒+𝑏()=(),𝑥1/𝑐𝑒𝑎−1𝑒+𝑏)(𝑐𝑒𝑎−1₃=𝑦₀−1+𝑦₀𝑥₁𝑦₂=𝑎((𝑒()=𝑎−1+𝑎𝑐/𝑐𝑒𝑎−1))𝑐𝑒𝑎−1,𝑦𝑐𝑒𝑎−1₃=𝑥₀−1+𝑥₀𝑦₁𝑥₂=𝑑𝑏=𝑑−1+𝑑𝑓/𝑓𝑏𝑑−1𝑓𝑏𝑑−1,𝑧𝑓𝑏𝑑−1₃=𝑥₀+𝑦₀−1+𝑥₀𝑦₁𝑥₂=𝑑+𝑎𝑏=−1+𝑑𝑓/𝑓𝑏𝑑−1𝑓𝑏𝑑−1𝑑+𝑎,𝑥𝑓𝑏𝑑−1₄=𝑦₁−1+𝑦₁𝑥₂𝑦₃=𝑓/𝑓𝑏𝑑−1𝑏=−1+𝑓/𝑓𝑏𝑑−1𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1𝑓/𝑓𝑏𝑑−1=−1+𝑓𝑏𝑑/𝑓𝑏𝑑−1𝑓/𝑓𝑏𝑑−1𝑦1/𝑓𝑏𝑑−1=𝑓,₄=𝑥₁−1+𝑥₁𝑦₂𝑥₃=𝑐/(𝑐𝑒𝑎−1)((𝑒((=()−1+𝑐/𝑐𝑒𝑎−1))𝑐𝑒𝑎−1)(𝑎/𝑐𝑒𝑎−1))𝑐/𝑐𝑒𝑎−1((=()−1+𝑐𝑒𝑎/𝑐𝑒𝑎−1))𝑐/𝑐𝑒𝑎−1()𝑧1/𝑐𝑒𝑎−1=𝑐,₄=𝑥₁+𝑦₁−1+𝑥₁𝑦₂𝑥₃=(𝑐/(𝑐𝑒𝑎−1))+𝑓/𝑓𝑏𝑑−1((𝑒((=𝑐()−1+𝑐/𝑐𝑒𝑎−1))𝑐𝑒𝑎−1)(𝑎/𝑐𝑒𝑎−1))𝑓𝑏𝑑−1+𝑓𝑐𝑒𝑎−1,𝑥𝑓𝑏𝑑−1₅=𝑦₂−1+𝑦₂𝑥₃𝑦₄=𝑒(𝑐𝑒𝑎−1)((𝑐𝑦−1+𝑒𝑐𝑒𝑎−1)(𝑎/𝑐𝑒𝑎−1))=𝑒,₅=𝑥₂−1+𝑥₂𝑦₃𝑥₄=𝑏𝑓𝑏𝑑−1𝑓𝑧−1+𝑏𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1=𝑏,₅=𝑥₂+𝑦₂−1+𝑥₂𝑦₃𝑥₄=𝑏𝑓𝑏𝑑−1+𝑒(𝑐𝑒𝑎−1)𝑓=𝑏()−1+𝑏𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1𝑓𝑏𝑑−1+𝑒𝑐𝑒𝑎−1,𝑥𝑓𝑏𝑑−1₆=𝑦₃−1+𝑦₃𝑥₄𝑦₅=𝑑/𝑓𝑏𝑑−1𝑦−1+𝑑/𝑓𝑏𝑑−1𝑓𝑏=𝑑,₆=𝑥₃−1+𝑥₃𝑦₄𝑥₅=𝑎/(𝑐𝑒𝑎−1)((𝑧−1+𝑎/𝑐𝑒𝑎−1))𝑐𝑒=𝑎,₆=𝑥₃+𝑦₃−1+𝑥₃𝑦₄𝑥₅=(𝑎/(𝑐𝑒𝑎−1))+𝑑/𝑓𝑏𝑑−1((=𝑎()−1+𝑎/𝑐𝑒𝑎−1))𝑐𝑒𝑓𝑏𝑑−1+𝑑𝑐𝑒𝑎−1.