Existence of a Period-Two Solution in Linearizable Difference Equations

Abstract

Consider the difference equation 𝑥_𝑛+1=𝑓(𝑥_𝑛,…,𝑥_𝑛−𝑘),𝑛=0,1,…, where 𝑘∈{1,2,…} and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it can be rewritten as the nonautonomous linear equation 𝑥_𝑛+𝑙=∑^𝑘_{𝑖=1−𝑙}𝑔_𝑖𝑥_𝑛−𝑖, 𝑛=0,1,…, where 𝑙,𝑘∈{1,2,…} and the functions 𝑔_𝑖∶ℝ^𝑘+𝑙→ℝ. We give some necessary and sufficient conditions for the equation to have a minimal period-two solution when 𝑙=1.

1. Introduction

Consider the difference equation 𝑥_𝑛+1=𝑓𝑥_𝑛,…,𝑥_𝑛−𝑘,𝑛=0,1,…, (1) where 𝑘∈{1,2,…} and the initial conditions are real numbers. We investigate the existence and nonexistence of the minimal period-two solution of this equation when it has the linearization of the form 𝑥_𝑛+𝑙=^𝑘_{𝑖=1−𝑙}𝑔_𝑖𝑥_𝑛−𝑖,𝑛=0,1,…, (2) where 𝑘,𝑙∈{1,2,…} and the functions 𝑔_𝑖∶ℝ^𝑘+𝑙→ℝ. By “(1) has the linearization (2)” we mean that (1) can be rewritten as the nonautonomous linear equation (2); see [1, 2].

The importance of a period-two solution is well known in the case of first order difference equations of the form of (1) with 𝑘=0, where the periods of the solutions appear in the well-known Sharkovsky ordering starting with period two. As a consequence of the results on Sharkovsky ordering the nonexistence of the period-two solution implies the nonexistence of periodic solutions of any period; see [3–5].

In the case of second order difference equations the following result has been obtained in [6].

Theorem 1. Let 𝐼⊆𝑅 and let 𝑓∈𝐶[𝐼×𝐼,𝐼] be a function which either increases in both variables or decreases in the first variable and increases in the second variable. Then for every solution of (1) with 𝑘=1 the subsequences {𝑥_2𝑛}^∞_𝑛=0 and {𝑥_2𝑛+1}^∞_𝑛=−1 of even and odd terms of the solution do exactly one of the following.(i)Eventually they are both monotonically increasing.(ii)Eventually they are both monotonically decreasing.(iii)One of them is monotonically increasing and the other is monotonically decreasing.

As a consequence of Theorem 1 every bounded solution of (1) with 𝑘=1 approaches either an equilibrium solution, a period-two solution, or a finite point at the boundary, and every unbounded solution is asymptotic to the point at infinity in a monotonic way. In view of Theorem 1 the results on the nonexistence of period-two solutions are as important as the results on the existence of these solutions. The importance of the existence or nonexistence of period-two solutions is also clear from the fact that one of the two most common local bifurcations for second order monotone autonomous difference equations is period-doubling bifurcation; see [3, 4, 7–9] for related results. Also the only known global bifurcation for second order monotone autonomous difference equations is period-doubling bifurcation [10]. See [11–13] for related results. The nonexistence results for periodic solutions which are the discrete analogue of the Bendixson’s nonexistence result for periodic solutions of differential equations have been obtained in [14, 15].

The results obtained in this paper are applicable to both autonomous and nonautonomous difference equations as the coefficients 𝑔_𝑖 in (2) are in general functions of 𝑛 and 𝑥_𝑛−𝑖,𝑖=0,1,…. Some of our examples will reflect this situation. The method of finding period-two solutions in the autonomous case consists of finding the fixed points of the second iterate of the corresponding map. However, in the nonautonomous case this method does not work and the results which will be presented in this paper can be used to find period-two solutions.

Some interesting points of our results can be demonstrated by the following example.

