Periodic Solutions to a Third-Order Conditional Difference Equation over the Integers

Abstract

This paper studies a third-order conditional difference equation which is a generalization from the literature. We investigate this equation by transforming it into a first-order system. Finally it is proved that the equation has no period-two (or three) integer solutions. Besides, its all period-four (and five) integer solutions are derived under appropriate rational parameters.

1. Introduction

As mentioned in [1], difference equations appear naturally as a discrete analogue and as a numerical solution of differential and delay differential equations having applications in various scientific branches, such as biology, ecology, physics, economy, technics, and sociology. The stability, asymptotic behavior, and periodic property of solutions to difference equations had been widely investigated, such as [2–14]. Recently, the study of max-type difference equation attracted a considerable attention, for example, [7, 11, 15–25]. Although max-type difference equations are relatively simple in form, it is unfortunately extremely difficult to understand thoroughly the behavior of their solutions. The max operator arises naturally in certain models in automatic control theory. On the other hand, there exists another kind of difference equations called conditional difference equations, which also have simple forms, but it is difficult to understand clearly the behavior of their solutions.

From [2, 5], we know that the following conditional difference equation 𝑎_𝑛=𝑎_𝑛−1+𝑎_𝑛−22,if2∣𝑎_𝑛−1+𝑎_𝑛−2,𝑎_𝑛−1+𝑎_𝑛−2,otherwise, (1.1) with positive initial integers 𝑎₀,𝑎₁, has the property that each positive integer solution (𝑎_𝑛) is either stationary or unbounded.

Later, Ladas [26] gave the conjecture (also mentioned in [5]) that all solutions to the conditional difference equation𝑎_𝑛=𝑎_𝑛−1+𝑎_𝑛−23,if3∣𝑎_𝑛−1+𝑎_𝑛−2,𝑎_𝑛−1+𝑎_𝑛−2,otherwise, (1.2) are unbounded except for certain obvious periodic solutions, such as the solutions 1,1,2,1,1,2,… or 7,14,7,7,14,7,…. For such kind of conditional difference equations, any solution which is not eventually periodic must be unbounded; hence, the only problem is to classify the periodic solutions. However, this problem seems extremely difficult.

Equations such as (1.1) and (1.2) have many other generalizations, see [6, 26]. For instance, Clark [3] studied periodic solutions to 𝑎_𝑛=⌈𝑐𝑎_𝑛−1⌉−𝑎_𝑛−2 for various real 𝑐.

Greene and Niedzielski [5] considered a generalization of (1.1) and (1.2) and studied the following conditional difference equation:𝑎_𝑛=𝑟𝑎_𝑛−1+𝑎_𝑛−2,if𝑟𝑎_𝑛−1+𝑎_𝑛−2𝑎∈ℤ,_𝑛−1+𝑎_𝑛−2,otherwise, (1.3) where 𝑟 is some fixed rational number. However, they pointed out that it was also very hard to characterize the periodic solutions to (1.3). Hence, they addressed a different, easier question, and then did some research.

Motivated by the above works, in this paper, we study the following conditional third-order difference equation, which is another generalization of (1.1), (1.2), and (1.3):𝑦_𝑛=𝑟𝑦_𝑛−1+𝑦_𝑛−2+𝑦_𝑛−3,if𝑟𝑦_𝑛−1+𝑦_𝑛−2+𝑦_𝑛−3𝑦∈ℤ,_𝑛−1+𝑦_𝑛−2+𝑦_𝑛−3,otherwise, (1.4) where 𝑟 is some appropriate rational number. We study this equation by transforming it into a first-order system. It is eventually proved that the equation has no period-two (or three) positive integer solutions. Besides, its all period-four (and five) positive integer solutions are derived under appropriate rational parameters. The main results are presented in Section 4.