𝑓𝑏𝑑−1 (7) For 𝑛=6,7,8,9,10,11, assume that 𝑥₇=𝑦₄−1+𝑦₄𝑥₅𝑦₆=𝑐−1+𝑐𝑒𝑎=𝑥₁,𝑦₇=𝑥₄−1+𝑥₄𝑦₅𝑥₆=𝑓−1+𝑓𝑏𝑑=𝑦₁,𝑧₇=𝑥₄+𝑦₄−1+𝑥₄𝑦₅𝑥₆=𝑓+𝑐−1+𝑓𝑏𝑑=𝑧₁,𝑥₈=𝑦₅−1+𝑦₅𝑥₆𝑦₇=𝑏−1+𝑏𝑑𝑓/−1+𝑓𝑏𝑑=𝑏𝑓𝑏𝑑−1=𝑥₂,𝑦₈=𝑥₅−1+𝑥₅𝑦₆𝑥₇=𝑒((()−1+𝑒𝑎𝑐/−1+𝑐𝑒𝑎))=𝑒𝑐𝑒𝑎−1=𝑦₂,𝑧₈=𝑥₅+𝑦₅−1+𝑥₅𝑦₆𝑥₇=𝑒+𝑏((=()−1+𝑒𝑎𝑐/−1+𝑐𝑒𝑎))𝑒+𝑏)(𝑐𝑒𝑎−1=𝑧₂,𝑥₉=𝑦₆−1+𝑦₆𝑥₇𝑦₈=𝑎((𝑒()=𝑎−1+𝑎𝑐/−1+𝑐𝑒𝑎))𝑐𝑒𝑎−1𝑐𝑒𝑎−1=𝑥₃,𝑦₉=𝑥₆−1+𝑥₆𝑦₇𝑥₈=𝑑𝑏=𝑑−1+𝑑𝑓/−1+𝑓𝑏𝑑𝑓𝑏𝑑−1𝑓𝑏𝑑−1=𝑦₃,𝑧₉=𝑥₆+𝑦₆−1+𝑥₆𝑦₇𝑥₈=𝑑+𝑎𝑏=−1+𝑑𝑓/−1+𝑓𝑏𝑑𝑓𝑏𝑑−1𝑑+𝑎𝑓𝑏𝑑−1=𝑧₃,𝑥₁₀=𝑦₇−1+𝑦₇𝑥₈𝑦₉=𝑓/−1+𝑓𝑏𝑑𝑏=−1+𝑓/−1+𝑓𝑏𝑑𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1𝑓/−1+𝑓𝑏𝑑−1+𝑓𝑏𝑑/−1+𝑓𝑏𝑑=𝑓=𝑥₄,𝑦₁₀=𝑥₇−1+𝑥₇𝑦₈𝑥₉=𝑐/(−1+𝑐𝑒𝑎)((𝑒((=()−1+𝑐/−1+𝑐𝑒𝑎))𝑐𝑒𝑎−1)(𝑎/𝑐𝑒𝑎−1))𝑐/−1+𝑐𝑒𝑎−1+(𝑐𝑒𝑎/(−1+𝑐𝑒𝑎))=𝑐=𝑦₄,𝑧₁₀=𝑥₇+𝑦₇−1+𝑥₇𝑦₈𝑥₉=(𝑐/(−1+𝑐𝑒𝑎))+𝑓/−1+𝑓𝑏𝑑((𝑒((=𝑐()−1+𝑐/−1+𝑐𝑒𝑎))𝑐𝑒𝑎−1)(𝑎/𝑐𝑒𝑎−1))𝑓𝑏𝑑−1+𝑓𝑐𝑒𝑎−1𝑓𝑏𝑑−1=𝑧₄,𝑥₁₁=𝑦₈−1+𝑦₈𝑥₉𝑦₁₀=𝑒(𝑐𝑒𝑎−1)−1+𝑒(𝑐𝑒𝑎−1)(𝑎/(𝑐𝑒𝑎−1))𝑐=𝑒=𝑥₅,𝑦₁₁=𝑥₈−1+𝑥₈𝑦₉𝑥₁₀=𝑏𝑓𝑏𝑑−1−1+𝑏𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