Example 2. The period-two solution {Φ,Ψ}, Φ≠Ψ of the difference equation 𝑥_𝑛+1=𝑎_𝑛𝑥_𝑛+𝑏_𝑛𝑥_𝑛−1,𝑛=0,1,…, (3) where {𝑎_𝑛} and {𝑏_𝑛} are two real sequences, satisfies Φ=𝑎_𝑛Ψ+𝑏_𝑛Φ,Ψ=𝑎_𝑛Φ+𝑏_𝑛Ψ,𝑛=0,1,…, (4) which implies 𝑏_𝑛−𝑎_𝑛=1,𝑛=0,1,…. (5) Conversely, if condition (5) holds, then any possible period-two solution {Φ,Ψ}, Φ≠Ψ of (3) must satisfy 𝑎_𝑛(Φ+Ψ)=0,𝑛=0,1,…. (6) If 𝑎_𝑁≠0 for some 𝑁=0,1,… then Ψ=−Φ and (3) has an infinite number of period-two solutions of the form {Φ,−Φ}, Φ≠0. If 𝑎_𝑛=0 for every 𝑛=0,1,… then 𝑏_𝑛=1 for every 𝑛=0,1,… in which case every nonequilibrium solution of (3) is a period-two solution.
Thus, condition (5) is a necessary and sufficient condition for the existence of a period-two solution. This condition is clearly satisfied if, for instance, 𝑎_𝑛=1𝑛+1−1,𝑏_𝑛=1𝑛+1,𝑛=0,1,…, (7) or if {𝑎_𝑛} and {𝑏_𝑛} are period-two sequences which satisfy 𝑏_2𝑛=1,𝑏_2𝑛+1𝑎=𝑏,_2𝑛=0,𝑎_2𝑛+1=𝑏−1,𝑛=0,1,…, (8) or if {𝑎_𝑛} and {𝑏_𝑛} are period-𝑘 sequences which satisfy 𝑏_𝑛+𝑖−𝑎_𝑛+𝑖=1,𝑖=0,1,…,𝑘−1,𝑛=0,1,…. (9) An example of a nonautonomous nonlinear difference equation for which one can find a period-two solution is the following equation 𝑥_𝑛+1=𝑐_𝑛+𝑥_𝑛−2+𝑥_𝑛−4−1𝑥_𝑛+𝑐_𝑛+𝑥_𝑛−2+𝑥_𝑛−4𝑥_𝑛−1,𝑛=0,1,…. (10) The quadratic second order difference equation 𝑥_𝑛+1=𝐴𝑥²_𝑛+𝐶𝑥²_𝑛−1+𝐷𝑥_𝑛+𝐸𝑥_𝑛−1,𝑛=0,1,…, (11) where 𝐴,𝐶,𝐷,and𝐸 are constants, can be linearized as 𝑥_𝑛+1=𝐴𝑥_𝑛+𝐷𝑥_𝑛+𝐶𝑥_𝑛−1+𝐸𝑥_𝑛−1,𝑛=0,1,…, (12) which is of the form of (3) where 𝑎_𝑛=𝐴𝑥_𝑛+𝐷,𝑏_𝑛=𝐶𝑥_𝑛−1+𝐸 and condition (5) becomes the first order linear difference equation 𝐶𝑥_𝑛−1+𝐸−𝐴𝑥_𝑛−𝐷=1,𝑛=0,1,…. (13) Since 𝑎_𝑛≠0,𝑛≥0 then the period-two solution of (11) has the form {𝑥₋₁,−𝑥₋₁}, 𝑥₋₁≠0.

2. The Constant Case

In this section we consider the case when the sums of the even indexed functions 𝑔_𝑖 and the odd indexed functions 𝑔_𝑖 are both constants.

The following simple result will be a useful technical tool.

Lemma 3. Suppose that (1) has the linearization (2). Let 𝛼=_{{}𝑖∈1−𝑙,…,𝑘∶𝑖𝑒𝑣𝑒𝑛}𝑔_𝑖,𝛽=_{𝑖∈{1−𝑙,…,𝑘∶𝑖𝑜𝑑𝑑}}𝑔_𝑖,𝑛=0,1,…. (14) Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,  𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (a)For 𝑙∈{1,3,…}, 𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛,𝑡ℎ𝑒𝑛Φ1−𝛽=Ψ𝛼;𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑,𝑡ℎ𝑒𝑛Ψ1−𝛽=Φ𝛼. (15)(16)(b)For 𝑙∈{2,4,…}, ()()𝑖𝑓𝑛𝑖𝑠𝑒𝑣𝑒𝑛,𝑡ℎ𝑒𝑛Ψ1−𝛼=Φ𝛽;𝑖𝑓𝑛𝑖𝑠𝑜𝑑𝑑,𝑡ℎ𝑒𝑛Φ1−𝛼=Ψ𝛽. (17)(18)

Proof. By plugging Ψ=𝑥_2𝑛 and Φ=𝑥_2𝑛−1 for 𝑛≥0 in (2) and assuming that 𝑙 is odd we obtain immediately that when 𝑛 is even (15) holds, while in the case when 𝑛 is odd (16) holds. Similarly assuming that 𝑙 is even we obtain immediately that in the case when 𝑛 is even (17) holds, while when 𝑛 is odd (18) holds.

Theorem 4. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝛼,𝛽 are given by (14). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0 if and only if the following hold: (a)if 𝑛 is even, then Φ(1−𝛽)=Ψ𝛼;(b)if 𝑛 is odd, then Ψ(1−𝛽)=Φ𝛼.

Proof. The necessary part of the proof follows from part (a) of Lemma 3.
For the proof of the sufficient part choose the initial conditions Ψ=𝑥₀= 𝑥₋₂=⋯,Φ=𝑥₋₁=𝑥₋₃=⋯ and 𝑥₀≠𝑥₋₁. Setting 𝑛=0,1,… we get that, by part (a), 𝑥₁𝑥=𝛼Ψ+𝛽Φ=Φ,₂𝑥=𝛼Φ+𝛽Ψ=Ψ,₃𝑥=𝛼Ψ+𝛽Φ=Φ,₄=𝛼Φ+𝛽Ψ=Ψ. (19) By using induction we get that Φ≠Ψ,Ψ=𝑥_2𝑛,andΦ=𝑥_2𝑛−1 for 𝑛≥0.

Theorem 5. Suppose that (1) has the linearization (2) and that 𝛼,𝛽 are given by (14). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (1)If  𝑙∈{1,3,…}, then (a)𝛽=1 if and only if 𝛼=0; (b)if 𝛽≠1, then 𝛽−𝛼=1 and Ψ=−Φ. (2)If  𝑙∈{2,4,…}, then (a)𝛼=1 if and only if 𝛽=0; (b)if 𝛼≠1, then 𝛼−𝛽=1 and Ψ=−Φ.