2. Auxiliary Results

For the convenience of investigation, higher-order difference equations are usually converted into first- or lower-order difference equation systems. As we all know, second-order difference equations can be transformed into first-order difference equation system. First, we define a mapping ℱ∶ℤ³→ℤ³ such thatℱ(𝐱)=𝐃𝐱,if𝑟𝑥+𝑦+𝑧∈ℤ,𝐂𝐱,otherwise, (2.1) where 𝐱=(𝑥,𝑦,𝑧)^𝑇∈ℤ³, 𝑟 is a rational parameter, and𝐃=010001𝑟𝑟𝑟,𝐂=010001111. (2.2)

Applying the defined mapping ℱ(𝐱), the third-order difference equation (1.4) can be converted into a corresponding first-order difference equation system as follows. Let 𝐱_𝑛=(𝑥_𝑛,𝑦_𝑛,𝑧_𝑛)^𝑇∈ℤ³, 𝑛∈ℕ₀ and 𝐱_𝑛=ℱ^𝑛(𝐱₀), 𝑛∈ℕ, then we have the following first-order difference equation system:𝐱_𝑛+1=ℱ𝐱_𝑛=𝐃𝐱_𝑛,if𝑟𝑥_𝑛+𝑦_𝑛+𝑧_𝑛∈ℤ,𝐂𝐱_𝑛,otherwise, (2.3) where 𝐱₀=(𝑦₀,𝑦₁,𝑦₂)^𝑇, 𝑦₀,𝑦₁,𝑦₂ are initial integers in (1.4). Note that the condition 𝑟(𝑥_𝑛+𝑦_𝑛+𝑧_𝑛)∈ℤ can be replaced by the equivalent matrix product condition 𝑟((1,1,1)𝐱_𝑛)∈ℤ.

For instance, when 𝑟=1/5, then the periodic solution (1,1,1,3,1,1,1,3,…) to (1.4) corresponds to the following periodic solution to system (2.3):111,113,131,311,111,…. (2.4) Given an initial vector 𝐱₀, each subsequent 𝐱_𝑛 can be got from 𝐱₀ via a formula 𝐱_𝑛=𝐋_𝑛𝐱₀, where 𝐋_𝑛 is an appropriate product of 𝑛 matrices, each of which is 𝐂 or 𝐃. Hence, for the above example, if 𝐱₀=(1,1,1)^𝑇, then 𝐱₁=𝐂𝐱₀, 𝐱₂=𝐃𝐱₁=𝐃𝐂𝐱₀, 𝐱₃=𝐃²𝐂𝐱₀, 𝐱₄=𝐃³𝐂𝐱₀, and so forth. Note that the matrices multiply 𝐱₀ from right to left and that 𝐱₄=𝐱₀, so 𝐃³𝐂𝐱₀=𝐱₀. We have the following obvious result.

Lemma 2.1. If system (2.3) has a periodic solution with period 𝑘 and 𝐱 is a vector in that periodic solution, then there exists a corresponding matrix 𝐋, which is a product of 𝑘 matrices, each of which is 𝐂 or 𝐃, such that 𝐱 is an eigenvector of 𝐋 with eigenvalue 1.

The proof of Lemma 2.1 is simple; here we point out that the converse of Lemma 2.1 is not true. The problem is that the sum of the entries in some 𝐱_𝑖 may be divisible by the denominator of 𝑟 at a time, when multiplication by 𝐂 is called for.

Next, we present the following linear algebra facts, for example, [27, Chapter  7], (similar results presented in [5]), followed by the form most convenient to us.

Lemma 2.2. (i) If 𝐯 is an eigenvector with eigenvalue 𝜆 for a matrix 𝐋, then 𝐯 is an eigenvector with eigenvalue 𝜆^𝑘 for 𝐋^𝑘.
(ii) The characteristic polynomial for a 3×3 matrix 𝐋 has the form 𝑓_𝐋(𝜆)=det(𝜆𝐈−𝐋)=𝜆³−tr(𝐋)𝜆²+det𝐋_1,1+det𝐋_2,2+det𝐋_3,3𝜆−det(𝐋), (2.5) where tr(𝐋) is the trace of 𝐋.
(iii) The characteristic polynomials for 𝐋𝐌 and 𝐌𝐋 are identical for any 𝑛×𝑛 matrices 𝐋 and 𝐌. In particular, tr(𝐋𝐌)=tr(𝐌𝐋).
(iv) The trace is linear. That is, tr(𝛼𝐋+𝛽𝐌)=𝛼tr(𝐋)+𝛽tr(𝐌), for any 𝑛×𝑛 matrices 𝐋,𝐌 and scalars 𝛼,𝛽.
(v) Every matrix satisfies its characteristic polynomial. In particular, for a 3×3 matrices 𝐋, one has 𝐋³−tr(𝐋)𝐋²+det𝐋_1,1+det𝐋_2,2+det𝐋_3,3𝐋−det(𝐋)𝐈=𝟎. (2.6)