1𝑓=𝑏=𝑦₅,𝑧₁₁=𝑥₈+𝑦₈−1+𝑥₈𝑦₉𝑥₁₀=𝑏𝑓𝑏𝑑−1+𝑒(𝑐𝑒𝑎−1)𝑓=𝑏()−1+𝑏𝑓𝑏𝑑−1𝑑/𝑓𝑏𝑑−1𝑓𝑏𝑑−1+𝑒𝑐𝑒𝑎−1𝑓𝑏𝑑−1=𝑧₅,𝑥₁₂=𝑦₉−1+𝑦₉𝑥₁₀𝑦₁₁=𝑑/𝑓𝑏𝑑−1−1+𝑑/𝑓𝑏𝑑−1𝑓𝑏=𝑑=𝑥₆,𝑦₁₂=𝑥₉−1+𝑥₉𝑦₁₀𝑥₁₁=𝑎/(𝑐𝑒𝑎−1)−1+(𝑎/(𝑐𝑒𝑎−1))𝑐𝑒=𝑎=𝑦₆,𝑧₁₂=𝑥₉+𝑦₉−1+𝑥₉𝑦₁₀𝑥₁₁=(𝑎/(𝑐𝑒𝑎−1))+𝑑/𝑓𝑏𝑑−1((=𝑎()−1+𝑎/𝑐𝑒𝑎−1))𝑐𝑒𝑓𝑏𝑑−1+𝑑𝑐𝑒𝑎−1𝑓𝑏𝑑−1=𝑧₆ (8) are true. Also, we have 𝑥₁=𝑐𝑐𝑒𝑎−1=𝑥₇=𝑥₁₃=⋯=𝑥_6𝑛+1,𝑛∈ℕ₀,𝑥₂=𝑏𝑓𝑏𝑑−1=𝑥₈=𝑥₁₄=⋯=𝑥_6𝑛+2,𝑛∈ℕ₀,𝑥₃=𝑎𝑐𝑒𝑎−1=𝑥₉=𝑥₁₅=⋯=𝑥_6𝑛+3,𝑛∈ℕ₀,𝑥₄=𝑓=𝑥₁₀=𝑥₁₆=⋯=𝑥_6𝑛+4,𝑛∈ℕ₀,𝑥₅=𝑒=𝑥₁₁=𝑥₁₇=⋯=𝑥_6𝑛+5,𝑛∈ℕ₀,𝑥₆=𝑑=𝑥₁₂=𝑥₁₈=⋯=𝑥_6𝑛+6,𝑛∈ℕ₀,𝑦₁=𝑓𝑓𝑏𝑑−1=𝑦₇=𝑦₁₃=⋯=𝑦_6𝑛+1,𝑛∈ℕ₀,𝑦₂=𝑒(𝑐𝑒𝑎−1)=𝑦₈=𝑦₁₄=⋯=𝑦_6𝑛+2,𝑛∈ℕ₀,𝑦₃=𝑑𝑓𝑏𝑑−1=𝑦₉=𝑦₁₅=⋯=𝑦_6𝑛+3,𝑛∈ℕ₀,𝑦₄=𝑐=𝑦₁₀=𝑦₁₆=⋯=𝑦_6𝑛+4,𝑛∈ℕ₀,𝑦₅=𝑏=𝑦₁₁=𝑦₁₇=⋯=𝑦_6𝑛+5,𝑛∈ℕ₀,𝑦₆=𝑎=𝑦₁₂=𝑦₁₈=⋯=𝑦_6𝑛+6,𝑛∈ℕ₀,𝑧₁=𝑓+𝑐𝑓𝑏𝑑−1=𝑧₇=𝑧₁₃=⋯=𝑧_6𝑛+1,𝑛∈ℕ₀,𝑧₂=(𝑒+𝑏)(𝑐𝑒𝑎−1)=𝑧₈=𝑧₁₄=⋯=𝑧_6𝑛+2,𝑛∈ℕ₀,𝑧₃=𝑑+𝑎𝑓𝑏𝑑−1=𝑧₉=𝑧₁₅=⋯=𝑧_6𝑛+3,𝑛∈ℕ₀,𝑧₄=𝑐𝑓𝑏𝑑−1+𝑓(𝑐𝑒𝑎−1)𝑓𝑏𝑑−1=𝑧₁₀=𝑧₁₆=⋯=𝑧_6𝑛+4,𝑛∈ℕ₀,𝑧₅=𝑏𝑓𝑏𝑑−1+𝑒(𝑐𝑒𝑎−1)𝑓𝑏𝑑−1=𝑧₁₁=𝑧₁₇=⋯=𝑧_6𝑛+5,𝑛∈ℕ₀,𝑧₆=𝑎𝑓𝑏𝑑−1+𝑑(𝑐𝑒𝑎−1)𝑓𝑏𝑑−1=𝑧₁₂=𝑧₁₈=⋯=𝑧_6𝑛+6,𝑛∈ℕ₀. (9)