Proof. The proof is as follows. (1) Let 𝑙∈{1,3,…}. In view of Lemma 3 part (a) the identities (15) and (16) are satisfied. (a) Assume that 𝛽=1. Then by (15) Ψ𝛼=0. Hence either Ψ=0 or 𝛼=0. If Ψ=0, then by (16) Φ𝛼=0, and since Φ≠0, we have 𝛼=0.  Assume that 𝛼=0. Then by (15) Φ(1−𝛽)=0. Hence either Φ=0 or 𝛽=1. If Φ=0, then by (16) Ψ(1−𝛽)=0, and since Ψ≠0, we obtain that 𝛽=1. (b) Assume that 𝛽≠1. Then by (15) and (16) Φ²(1−𝛽)=ΦΨ𝛼=Ψ²(1−𝛽). Thus Φ²=Ψ² and so Ψ=−Φ. By (15) Φ(1−𝛽)=−Φ𝛼, and since Φ≠0, then 𝛽−𝛼=1. (2) Let 𝑙∈{2,4,…}. In view of Lemma 3 part (b) the identities (17) and (18) are satisfied. (a) Assume that 𝛼=1. Then by (17) Φ𝛽=0. Hence either Φ=0 or 𝛽=0. If Φ=0, then by (18) Ψ𝛽=0, and since Ψ≠0, we have 𝛽=0.  Conversely, assume that 𝛽=0. Then by (17) Ψ(1−𝛼)=0. Hence either Ψ=0 or 𝛼=1. If Ψ=0, then by (18) Φ(1−𝛼)=0, and since Φ≠0, we obtain that 𝛼=1. (b) Assume that 𝛼≠1. Then by (17) and (18) Φ²(1−𝛼)=ΦΨ𝛽=Ψ²(1−𝛼). Thus Φ²=Ψ² and so Ψ=−Φ. By (17) −Φ(1−𝛼)=Φ𝛽, and since Φ≠0, then 𝛼−𝛽=1.

Theorem 6. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝛼,𝛽 are given by (14). Then (1) has a minimal period-two solution if and only if 𝛽−𝛼=1. (20)

Proof. If (1) has a minimal period-two solution then by Theorem 5 part (1) the necessary condition follows.
Conversely, assume that 𝛽=1. Then 𝛼=0. Choose the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯,𝑥₀≠𝑥₋₁. (21) Then for 𝑛=0 we get 𝑥₁=𝛼𝑥₀+𝛽𝑥₋₁=𝑥₋₁ and for 𝑛=1 we get 𝑥₂=𝛼𝑥₋₁+𝛽𝑥₀=𝑥₀, which shows that {𝑥_𝑛} is a minimal period-two solution.
Now suppose that 𝛽≠1. Choose the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯,𝑥₀=−𝑥₋₁. (22) Then for 𝑛=0 we get 𝑥₁=𝛼𝑥₀+𝛽𝑥₋₁=(𝛽−𝛼)𝑥₋₁=𝑥₋₁ and for 𝑛=1 we get 𝑥₂=𝛼𝑥₋₁+𝛽𝑥₀=(𝛽−𝛼)𝑥₀=𝑥₀, which shows that {𝑥_𝑛} is a minimal period-two solution.

An immediate consequence of Theorem 6 is this result.

Corollary 7. (a) If 𝛽=1 and 𝛼=0, then the minimal period-two solution of (1) is …,Ψ,Φ,….
(b) If 𝛽≠1 and 𝛽−𝛼=1, then the minimal period-two solution of (1) is either …,−Φ,Φ,… or …,−Ψ,Ψ,….

3. The Nonconstant Case

In this section we consider the case when the sums of the even indexed functions 𝑔_𝑖 and the odd indexed functions 𝑔_𝑖 are both nonconstants.

Theorem 8. Suppose that (1) has the linearization (2) with 𝑙=1. Let 𝐴_𝑛=_{{}𝑖∈1−𝑙,…,𝑘;𝑖𝑒𝑣𝑒𝑛}𝑔_𝑖,𝐵_𝑛=_{𝑖∈{1−𝑙,…,𝑘;𝑖𝑜𝑑𝑑}}𝑔_𝑖,𝑛=0,1,…. (23) Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0 if and only if 𝐵_𝑛−𝐴_𝑛=1,𝑛=0,1,….

Proof. The proof follows from the same reasoning as in Example 2.

Note that it is possible for (3) to have a minimal period-two solution other than {𝑥₋₁,−𝑥₋₁},𝑥₋₁≠0, when 𝑎_𝑛≠0 for all 𝑛≥0 (see Example 12). In order to handle the cases not covered by Theorem 8 we establish the following results.

The following simple result will be a useful technical tool.

Lemma 9. Suppose that (1) has the linearization (2) and that 𝐴_𝑛,𝐵_𝑛 are given by (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. Then for 𝑛=0,1,…,(a)if  𝑙∈{1,3,…}, then Φ1−𝐵_2𝑛=Ψ𝐴_2𝑛,Ψ1−𝐵_2𝑛+1=Φ𝐴_2𝑛+1; (24)(25)(b)if  𝑙∈{2,4,…}, then Ψ1−𝐴_2𝑛=Φ𝐵_2𝑛,Φ1−𝐴_2𝑛+1=Ψ𝐵_2𝑛+1. (26)(27)

Proof. Assume that 𝑙 is odd. By plugging Ψ=𝑥_2𝑛,Φ=𝑥_2𝑛−1 for 𝑛≥0 in (2) and setting 𝑛=0,1,… we obtain immediately that 𝑥_𝑙=𝐴₀Ψ+𝐵₀𝑥Φ=Φ,_𝑙+1=𝐴₁Φ+𝐵₁𝑥Ψ=Ψ,_𝑙+2=𝐴₂Ψ+𝐵₂𝑥Φ=Φ,_𝑙+3=𝐴₃Φ+𝐵₃Ψ=Ψ. (28) Now simple induction completes the proof of (24) and (25) and so of part (a). The proof of part (b) is similar.

Theorem 10. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝐴_𝑛,𝐵_𝑛 are given by (23). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0 if and only if (24) and (25) hold.