Let the sequences (𝐻_𝑛)_𝑛≥0, (𝐽_𝑛)_𝑛≥0 be two solutions to the difference equation𝑦_𝑛=𝑦_𝑛−1+𝑦_𝑛−2+𝑦_𝑛−3,𝑛≥3 (2.7) with initial values 𝐻₀=𝐻₁=0, 𝐻₂=1 and 𝐽₀=𝐽₂=1, 𝐽₁=0, respectively.

Then, we get the following result about the matrix 𝐂 in (2.2).

Lemma 2.3. For the matrix  𝐂  in (2.2), one has 𝐂^𝑛=𝐻_𝑛−1𝐽_𝑛−1𝐻_𝑛𝐻_𝑛𝐽_𝑛𝐻_𝑛+1𝐻_𝑛+1𝐽_𝑛+1𝐻_𝑛+2, (2.8) for 𝑛∈ℕ.

Proof. This result can be proved by induction. By (2.2) and the definitions of the sequences (𝐻_𝑛)_𝑛≥0, (𝐽_𝑛)_𝑛≥0, we have =𝐻𝐂=010001111⁰𝐽₀𝐻₁𝐻₁𝐽₁𝐻₂𝐻₂𝐽₂𝐻₃. (2.9) Therefore, (2.8) holds for 𝑛=1. Now, we assume that (2.8) holds for 𝑛=𝜔, 𝜔∈ℕ. In the following, it suffices to prove that (2.8) holds for 𝑛=𝜔+1. By (2.2) and the associative law of matrix multiplication, we have 𝐂^𝜔+1=𝐂𝐂^𝜔=010𝐻001111^𝜔−1𝐽_𝜔−1𝐻_𝜔𝐻_𝜔𝐽_𝜔𝐻_𝜔+1𝐻_𝜔+1𝐽_𝜔+1𝐻_𝜔+2=𝐻_𝜔𝐽_𝜔𝐻_𝜔+1𝐻_𝜔+1𝐽_𝜔+1𝐻_𝜔+2𝐻_𝜔+2𝐽_𝜔+2𝐻_𝜔+3. (2.10) Therefore, (2.8) holds for 𝑘=𝜔+1. The proof is complete.

Given a 3×3 matrix 𝐋=𝐍₁𝐍₂⋯𝐍_𝑘, 𝑘∈ℕ, where each 𝐍_𝑗 is either 𝐂 or 𝐃, by Lemma 2.2 we define the following polynomial:𝒫_𝐋(𝑟)=−det(𝐈−𝐋)=−1+tr(𝐋)+det(𝐋)−det𝐋_1,1+det𝐋_2,2+det𝐋_3,3, (2.11) where 𝐈 is the identity matrix, and seek 𝐋 for which 𝒫_𝐋(𝑟) has rational zeros.

3. Some Other Properties

In this section, several properties of 𝒫_𝐋(𝑟) defined in (2.11) are derived. They are used to derive restrictions on values of 𝑟 that allow periodic solutions to (1.4).

Lemma 3.1. For the polynomial  𝒫_𝐋(𝑟)  in (2.11),  if 𝒫_𝐋(𝑟)=0,  then  𝒫_𝐋^𝑘(𝑟)=0,  for each  𝑘∈ℕ.

Proof. Since 𝒫_𝐋^𝑘(𝑟)=−det(𝐈−𝐋^𝑘)=−det(𝐈−𝐋)det(𝐋^𝑘−1+𝐋^𝑘−1+⋯+𝐈), then 𝒫_𝐋^𝑘(𝑟)=𝒫_𝐋(𝑟)𝑞(𝑟) for some polynomial 𝑞(𝑟)=det𝐋^𝑘−1+𝐋^𝑘−1+⋯+𝐈, (3.1) and the result follows.

Lemma 3.2. If 𝐂=⁰¹⁰⁰⁰¹₁₁₁,𝐄=⁰¹⁰⁰⁰¹_𝑥𝑥𝑥, then tr(𝐂^𝑛1𝐄𝐂^𝑛2𝐄⋯𝐂^𝑛𝑘𝐄) is a polynomial in 𝑥 of degree 𝑘 with nonnegative integer coefficients, with leading coefficient ∏^𝑘_𝑗=1𝐻_𝑛𝑗+3, where 𝑛_𝑗∈ℕ.