Theorem 2. Let 𝑦₀=𝑎, 𝑦₋₁=𝑏, 𝑦₋₂=𝑐, 𝑥₀=𝑑, 𝑥₋₁=𝑒, 𝑥₋₂=𝑓, 𝑧₀=𝑘, 𝑧₋₁=𝑝, and 𝑧₋₂=𝑞 be arbitrary real numbers, and let {𝑥_𝑛,𝑦_𝑛,𝑧_𝑛} be a solution of the system 𝑥_𝑛+1=𝑦_𝑛−2−1−𝑦_𝑛−2𝑥_𝑛−1𝑦_𝑛,𝑦_𝑛+1=𝑥_𝑛−2−1−𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑧_𝑛+1=𝑥_𝑛−2+𝑦_𝑛−2−1−𝑥_𝑛−2𝑦_𝑛−1𝑥_𝑛,𝑛∈ℕ₀. (10) Also, assume that 𝑏≠0, 𝑒≠0, 𝑓𝑏𝑑≠−1, and 𝑐𝑒𝑎≠−1. Then, all six-period solutions of (10) are as follows: 𝑥_6𝑛+1𝑐=−1+𝑐𝑒𝑎,𝑦_6𝑛+1𝑓=−,𝑧𝑓𝑏𝑑+1_6𝑛+1=−−𝑓1+𝑓𝑏𝑑+𝑐,𝑥𝑓𝑏𝑑+1_6𝑛+1=−𝑏𝑓𝑏𝑑+1,𝑦_6𝑛+2(),𝑧=−𝑒𝑐𝑒𝑎+1_6𝑛+2(),𝑥=𝑒−𝑏1+𝑐𝑒𝑎_6𝑛+1𝑎=−1+𝑐𝑒𝑎,𝑦_6𝑛+3𝑑=−,𝑧𝑓𝑏𝑑+1_6𝑛+3=−−𝑑𝑓𝑏𝑑+1+𝑎,𝑥𝑓𝑏𝑑+1_6𝑛+1=𝑓,𝑦_6𝑛+4𝑧=𝑐,_6𝑛+4=−𝑐𝑓𝑏𝑑+1+𝑓1+𝑐𝑒𝑎+𝑐²𝑒²𝑎²(),𝑥𝑓𝑏𝑑+1𝑐𝑒𝑎+1_6𝑛+1=𝑒,𝑦_6𝑛+5𝑧=𝑏,_6𝑛+5=−𝑏1+2𝑓𝑏𝑑+𝑓²𝑏²𝑑²+𝑒(1+𝑐𝑒𝑎),𝑥𝑓𝑏𝑑+1_6𝑛+1=𝑑,𝑦_6𝑛+6𝑧=𝑎,_6𝑛+6=−𝑎𝑓𝑏𝑑+1+𝑑1+𝑐𝑒𝑎+𝑐²𝑒²𝑎²(,1+𝑐𝑒𝑎)𝑓𝑏𝑑+1𝑛∈ℕ₀. (11)