Proof. The necessary part of the proof follows from part (a) of Lemma 9.
For the proof of sufficient part choose the initial conditions Ψ=𝑥₀=𝑥₋₂=⋯,Φ=𝑥₋₁=𝑥₋₃=⋯, and 𝑥₀≠𝑥₋₁.(1)Assume Φ,Ψ≠0. Setting 𝑛=0,1,… we get that by(24)𝑥₁=𝐴₀Ψ+𝐵₀()𝑥Φ=Φ,by25₂=𝐴₁Φ+𝐵₁()𝑥Ψ=Ψ,by24₃=𝐴₂Ψ+𝐵₂()𝑥Φ=Φ,by25₄=𝐴₃Φ+𝐵₃Ψ=Ψ. (29) By using induction we get that Φ≠Ψ,Ψ=𝑥_2𝑛,andΦ=𝑥_2𝑛−1 for 𝑛≥0. (2)Assume Ψ=0. We obtain from (24) and (25) that 𝐵_2𝑘=1 and 𝐴_2𝑘+1=0 for 𝑘=0,1,…,   which implies that 𝑥_2𝑛=0 and 𝑥_2𝑛+1=Φ,𝑛=0,1,….(3)Assume Φ=0. We obtain from (24) and (25) that 𝐴_2𝑘=0 and 𝐵_2𝑘+1=1 for 𝑘=0,1,…,  which implies that 𝑥_2𝑛+1=0 and 𝑥_2𝑛=Ψ,𝑛=0,1,….

When (1) has been embedded into a higher order equation, the following results can be used to establish either the nonexistence or the necessary conditions for existence of a minimal period-two solution.

Theorem 11. Let 𝑙∈{1,3,…}. Suppose that (1) has the linearization (2) and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ, Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1  for 𝑛≥0.(1) Let either 𝐵_𝑛≠1 or 𝐴_𝑛≠0  for 𝑛≥0; then (a)Φ/Ψ=𝐴₀/(1−𝐵₀)=𝐴_2𝑛/(1−𝐵_2𝑛),𝑛=0,1,…,(b)Ψ/Φ=𝐴₁/(1−𝐵₁)=𝐴_2𝑛+1/(1−𝐵_2𝑛+1),𝑛=0,1,….(2) Let 𝐵_𝑛=1 for 𝑛≥0.(a) If 𝐴_2𝑛=𝐴_2𝑛+1,𝑛=0,1,…,  then 𝐴_2𝑛=𝐴_2𝑛+1=0,𝑛=0,1,….(b) If 𝐴_2𝑛≠𝐴_2𝑛+1,𝑛=0,1,…,  then the following are true: (i)𝐴_2𝑛≠0 if and only if 𝐴_2𝑛+1=0 for 𝑛=0,1,…;(ii)𝐴_2𝑛=0 if and only if 𝐴_2𝑛+1≠0 for 𝑛=0,1,…;(iii) either Ψ=0 or Φ=0. (3) Let 𝐴_𝑛=0 for 𝑛≥0. (a) If 𝐵_2𝑛=𝐵_2𝑛+1,𝑛=0,1,…,  then 𝐵_2𝑛=𝐵_2𝑛+1=1,𝑛=0,1,….(b) If 𝐵_2𝑛≠𝐵_2𝑛+1,𝑛=0,1,…,  then the following are true: (i)𝐵_2𝑛≠1 if and only if 𝐵_2𝑛+1=1 for 𝑛=0,1,…;(ii)𝐵_2𝑛=1 if and only if 𝐵_2𝑛+1≠1 for 𝑛=0,1,…;(iii) either Ψ=0 or Φ=0.