Proof. By Lemma 2.3, we get that 𝐂^𝑛𝐄=𝑥𝐻_𝑛𝑥𝐻_𝑛+𝐻_𝑛−1𝑥𝐻_𝑛+𝐽_𝑛−1𝑥𝐻_𝑛+1𝑥𝐻_𝑛+1+𝐻_𝑛𝑥𝐻_𝑛+1+𝐽_𝑛𝑥𝐻_𝑛+2𝑥𝐻_𝑛+2+𝐻_𝑛+1𝑥𝐻_𝑛+2+𝐽_𝑛+1𝐻=𝑥^𝑛𝐻_𝑛𝐻_𝑛𝐻_𝑛+1𝐻_𝑛+1𝐻_𝑛+1𝐻_𝑛+2𝐻_𝑛+2𝐻_𝑛+2+0𝐻_𝑛−1𝐽_𝑛−10𝐻_𝑛𝐽_𝑛0𝐻_𝑛+1𝐽_𝑛+1=𝑥𝐻_𝑛,𝐻_𝑛+1,𝐻_𝑛+2^𝑇𝟏+𝐅_𝑛, (3.2) where 𝟏 represents the vector (1,1,1), and 𝐅_𝑛=0𝐻_𝑛−1𝐽_𝑛−10𝐻_𝑛𝐽_𝑛0𝐻_𝑛+1𝐽_𝑛+1. (3.3) Hence, 𝐂^𝑛1𝐄𝐂^𝑛2𝐄⋯𝐂^𝑛𝑘𝐄=𝑥𝐻_𝑛1,𝐻_𝑛1+1,𝐻_𝑛1+2^𝑇𝟏+𝐅_𝑛1⋯𝑥𝐻_𝑛𝑘,𝐻_𝑛𝑘+1,𝐻_𝑛𝑘+2^𝑇𝟏+𝐅_𝑛𝑘=𝑥^𝑘𝐻_𝑛1,𝐻_𝑛1+1,𝐻_𝑛1+2^𝑇𝟏𝐻_𝑛2,𝐻_𝑛2+1,𝐻_𝑛2+2^𝑇𝟏⋯𝐻_𝑛𝑘,𝐻_𝑛𝑘+1,𝐻_𝑛𝑘+2^𝑇𝟏+⋯+𝐅_𝑛1𝐅_𝑛2⋯𝐅_𝑛𝑘=𝑥^𝑘𝑘_𝑗=1𝐻_𝑛𝑗,𝐻_𝑛𝑗+1,𝐻_𝑛𝑗+2^𝑇𝟏+⋯+^k_𝑗=1𝐅_𝑛𝑗. (3.4) To obtain the leading coefficient, we may induct on the simple calculation (𝑎,𝑏,𝑐)^𝑇𝟏𝑑,𝑒,𝑓^𝑇𝟏=𝑑+𝑒+𝑓(𝑎,𝑏,𝑐)^𝑇𝟏 (3.5) to show that the trace of the product is the product of the traces, such as the following. Easily, the trace of each individual matrix is 𝐻_𝑛𝑗+3,𝑗=1,2,…,𝑘. By (2.7) and (3.5), we have that ^𝑘_𝑗=1𝐻_𝑛𝑗,𝐻_𝑛𝑗+1,𝐻_𝑛𝑗+2^𝑇𝟏=𝐻_𝑛𝑘+𝐻_𝑛𝑘+1+𝐻_𝑛𝑘+2^𝑘−1_𝑗=1𝐻_𝑛𝑗,𝐻_𝑛𝑗+1,𝐻_𝑛𝑗+2^𝑇𝟏=𝐻_𝑛𝑘+3^𝑘−1_𝑗=1𝐻_𝑛𝑗,𝐻_𝑛𝑗+1,𝐻_𝑛𝑗+2^𝑇𝟏=⋯=𝐻_𝑛𝑘+3𝐻_{𝑛𝑘−1+3}⋯𝐻_𝑛2+3𝐻_𝑛1,𝐻_𝑛1+1,𝐻_𝑛1+2^𝑇𝟏=^𝑘_𝑗=2𝐻_𝑛𝑗+3𝐻_𝑛1,𝐻_𝑛1+1,𝐻_𝑛1+2^𝑇𝟏. (3.6) Thus the leading coefficient is ^𝑘_𝑗=2𝐻_𝑛𝑗+3𝐻_𝑛1+𝐻_𝑛1+1+𝐻_𝑛1+2=^𝑘_𝑗=1𝐻_𝑛𝑗+3. (3.7) The proof is complete.