Proof. For 𝑛=0,1,2,3,4,5, we have 𝑥₁=𝑦₋₂−1−𝑦₋₂𝑥₋₁𝑦₀=𝑐𝑐−1−𝑐𝑒𝑎=−,𝑦1+𝑐𝑒𝑎₁=𝑥₋₂−1−𝑥₋₂𝑦₋₁𝑥₀=𝑓𝑓−1−𝑓𝑏𝑑=−,𝑧1+𝑓𝑏𝑑₁=𝑥₋₂+𝑦₋₂−1−𝑦₋₂𝑥₋₁𝑦₀=𝑓+𝑐−1−𝑓𝑏𝑑=−𝑓+𝑐,𝑥1+𝑓𝑏𝑑₂=𝑦₋₁−1−𝑦₋₁𝑥₀𝑦₁=𝑏−=𝑏−1−𝑏𝑑𝑓/𝑓𝑏𝑑+1−,𝑦1/𝑓𝑏𝑑+1=−𝑏1+𝑓𝑏𝑑₂=𝑥₋₁−1−𝑥₋₁𝑦₀𝑥₁=𝑒(−((=𝑒−1−𝑒𝑎𝑐/1+𝑐𝑒𝑎)))−(((),𝑧1/1+𝑐𝑒𝑎))=−𝑒1+𝑐𝑒𝑎₂=𝑥₋₁+𝑦₋₁−1−𝑥₋₁𝑦₀𝑥₁=𝑒+𝑏(−((=−1−𝑒𝑎𝑐/1+𝑐𝑒𝑎)))𝑒+𝑏−(((),𝑥1/1+𝑐𝑒𝑎))=−𝑒+𝑏)(1+𝑐𝑒𝑎₃=𝑦₀−1−𝑦₀𝑥₁𝑦₂=𝑎(−(((𝑎−1−𝑎𝑐/1+𝑐𝑒𝑎)))(−𝑒1+𝑐𝑒𝑎))=−,𝑦1+𝑐𝑒𝑎₃=𝑥₀−1−𝑥₀𝑦₁𝑥₂=𝑑−𝑑−1−𝑑𝑓/1+𝑓𝑏𝑑−𝑏1+𝑓𝑏𝑑=−,𝑧1+𝑓𝑏𝑑₃=𝑥₀+𝑦₀−1−𝑥₀𝑦₁𝑥₂=𝑑+𝑎−1−𝑑−𝑓/1+𝑓𝑏𝑑−𝑏1+𝑓𝑏𝑑=−𝑑+𝑎,𝑥1+𝑓𝑏𝑑₄=𝑦₁−1−𝑦₁𝑥₂𝑦₃=−𝑓/1+𝑓𝑏𝑑𝑏=−−1+𝑓/1+𝑓𝑏𝑑1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑𝑓/1+𝑓𝑏𝑑=−−1+𝑓𝑏𝑑/1+𝑓𝑏𝑑𝑓/1+𝑓𝑏𝑑−𝑦1/1+𝑓𝑏𝑑=𝑓,₄=𝑥₁−1−𝑥₁𝑦₂𝑥₃=−(𝑐/(1+𝑐𝑒𝑎))((𝑒((=−((−1+𝑐/1+𝑐𝑒𝑎))1+𝑐𝑒𝑎)(𝑎/1+𝑐𝑒𝑎))𝑐/1+𝑐𝑒𝑎))((=−((−1+𝑐𝑒𝑎/1+𝑐𝑒𝑎))𝑐/1+𝑐𝑒𝑎))−((𝑧1/1+𝑐𝑒𝑎))=𝑐,₄=𝑥₁+𝑦₁−1−𝑥₁𝑦₂𝑥₃=−(𝑐/(1+𝑐𝑒𝑎))−𝑓/1+𝑓𝑏𝑑((𝑒((=𝑐()−1+𝑐/1+𝑐𝑒𝑎))1+𝑐𝑒𝑎)(𝑎/1+𝑐𝑒𝑎))1+𝑓𝑏𝑑+𝑓1+𝑐𝑒𝑎,𝑥1+𝑓𝑏𝑑₅=𝑦₂−1−𝑦₂𝑥₃𝑦₄=−𝑒(1+𝑐𝑒𝑎)((𝑐𝑦−1−𝑒1+𝑐𝑒𝑎)(𝑎/1+𝑐𝑒𝑎))=𝑒,₅=𝑥₂−1−𝑥₂𝑦₃𝑥₄=−𝑏1+𝑓𝑏𝑑𝑓𝑧−1−𝑏1