Proof. The proof is as follows.(1) Assume that 𝐵_𝑛≠1 for 𝑛≥0. Then Φ,Ψ≠0. Otherwise, suppose that Ψ=0. Then Φ≠0 and in view of (24) Φ(1−𝐵_2𝑛)=0. This implies 𝐵_2𝑛=1 for 𝑛≥0, which is a contradiction. Now, suppose that Φ=0. Then Ψ≠0 and in view of (25) Ψ(1−𝐵_2𝑛+1)=0. This implies 𝐵_2𝑛+1=1 for 𝑛≥0, which is a contradiction.  Thus Φ,Ψ≠0, which in view of Lemma 9, implies 𝐴_𝑛≠0 for 𝑛≥0. (a) By (24) Φ/Ψ=𝐴₀/(1−𝐵₀)=𝐴_2𝑛/(1−𝐵_2𝑛),𝑛=0,1,…. (b) By (25) Ψ/Φ=𝐴₁/(1−𝐵₁)=𝐴_2𝑛+1/(1−𝐵_2𝑛+1),𝑛=0,1,….  Next, assume that 𝐴_𝑛≠0 for 𝑛≥0. Then Φ,Ψ≠0. Otherwise, suppose that Ψ=0. Then Φ≠0 and in view of (25) Φ𝐴_2𝑛+1=0. This implies 𝐴_2𝑛+1=0 for 𝑛≥0, which is a contradiction. Now, suppose that Φ=0. Then Ψ≠0 and in view of (24) Ψ𝐴_2𝑛=0. This implies 𝐴_2𝑛=0 for 𝑛≥0, which is a contradiction.  Thus Φ,Ψ≠0, which in view of Lemma 9, implies 𝐵_𝑛≠1 for 𝑛≥0.  The rest of the proof is similar to the first part of the proof and will be omitted. (2) In view of (24) and (25) the condition 𝐵_𝑛=1 for 𝑛≥0 implies 𝐴_2𝑛Ψ=𝐴_2𝑛+1Φ=0 for 𝑛=0,1,….(a) Assume that 𝐴_2𝑛=𝐴_2𝑛+1,𝑛=0,1,…. If 𝐴_2𝑛≠0 for some 𝑛≥0, then 𝐴_2𝑛+1≠0 for this 𝑛 and so Φ=Ψ=0, which is a contradiction. Thus 𝐴_2𝑛=0 for 𝑛≥0, and so 𝐴_2𝑛=𝐴_2𝑛+1=0,𝑛=0,1,….(b) Assume that 𝐴_2𝑛≠𝐴_2𝑛+1 for 𝑛≥0.(i) If 𝐴_2𝑛+1=0 for 𝑛≥0, then 𝐴_2𝑛≠0 for 𝑛≥0. Now suppose that 𝐴_2𝑛≠0 for 𝑛≥0; then from 𝐴_2𝑛Ψ=𝐴_2𝑛+1Φ=0 we get that Ψ=Φ𝐴_2𝑛+1/𝐴_2𝑛=0 for 𝑛≥0.  Since Ψ=0, then Φ≠0. Thus 𝐴_2𝑛+1=0 for 𝑛≥0. (ii) The proof is similar to part (i) and will be omitted. (iii) By (24) 𝐴_2𝑛Ψ=0 for 𝑛≥0. Now suppose that 𝐴_2𝑛≠0 for 𝑛≥0, then Ψ=0.  If 𝐴_2𝑛=0 for 𝑛≥0; then 𝐴_2𝑛+1≠0 for 𝑛≥0. In view of (25) 𝐴_2𝑛+1Φ=0 for 𝑛≥0 and so Φ=0. (3) In view of (24) and (25) condition 𝐴_𝑛=0 for 𝑛≥0 implies (1−𝐵_2𝑛)Φ=(1−𝐵_2𝑛+1)Ψ=0 for 𝑛=0,1,…. (a) Assume that 𝐵_2𝑛=𝐵_2𝑛+1,𝑛=0,1,…. If 𝐵_2𝑛≠1 for some 𝑛≥0, then 𝐵_2𝑛+1≠1 for this 𝑛 and so Φ=Ψ=0, which is a contradiction. Thus 𝐵_2𝑛=1 for 𝑛=0,1,… and consequently 𝐵_2𝑛=𝐵_2𝑛+1=1,𝑛=0,1,…. (b) Assume that 𝐵_2𝑛≠𝐵_2𝑛+1 for 𝑛≥0. (i) If 𝐵_2𝑛+1=1 for 𝑛≥0, then 𝐵_2𝑛≠1 for 𝑛≥0. Now suppose that 𝐵_2𝑛≠1 for 𝑛≥0; then from (1−𝐵_2𝑛)Φ=(1−𝐵_2𝑛+1)Ψ=0 we get that Φ=Ψ(1−𝐵_2𝑛+1)/(1−𝐵_2𝑛)=0 for 𝑛≥0. Since Φ=0, then Ψ≠0. Thus 𝐵_2𝑛+1=1 for 𝑛≥0. (ii) The proof is similar to part (i) and will be omitted. (iii) By (24) (1−𝐵_2𝑛)Φ=0 for 𝑛≥0. Now suppose that 𝐵_2𝑛≠1 for 𝑛≥0, then Φ=0. If 𝐵_2𝑛=1 for 𝑛≥0, then 𝐵_2𝑛+1≠1 for 𝑛≥0. In view of (25) Ψ(1−𝐵_2𝑛+1)=0 for 𝑛≥0 and so Ψ=0.(i) If 𝐵_2𝑛+1=0 for 𝑛≥0; then 𝐵_2𝑛≠0 for 𝑛≥0. Now suppose that 𝐵_2𝑛≠0 for 𝑛≥0; then from 𝐵_2𝑛Φ=𝐵_2𝑛+1Ψ=0 we get that Φ=Ψ𝐵_2𝑛+1/𝐵_2𝑛=0 for 𝑛≥0. Since Φ=0, then Ψ≠0. Thus 𝐵_2𝑛+1=0 for 𝑛≥0.(ii) The proof is similar to the proof of part (i).(iii) By (26) 𝐵_2𝑛Φ=0 for 𝑛≥0. If 𝐵_2𝑛≠0 for 𝑛≥0, then Φ=0. Now if 𝐵_2𝑛=0 for 𝑛≥0, then 𝐵_2𝑛+1≠0 for 𝑛≥0. In view of (27) Ψ𝐵_2𝑛+1=0 for 𝑛≥0 and so Ψ=0.

Example 12. The difference equation 𝑥_𝑛+1=𝑐_𝑛𝑥𝑛+1^𝑛+11−𝑥𝑛+1^𝑛−1,𝑛=0,1,…, (30) where {𝑐_𝑛} is a period-two sequence such that 𝑐_2𝑛=𝛾≠0,1, 𝑐_2𝑛+1=1/𝛾, has an infinite number of period-two solutions of the form {𝑥₋₁,𝑥₋₁/𝛾},𝑥₋₁≠0, which can be seen by immediate checking.
This equation is an illustration of Theorem 11 part 1. In this case 𝐴_𝑛={𝛾,1/2𝛾,𝛾/3,1/4𝛾,𝛾/5,…} and 𝐵_𝑛={0,1/2,2/3,3/4,4/5,…}.