4. Periodic Solutions

In this section, we prove that (1.4) has no periodic solution with prime period two or three and derive all periodic solutions to (1.4) with prime period four and five.

Theorem 4.1. Suppose that 𝑟 is a rational number and 𝑟∉ℤ, then (1.4) has no periodic solutions with prime period two.

Proof. Suppose that 𝐱∈ℤ³ be a vector of a period-two solution to system (2.3), then, for some matrix 𝐋₂, we have 𝐱=𝐋₂𝐱; here the matrix 𝐋₂ has four possible cases 𝐂², 𝐂𝐃, 𝐃𝐂 and 𝐃².
Take 𝐋₂=𝐃², for example, through some calculations, we get that 𝐃²=𝑟001𝑟𝑟𝑟₂𝑟()𝑟(),𝒫𝑟+1𝑟+1_𝐃2(𝑟)=3𝑟²+2𝑟−1. (4.1)
By solving the equation 𝒫_𝐃²(𝑟)=3𝑟²+2𝑟−1=0, we get two real roots 𝑟₁=1/3 and 𝑟₂=−1. Through similar calculations, we can get Table 1.
The only value of 𝑟 which may lead to period-two solutions to system (2.3) is 𝑟=1/3 since 𝑟∉ℤ. Hence, by solving the following matrix equation (with 𝑟=1/3) 𝐃²−𝐈𝐱=𝟎, (4.2) we get its all integer solutions 𝐱=(𝑡,𝑡,𝑡)^𝑇, 𝑡∈ℤ. Obviously, the solutions are equilibrium points which contradicts the assumption. The proof is complete.

Table 1 

Theorem 4.2. Suppose that 𝑟 is a rational number and 𝑟∉ℤ, then (1.4) has no periodic solutions with prime period three.

Proof. Assume that 𝐱∈ℤ³ be a vector of a period-three solution to system (2.3), then we get 𝐱=𝐋₃𝐱 where 𝐋₃ is a matrix product of three matrices, each of which is 𝐂 or 𝐃.
Obviously, 𝐋₃ has eight possible cases which can be divided into four categories such as three 𝐂s, two 𝐂s and one 𝐃, one 𝐂 and two 𝐃s, and three 𝐃s. By (2.2) and through certain calculations, the matrices 𝐂²𝐃,𝐂𝐃𝐂,𝐃𝐂² are similar, and; thus, they generate the same polynomials 𝒫_𝐋3(𝑟), so do the matrices 𝐂𝐃²,𝐃𝐂𝐃,𝐃²𝐂.
In the following, we take 𝐋₃=𝐃³, for example. Through some calculations, we have 𝐃³=𝑟𝑟𝑟𝑟₂r()𝑟()𝑟𝑟+1𝑟+1₂(𝑟+1)𝑟²(𝑟+2)𝑟𝑟²,𝒫+2𝑟+1_𝐃3(𝑟)=3𝑟−1. (4.3) By solving the equation 𝒫_𝐃³(𝑟)=3𝑟−1=0, we get the only root 𝑟=1/3. Through similar calculations, we can obtain Table 2.
The only value of 𝑟 which may lead to period-three solutions to system (2.3) is 𝑟=1/3 since 𝑟∉ℤ. Hence, by solving the following matrix equation (with 𝑟=1/3) 𝐃³−𝐈𝐱=𝟎, (4.4) we get its all integer solutions 𝐱=(𝑡,𝑡,𝑡)^𝑇,𝑡∈ℤ. Obviously, the solutions are equilibrium points which contradicts the assumption. The proof is complete.

Table 2 

In the following, denote by 𝒮(𝑘,𝑟) the set of initial values (𝑦₀,𝑦₁,𝑦₂) which lead to period-𝑘 solutions to (1.4) with the parameter 𝑟.