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑=𝑏,₅=𝑥₂+𝑦₂−1−𝑥₂𝑦₃𝑥₄=−𝑏1+𝑓𝑏𝑑−𝑒(1+𝑐𝑒𝑎)𝑓=𝑏()−1−𝑏1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑1+𝑓𝑏𝑑+𝑒1+𝑐𝑒𝑎,𝑥1+𝑓𝑏𝑑₆=𝑦₃−1−𝑦₃𝑥₄𝑦₅=−𝑑/1+𝑓𝑏𝑑𝑦−1+𝑑/1+𝑓𝑏𝑑𝑓𝑏=𝑑,₆=𝑥₃−1−𝑥₃𝑦₄𝑥₅=−(𝑎/(1+𝑐𝑒𝑎))((𝑧−1+𝑎/1+𝑐𝑒𝑎))𝑐𝑒=𝑎,₆=𝑥₃+𝑦₃−1−𝑥₃𝑦₄𝑥₅=−(𝑎/(1+𝑐𝑒𝑎))−𝑑/1+𝑓𝑏𝑑((=𝑎()−1+𝑎/1+𝑐𝑒𝑎))𝑐𝑒1+𝑓𝑏𝑑+𝑑1+𝑐𝑒𝑎.1+𝑓𝑏𝑑 (12) For 𝑛=6,7,8,9,10,11, assume that 𝑥₇=𝑦₄−1−𝑦₄𝑥₅𝑦₆=𝑐𝑐−1−𝑐𝑒𝑎=−1+𝑐𝑒𝑎=𝑥₁,𝑦₇=𝑥₄−1−𝑥₄𝑦₅𝑥₆=𝑓𝑓−1−𝑓𝑏𝑑=−1+𝑓𝑏𝑑=𝑦₁,𝑧₇=𝑥₄+𝑦₄−1−𝑥₄𝑦₅𝑥₆=𝑓+𝑐−1−𝑓𝑏𝑑=−𝑓+𝑐1+𝑓𝑏𝑑=𝑧₁,𝑥₈=𝑦₅−1−𝑦₅𝑥₆𝑦₇=𝑏−1+𝑏𝑑𝑓/1+𝑓𝑏𝑑=−𝑏1+𝑓𝑏𝑑=𝑥₂,𝑦₈=𝑥₅−1−𝑥₅𝑦₆𝑥₇=𝑒((()−1+𝑒𝑎𝑐/1+𝑐𝑒𝑎))=−𝑒1+𝑐𝑒𝑎=𝑦₂,𝑧₈=𝑥₅+𝑦₅−1−𝑥₅𝑦₆𝑥₇=𝑒+𝑏((()−1+𝑒𝑎𝑐/1+𝑐𝑒𝑎))=−𝑒+𝑏)(1+𝑐𝑒𝑎=𝑧₂,𝑥₉=𝑦₆−1−𝑦₆𝑥₇𝑦₈=𝑎((𝑒()𝑎−1−𝑎𝑐/1+𝑐𝑒𝑎))1+𝑐𝑒𝑎=−1+𝑐𝑒𝑎=𝑥₃,𝑦₉=𝑥₆−1−𝑥₆𝑦₇𝑥₈=𝑑𝑏𝑑−1−𝑑𝑓/1+𝑓𝑏𝑑1+𝑓𝑏𝑑=−1+𝑓𝑏𝑑=𝑦₃,𝑧₉=𝑥₆+𝑦₆−1−𝑥₆𝑦₇𝑥₈=𝑑+𝑎−1−𝑑𝑓/1+𝑓𝑏𝑑𝑏1+𝑓𝑏𝑑=−𝑑+𝑎1+𝑓𝑏𝑑=𝑧₃,𝑥₁₀=𝑦₇−1−𝑦₇𝑥₈𝑦₉=−𝑓/1+𝑓𝑏𝑑𝑏=−−1+𝑓/1+𝑓𝑏𝑑1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑𝑓/1+𝑓𝑏𝑑−1/1+𝑓𝑏𝑑=𝑓=𝑥₄,𝑦₁₀=𝑥₇−1−𝑥₇𝑦₈𝑥₉=−(𝑐/(1+𝑐𝑒𝑎))((𝑒((=−((−1+𝑐/1+𝑐𝑒𝑎))1+𝑐𝑒𝑎)(𝑎/1+𝑐𝑒𝑎))𝑐/1+𝑐𝑒𝑎))−1/(1+𝑐𝑒𝑎)=𝑐=𝑦₄,𝑧₁₀=𝑥₇+𝑦₇−1−𝑥₇𝑦₈𝑥₉=−(𝑐/(1+𝑐𝑒𝑎))−𝑓/1+𝑓𝑏𝑑((𝑒((=𝑐()−1+𝑐/1+𝑐𝑒𝑎))1+𝑐𝑒𝑎)(𝑎/1+𝑐𝑒𝑎))1+𝑓𝑏𝑑+𝑓1+𝑐𝑒𝑎1+𝑓𝑏𝑑=𝑧₄,𝑥₁₁=𝑦₈−1−𝑦₈𝑥₉𝑦₁₀=−𝑒(1+𝑐𝑒𝑎)−1−𝑒(1+𝑐𝑒𝑎)(𝑎/(1+𝑐𝑒𝑎))𝑐=𝑒=𝑥₅,𝑦₁₁=𝑥₈−1−𝑥₈𝑦₉𝑥₁₀=−𝑏1+𝑓𝑏𝑑−1−𝑏1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑𝑓=𝑏=𝑦₅,𝑧₁₁=𝑥₈+𝑦₈−1−𝑥₈𝑦₉𝑥₁₀=−𝑏1+𝑓𝑏𝑑−𝑒(1+𝑐𝑒𝑎)𝑓=𝑏()−1−𝑏1+𝑓𝑏𝑑𝑑/1+𝑓𝑏𝑑1+𝑓𝑏𝑑+𝑒1+𝑐𝑒𝑎1+𝑓𝑏𝑑=𝑧₅,𝑥₁₂=𝑦₉−1−𝑦₉𝑥₁₀𝑦₁₁=−𝑑/1+𝑓𝑏𝑑−1+𝑑/1+𝑓𝑏𝑑𝑓𝑏=𝑑=𝑥₆,𝑦₁₂=𝑥₉−1−𝑥₉𝑦₁₀𝑥₁₁=−(𝑎/(1+𝑐𝑒𝑎))−1+(𝑎/(1+𝑐𝑒𝑎))𝑐𝑒=𝑎=𝑦₆,𝑧₁₂=𝑥₉+𝑦₉−1−𝑥₉𝑦₁₀𝑥₁₁=−(𝑎/(1+𝑐𝑒𝑎))−𝑑/1+𝑓𝑏𝑑((=𝑎()−1+𝑎/1+𝑐𝑒𝑎))𝑐𝑒1+𝑓𝑏𝑑+𝑑1+𝑐𝑒𝑎1+𝑓𝑏𝑑=𝑧₆ (13) are true. Also, we have 𝑥₁𝑐=−1+𝑐𝑒𝑎=𝑥₇=𝑥₁₃=⋯=𝑥_6𝑛+1,𝑥𝑛=0,1,2,3,…,₂=−𝑏1+𝑓𝑏𝑑=𝑥₈=𝑥₁₄=⋯=𝑥_6𝑛+2,𝑥𝑛=0,1,2,3,…,₃𝑎=−1+𝑐𝑒𝑎=𝑥₉=𝑥₁₅=⋯=𝑥_6𝑛+3,𝑥𝑛=0,1,2,3,…,₄=𝑓=𝑥₁₀=𝑥₁₆=⋯=𝑥_6𝑛+4𝑥,𝑛=0,1,2,3,…,₅=𝑒=𝑥₁₁=𝑥₁₇=⋯=𝑥_6𝑛+5𝑥,𝑛=0,1,2,3,…,₆=𝑑=𝑥₁₂=𝑥₁₈=⋯=𝑥_6𝑛+6𝑦,𝑛=0