Theorem 13. Let 𝑙∈{2,4,…}. Suppose that (1) has the linearization (2) and 𝐴_𝑛,𝐵_𝑛 are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,  Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (1) Let either 𝐴_𝑛≠1 or 𝐵_𝑛≠0 for 𝑛≥0; then (a)Ψ/Φ=𝐵₀/(1−𝐴₀)=𝐵_2𝑛/(1−𝐴_2𝑛),𝑛=0,1,…,(b)Φ/Ψ=𝐵₁/(1−𝐴₁)=𝐵_2𝑛+1/(1−𝐴_2𝑛+1),𝑛=0,1,….(2) Let 𝐴_𝑛=1 for 𝑛≥0. (a) If 𝐵_2𝑛=𝐵_2𝑛+1,𝑛=0,1,…,  then 𝐵_2𝑛=𝐵_2𝑛+1=0,𝑛≥0. (b) If 𝐵_2𝑛≠𝐵_2𝑛+1,𝑛=0,1,…, then the following are true: (i)𝐵_2𝑛≠0 if and only if 𝐵_2𝑛+1=0 for 𝑛=0,1,…;(ii)𝐵_2𝑛=0 if and only if 𝐵_2𝑛+1≠0 for 𝑛=0,1,…;(iii) either Ψ=0 or Φ=0. (3) Let 𝐵_𝑛=0 for 𝑛≥0. (a) If 𝐴_2𝑛=𝐴_2𝑛+1,𝑛=0,1,…, then 𝐴_2𝑛=𝐴_2𝑛+1=1,𝑛≥0. (b) If 𝐴_2𝑛≠𝐴_2𝑛+1,𝑛=0,1,…, then the following are true: (i)𝐴_2𝑛≠1 if and only if 𝐴_2𝑛+1=1 for 𝑛=0,1,…;(ii)𝐴_2𝑛=1 if and only if 𝐴_2𝑛+1≠1 for 𝑛=0,1,…;(iii)either Ψ=0 or Φ=0.

Proof. The proof is as follows. (1) Assume that 𝐴_𝑛≠1 for 𝑛≥0. Then ΦΨ≠0. Otherwise, if Φ=0 then Ψ≠0. By (26) Ψ(1−𝐴_2𝑛)=0 and so 𝐴_2𝑛=1 for 𝑛≥0, which is a contradiction. In the other case, if Ψ=0 then Φ≠0. By (27) Φ(1−𝐴_2𝑛+1)=0 and so 𝐴_2𝑛+1=1 for 𝑛≥0, which is a contradiction. Hence 𝐵_𝑛≠0 for 𝑛≥0.(a) By (26) Ψ/Φ=𝐵₀/(1−𝐴₀)=𝐵_2𝑛/(1−𝐴_2𝑛) for 𝑛≥0. (b) By (27) Φ/Ψ=𝐵₁/(1−𝐴₁)=𝐵_2𝑛+1/(1−𝐴_2𝑛+1) for 𝑛≥0.  Now assume that 𝐵_𝑛≠0 for 𝑛≥0. Then ΦΨ≠0. Otherwise, if Ψ=0 then Φ≠0. By (26) Φ𝐵_2𝑛=0 and so 𝐵_2𝑛=0 for 𝑛≥0, which is a contradiction. In the other case, if Φ=0 then Ψ≠0. By (27) Ψ𝐵_2𝑛+1=0 and so 𝐵_2𝑛+1=0 for 𝑛≥0, which is a contradiction. Hence 𝐴_𝑛≠1 for 𝑛≥0 and the proof follows similarly to the previous part.(2) In view of (26) and (27) condition 𝐴_𝑛=1 for 𝑛≥0 implies 𝐵_2𝑛Φ=𝐵_2𝑛+1Ψ=0 for 𝑛=0,1,…. (a) Assume that 𝐵_2𝑛=𝐵_2𝑛+1,𝑛=0,1,….  If 𝐵_2𝑛≠0 for some 𝑛≥0, then 𝐵_2𝑛+1≠0 for this 𝑛 and so Φ=Ψ=0, which is a contradiction. Hence 𝐵_2𝑛=0 for 𝑛≥0. Thus 𝐵_2𝑛=𝐵_2𝑛+1=0,𝑛=0,1,…. (b) Assume that 𝐵_2𝑛≠𝐵_2𝑛+1 for 𝑛≥0. (i) If 𝐵_2𝑛+1=0 for 𝑛≥0; then 𝐵_2𝑛≠0 for 𝑛≥0. Now suppose that 𝐵_2𝑛≠0 for 𝑛≥0; then from 𝐵_2𝑛Φ=𝐵_2𝑛+1Ψ=0 we get that Φ=Ψ𝐵_2𝑛+1/𝐵_2𝑛=0 for 𝑛≥0. Since Φ=0, then Ψ≠0. Thus 𝐵_2𝑛+1=0 for 𝑛≥0.(ii) The proof is similar to the proof of part (i).(iii) By (26) 𝐵_2𝑛Φ=0 for 𝑛≥0. If 𝐵_2𝑛≠0 for 𝑛≥0, then Φ=0. Now if 𝐵_2𝑛=0 for 𝑛≥0, then 𝐵_2𝑛+1≠0 for 𝑛≥0. In view of (27) Ψ𝐵_2𝑛+1=0 for 𝑛≥0 and so Ψ=0. (3) In view of (26) and (27) condition 𝐵_𝑛=0 for 𝑛≥0 implies (1−𝐴_2𝑛)Ψ=(1−𝐴_2𝑛+1)Φ=0 for 𝑛=0,1,…. (a) Assume that 𝐴_2𝑛=𝐴_2𝑛+1,𝑛=0,1,….  If 𝐴_2𝑛≠1 for some 𝑛≥0, then 𝐴_2𝑛+1≠1 for this 𝑛 and so Φ=Ψ=0, which is a contradiction. Hence 𝐴_2𝑛=1 for 𝑛≥0. Thus 𝐴_2𝑛=𝐴_2𝑛+1=1,𝑛=0,1,….(b) Assume that 𝐴_2𝑛≠𝐴_2𝑛+1 for 𝑛≥0. (i) If 𝐴_2𝑛+1=1 for 𝑛≥0, then 𝐴_2𝑛≠1 for 𝑛≥0. Now assume that 𝐴_2𝑛≠1 for 𝑛≥0; then from (1−𝐴_2𝑛)Ψ=(1−𝐴_2𝑛+1)Φ=0 we get that Ψ=Φ(1−𝐴_2𝑛+1)/(1−𝐴_2𝑛)=0 for 𝑛≥0. Since Ψ=0, then Φ≠0. Thus 𝐴_2𝑛+1=1 for 𝑛≥0.(ii) The proof is similar to the proof of part (i).(iii) By (26) (1−𝐴_2𝑛)Ψ=0 for 𝑛≥0. If 𝐴_2𝑛≠1 for 𝑛≥0, then Ψ=0. Now if 𝐴_2𝑛=1 for 𝑛≥0, then 𝐴_2𝑛+1≠1 for 𝑛≥0. In view of (27) (1−𝐴_2𝑛+1)Φ=0 for 𝑛≥0 and so Φ=0.