Theorem 4.3. Suppose that 𝑟 is a rational number and 𝑟∉ℤ, then (1.4) has periodic solutions with prime period four if and only if 𝑟=−1/3 or 1/5. Moreover, 𝒮(4,−1/3) = {(𝑡,𝑡,𝑡),(−𝑡,𝑡,𝑡),(𝑡,−𝑡,𝑡),(𝑡,𝑡,−𝑡)∣𝑡∈ℤ,3∤𝑡}, and 𝒮(4,1/5) = {(𝑡,𝑡,3𝑡),(𝑡,3𝑡,𝑡),(3𝑡,𝑡,𝑡),(𝑡,𝑡,𝑡)∣𝑡∈ℤ,5∤𝑡}.

Proof. Let 𝐱∈ℤ³ be a vector of a period-four solution to system (2.3), then we get 𝐱=𝐋₄𝐱, where 𝐋₄ is an appropriate product of four matrices, each of which is 𝐂 or 𝐃. Through similar calculation to those in Theorems 4.1 and 4.2, Table 3 is derived. The only values of 𝑟 which possibly lead to period-four solutions to system (2.3) are 𝑟=1/3,−1/3, or 1/5 because of 𝑟∉ℤ.
Case 1 (𝑟=1/3). By solving the following matrix equation 𝐃⁴−𝐈𝐱=𝟎, (4.5) we get its all integer solutions 𝐱=(𝑡,𝑡,𝑡)^𝑇,𝑡∈ℤ. Obviously, the solutions are equilibrium points.Case 2 (𝑟=−1/3). In this case, the matrix 𝐋₄ has four possible cases 𝐂³𝐃,𝐂²𝐃𝐂,𝐂𝐃𝐂², and 𝐃𝐂³.
By solving the matrix equation 𝐂³𝐃−𝐈𝐱=𝟎, (4.6) we get its all integer solutions 𝐱=(𝑡,𝑡,𝑡)^𝑇,𝑡∈ℤ. On the condition of that 3∤𝑡, then we can verify that the initial vector 𝐱₀=(𝑡,𝑡,𝑡)^𝑇 leads to a period-four solution to system (2.3), such as the following: 𝑡𝑡𝑡,𝑡𝑡,𝑡𝑡,𝑡𝑡,𝑡𝑡−𝑡𝑡−𝑡−𝑡,…. (4.7) Similarly, for the following three matrix equations 𝐂²𝐃𝐂−𝐈𝐱=𝟎,𝐂𝐃𝐂²−𝐈𝐱=𝟎,𝐃𝐂³−𝐈𝐱=𝟎, (4.8) we derive its all integer solutions 𝐱=(−𝑡,𝑡,𝑡)^𝑇,𝐱=(𝑡,−𝑡,𝑡)^𝑇,𝐱=(𝑡,𝑡,−𝑡)^𝑇,𝑡∈ℤ, respectively.
Note that, on the condition 3∤𝑡, the initial vectors 𝐱₀=(−𝑡,𝑡,𝑡)^𝑇,(𝑡,−𝑡,𝑡)^𝑇, or (𝑡,𝑡,−𝑡)^𝑇 also lead to period-four solutions to system (2.3).Case 3 (𝑟=1/5). In this case, the matrix 𝐋₄ also has four possible cases 𝐂𝐃³,𝐃𝐂𝐃²,𝐃²𝐂𝐃, and 𝐃³𝐂. By solving the matrix equation 𝐂𝐃³−𝐈𝐱=𝟎, (4.9) we get its all integer solutions 𝐱=(𝑡,𝑡,3𝑡)^𝑇, 𝑡∈ℤ. On the condition of that 5∤𝑡, then we can verify that the initial vector 𝐱₀=(𝑡,𝑡,3𝑡)^𝑇 leads to a period-four solution to system (2.3), such as the following: 𝑡𝑡,𝑡𝑡,𝑡𝑡,𝑡𝑡𝑡,𝑡𝑡3𝑡3𝑡3𝑡3𝑡,…. (4.10) Obviously, on the condition 5∤𝑡, the initial vectors 𝐱₀=(𝑡,3𝑡,𝑡)^𝑇, (3𝑡,𝑡,𝑡)^𝑇, or (𝑡,𝑡,𝑡)^𝑇 (which are integer solutions to appropriate matrix equations corresponding to 𝐃𝐂𝐃², 𝐃²𝐂𝐃, 𝐃³𝐂, resp.) also lead to period-four solutions to system (2.3). The proof is complete.