,1,2,3,…,₁𝑓=−1+𝑓𝑏𝑑=𝑦₇=𝑦₁₃=⋯=𝑦_6𝑛+1,𝑛∈ℕ₀,𝑦₂=−𝑒(1+𝑐𝑒𝑎)=𝑦₈=𝑦₁₄=⋯=𝑦_6𝑛+2,𝑛∈ℕ₀,𝑦₃𝑑=−1+𝑓𝑏𝑑=𝑦₉=𝑦₁₅=⋯=𝑦_6𝑛+3,𝑛∈ℕ₀,𝑦₄=𝑐=𝑦₁₀=𝑦₁₆=⋯=𝑦_6𝑛+4,𝑛∈ℕ₀,𝑦₅=𝑏=𝑦₁₁=𝑦₁₇=⋯=𝑦_6𝑛+5,𝑛∈ℕ₀,𝑦₆=𝑎=𝑦₁₂=𝑦₁₈=⋯=𝑦_6𝑛+6,𝑛∈ℕ₀,𝑧₁=−𝑓+𝑐1+𝑓𝑏𝑑=𝑧₇=𝑧₁₃=⋯=𝑧_6𝑛+1,𝑛∈ℕ₀,𝑧₃=−𝑑+𝑎1+𝑓𝑏𝑑=𝑧₉=𝑧₁₅=⋯=𝑧_6𝑛+3,𝑛∈ℕ₀,𝑧₄=𝑐1+𝑓𝑏𝑑+𝑓(1+𝑐𝑒𝑎)1+𝑓𝑏𝑑=𝑧₁₀=𝑧₁₆=⋯=𝑧_6𝑛+4,𝑛∈ℕ₀,𝑧₅=𝑏1+𝑓𝑏𝑑+𝑒(1+𝑐𝑒𝑎)1+𝑓𝑏𝑑=𝑧₁₁=𝑧₁₇=⋯=𝑧_6𝑛+5,𝑛∈ℕ₀,𝑧₆=𝑎1+𝑓𝑏𝑑+𝑑(1+𝑐𝑒𝑎)1+𝑓𝑏𝑑=𝑧₁₂=𝑧₁₈=⋯=𝑧_6𝑛+6,𝑛∈ℕ₀. (14)

The following corollary follows from Theorem 1.

Corollary 3. The following conclusions are valid for 𝑛∈ℕ:(i)𝑥_6𝑛+2𝑦_6𝑛+3=𝑥_6𝑛+6𝑦_6𝑛+5,(ii)𝑥_6𝑛+1𝑦_6𝑛+2=𝑥_6𝑛+5𝑦_6𝑛+6,(iii)𝑥_6𝑛+1𝑦_6𝑛+6=𝑥_6𝑛+3𝑦_6𝑛+4,(iv)𝑥_6𝑛+4𝑦_6𝑛+3=𝑥_6𝑛+6𝑦_6𝑛+1.

The following corollary follows from Theorem 2.

Corollary 4. The following conclusions are valid for 𝑛∈ℕ:(i)𝑥_6𝑛+1𝑦_6𝑛+6=𝑥_6𝑛+3𝑦_6𝑛+4,(ii)𝑥_6𝑛+6𝑦_6𝑛+1=𝑥_6𝑛+4𝑦_6𝑛+3,(iii)𝑥_6𝑛+3𝑦_6𝑛+2=𝑥_6𝑛+5𝑦_6𝑛+6,(iv)𝑥_6𝑛+1𝑦_6𝑛+2=𝑥_6𝑛+5𝑦_6𝑛+4.