Corollary 14. Suppose that (1) has the linearization (2) and that 𝐴_𝑛,𝐵_𝑛 are given by (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (1)Let 𝑙∈{1,3,…}. (a)𝐴_𝑛≠0 if and only if 𝐵_𝑛≠1,𝑛=0,1,….(b)If  𝐴₀/(1−𝐵₀)=𝐴₁/(1−𝐵₁) and either 𝐵_𝑛≠1 or 𝐴_𝑛≠0 for 𝑛=0,1,…,  then (i)𝐵_2𝑛−𝐴_2𝑛=1 and 𝐵_2𝑛+1−𝐴_2𝑛+1=1,𝑛=0,1,…;(ii)Ψ=−Φ. (2)Let 𝑙∈{2,4,…}. (a)𝐴_𝑛≠1 if and only if 𝐵_𝑛≠0,𝑛=0,1,….(b)If  𝐵₀/(1−𝐴₀)=𝐵₁/(1−𝐴₁) and either 𝐴_𝑛≠1 or 𝐵_𝑛≠0 for 𝑛=0,1,…,  then (i)𝐴_2𝑛−𝐵_2𝑛=1 and 𝐴_2𝑛+1−𝐵_2𝑛+1=1,𝑛=0,1,…;(ii)Ψ=−Φ.

Proof. Part (1a) follows from the proof of Theorem 11 part (1). Part (1b) follows from the proof of Theorem 11 part (1) and 𝐴₀/(1−𝐵₀)=𝐴₁/(1−𝐵₁), which implies Φ/Ψ=Ψ/Φ. Hence Φ²=Ψ² and so Φ=−Ψ. Thus −1=𝐴_2𝑛/(1−𝐵_2𝑛)=𝐴_2𝑛+1/(1−𝐵_2𝑛+1),𝑛≥0 which implies 𝐵_2𝑛−𝐴_2𝑛=1 and 𝐵_2𝑛+1−𝐴_2𝑛+1=1 for 𝑛≥0.
Part (2a) follows from the proof of Theorem 13 part (1) and the proof of part (2b) follows in a similar way as the proof of part (1b).

The next result gives a necessary and sufficient condition for the existence of a minimal period-two solution in the special case when 𝑙=1 in (2).

Theorem 15. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Then (1) has a minimal period-two solution if and only if one of the following holds. (1)All the conditions of Theorem 11 part (1) are satisfied. Furthermore the period-two solution is …,𝐴₀𝑥₀/(1−𝐵₀),𝐴₁𝑥₋₁/(1−𝐵₁),….(2)𝐵_𝑛=1 for 𝑛≥0 and one of the following conditions is satisfied for 𝑛≥0: (a)𝐴_2𝑛=𝐴_2𝑛+1=0,(b)𝐴_2𝑛=0 and 𝐴_2𝑛+1≠0,(c)𝐴_2𝑛≠0 and 𝐴_2𝑛+1=0. (3)𝐴_𝑛=0 for 𝑛≥0 and one of the following conditions is satisfied for 𝑛≥0: (a)𝐵_2𝑛=𝐵_2𝑛+1=1,(b)𝐵_2𝑛=1 and 𝐵_2𝑛+1≠1,(c)𝐵_2𝑛≠1 and 𝐵_2𝑛+1=1.