Table 3 

To illustrate the results, we give two orbits (see the following Figure 1) of period-four integer solutions to (1.4) of the particular cases 𝑟=−1/3, 𝑡=8 and 𝑟=1/5, 𝑡=7 in Theorem 4.3.

(a)

(b)

Figure 1 

Equation (1.4). (a) 𝑟=−1/3. Initial values 𝑦0=8,𝑦1=8, 𝑦2=8(𝑡=8). (b) 𝑟=1/5. Initial values 𝑦0=7, 𝑦1=7, 𝑦2=21(𝑡=7).

Theorem 4.4. Suppose that 𝑟 is a rational number and 𝑟∉ℤ, then (1.4) has periodic solutions with prime period five if and only if 𝑟=−7/15. Moreover, 𝒮(5,−7/15) = {(5𝑡,𝑡,9𝑡),(𝑡,9𝑡,−7𝑡),(9𝑡,−7𝑡,3𝑡),(−7𝑡,3𝑡,5𝑡),(3𝑡,5𝑡,𝑡)∣𝑡∈ℤ,3∤𝑡,5∤𝑡}.

Proof. Let 𝐱∈ℤ³ be a vector of a period-five solution to system (2.3), then we get 𝐱=𝐋₅𝐱 where 𝐋₅ is an appropriate product of five matrices, each of which is 𝐂 or 𝐃. Through similar calculation to those in Theorems 4.1 and 4.2, Table 4 is obtained. Apparently, the only values of 𝑟 which possibly lead to period-three solutions to system (2.3) are 𝑟=1/3 or −7/15 because of 𝑟∉ℤ.
Case 1 (𝑟=1/3). By solving the matrix equation 𝐃⁵−𝐈𝐱=𝟎, (4.11) we get its all integer solutions 𝐱=(𝑡,𝑡,𝑡)^𝑇, 𝑡∈ℤ. Obviously, the solutions are equilibrium points.Case 2 (𝑟=−7/15). In this case, the matrix 𝐋₅ also has five possible cases 𝐂⁴𝐃,𝐂³𝐃𝐂,𝐂²𝐃𝐂²,𝐂𝐃𝐂³, and 𝐃𝐂^𝟒. Solve the matrix equation 𝐂⁴𝐃−𝐈𝐱=𝟎, (4.12) we get its all integer solutions 𝐱=(5𝑡,𝑡,9𝑡)^𝑇, 𝑡∈ℤ. On the conditions that 3∤𝑡,5∤𝑡, then we can verify that the initial vector 𝐱₀=(5𝑡,𝑡,9𝑡)^𝑇 leads to a period-four solution to system (2.3), such as the following: 𝑡,𝑡,,,𝑡,𝑡5𝑡9𝑡9𝑡−7𝑡9𝑡−7𝑡3𝑡−7𝑡3𝑡5𝑡3𝑡5𝑡5𝑡9𝑡,…. (4.13)
Obviously, on the conditions 3∤𝑡,5∤𝑡, the initial vectors 𝐱₀=(𝑡,9𝑡,−7𝑡)^𝑇, (9𝑡,−7𝑡,3𝑡)^𝑇, (−7𝑡,3𝑡,5𝑡)^𝑇, or (3𝑡,5𝑡,𝑡)^𝑇 (which are integer solutions to appropriate matrix equations corresponding to 𝐃𝐂⁴, 𝐂𝐃𝐂³, 𝐂²𝐃𝐂², and 𝐂³𝐃𝐂, resp.) also lead to period-four solutions to system (2.3). The proof is complete.

Table 4 

To illustrate the results, we give two orbits (see Figure 2) of period-five integer solutions to (1.4) of the particular cases 𝑡=2,11 in Theorem 4.4.

(a)

(b)

Figure 2 

Equation (1.4) 𝑟=−7/15. (a): Initial values 𝑦0=10, 𝑦1=2, 𝑦2=18(𝑡=2); (b): Initial values 𝑦0=55, 𝑦1=11, 𝑦2=99(𝑡=11).