Proof. The necessary part follows from Theorem 11. We will now prove the sufficient part.(1)Assume that conditions (a) and (b) of Theorem 11 part (1) are satisfied. Then by (a) 𝑥₀=(1−𝐵₀)𝑥₋₁/𝐴₀=(1−𝐵_2𝑛)𝑥₋₁/𝐴_2𝑛 for 𝑛≥0 and by (b) 𝑥₋₁=(1−𝐵₁)𝑥₀/ 𝐴₁=(1−𝐵_2𝑛+1)𝑥₀/𝐴_2𝑛+1 for 𝑛≥0. Choose the initial conditions 𝑥₀=𝑥₋₂=⋯,  𝑥₋₁=𝑥₋₃=⋯, and 𝑥₀≠𝑥₋₁. Then by using these equalities we get 𝑥₁=𝐴₀𝑥₀+𝐵₀𝑥₋₁=1−𝐵₀𝑥₋₁+𝐵₀𝑥₋₁=𝑥₋₁,𝑥₂=𝐴₁𝑥₋₁+𝐵₁𝑥₀=1−𝐵₁𝑥₀+𝐵₁𝑥₀=𝑥₀,𝑥₃=𝐴₂𝑥₀+𝐵₂𝑥₋₁=1−𝐵₂𝑥₋₁+𝐵₂𝑥₋₁=𝑥₋₁,𝑥₄=𝐴₃𝑥₋₁+𝐵₃𝑥₀=1−𝐵₃𝑥₀+𝐵₃𝑥₀=𝑥₀. (31) Simple induction completes the proof.(2)Next, suppose that 𝐵_𝑛=1 for 𝑛≥0.(a)Assume that 𝐴_2𝑛=𝐴_2𝑛+1=0 for 𝑛≥0. By choosing the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯, and 𝑥₀≠𝑥₋₁, an immediate calculation shows that {𝑥₋₁,𝑥₀} is a minimal period-two solution.(b)Assume that 𝐴_2𝑛=0 and 𝐴_2𝑛+1≠0 for 𝑛≥0. By choosing the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯=0, and 𝑥₀≠0, we obtain 𝑥₁=𝐴₀𝑥₀+𝐵₀𝑥₋₁𝑥=0,₂=𝐴₁𝑥₁+𝐵₁𝑥₀=𝑥₀,𝑥₃=𝐴₂𝑥₂+𝐵₂𝑥₁𝑥=0,₄=𝐴₃𝑥₃+𝐵₃𝑥₂=𝑥₀, (32) and straightforward induction shows that {0,𝑥₀} is the minimal period-two solution.(c)Assume that 𝐴_2𝑛≠0 and 𝐴_2𝑛+1=0 for 𝑛≥0. Choose the initial conditions 𝑥₀=𝑥₋₂=⋯=0,𝑥₋₁=𝑥₋₃=⋯, and 𝑥₋₁≠0. By straightforward induction we obtain that {𝑥₋₁,0} is the minimal period-two solution. (3)Next, suppose that 𝐴_𝑛=0 for 𝑛≥0.(a)Assume that 𝐵_2𝑛=𝐵_2𝑛+1=1 for 𝑛≥0. By choosing the initial conditions 𝑥₀ = 𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯, and 𝑥₀≠𝑥₋₁, an immediate calculation shows that {𝑥₋₁,𝑥₀} is the minimal period-two solution.(b)Assume that 𝐵_2𝑛=1 and 𝐵_2𝑛+1≠1 for 𝑛≥0. By choosing the initial conditions 𝑥₀=𝑥₋₂=⋯=0,𝑥₋₁=𝑥₋₃=⋯, and 𝑥₋₁≠0, we obtain 𝑥₁=𝐴₀𝑥₀+𝐵₀𝑥₋₁=𝑥₋₁,𝑥₂=𝐴₁𝑥₁+𝐵₁𝑥₀𝑥=0,₃=𝐴₂𝑥₂+𝐵₂𝑥₁=𝑥₋₁,𝑥₄=𝐴₃𝑥₃+𝐵₃𝑥₂=0, (33) and straightforward induction shows that {𝑥₋₁,0} is the minimal period-two solution. (c)Assume that 𝐵_2𝑛≠1 and 𝐵_2𝑛+1=1 for 𝑛≥0. By choosing the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯=0, and 𝑥₀≠0, and using a straightforward induction we obtain that {0,𝑥₀} is the minimal period-two solution.

Example 16. The difference equation 𝑥_𝑛+1=𝐵𝑥_𝑛𝑥_𝑛−1𝑑𝑥_𝑛+𝑒𝑥_𝑛−1,𝑛=0,1,…, (34) where 𝐵,𝑑,𝑒>0, has a minimal period-two solutions of the form {𝑥₋₁,−𝑥₋₁},𝑥₋₁≠0, if and only if 𝐵=𝑑−𝑒 and 𝐵≠𝑑.
This equation is an illustration of Theorem 15. The linearization of (34) gives 𝑥_𝑛+1=𝑔₁𝑥_𝑛−1,𝑛≥0, (35) where 𝑔₁=𝐵𝑥_𝑛/(𝑑𝑥_𝑛+𝑒𝑥_𝑛−1),𝑛≥0. Since 𝐴_𝑛=0 for all 𝑛≥0, then from Theorem 15 part (3a) a period-two solution exists if and only if 𝑔₁=𝐵_𝑛=𝐵𝑥_𝑛𝑑𝑥_𝑛+𝑒𝑥_𝑛−1=1,∀𝑛=0,1,…. (36) Observe that if 𝐵=𝑑, then 𝑒𝑥_𝑛−1=0 for 𝑛≥0, which is a contradiction.
Equation (36) gives the first order equation 𝑥_𝑛=𝑒𝑥𝐵−𝑑^𝑛−1,𝑛=0,1,… (37) which has a period-two solution if and only if 𝐵=𝑑−𝑒 and 𝑥₀=−𝑥₋₁ as Theorem 15 part (1) is satisfied.
Therefore, (34) has a period-two solution of the form {𝑥₋₁,−𝑥₋₁},𝑥₋₁≠0 if and only if 𝐵=𝑑−𝑒,𝐵≠𝑑.
This example can be extended to a more general equation of the form 𝑥_𝑛+1=𝐵𝑥_𝑛𝑥_𝑛−1𝑥_𝑛−2𝑎₀𝑥_𝑛−1𝑥_𝑛−2+𝑎₁𝑥_𝑛𝑥_𝑛−2+𝑎₂𝑥_𝑛𝑥_𝑛−1,𝑛=0,1,…, (38) where 𝐵,𝑎_𝑖>0,𝑖=0,1,2. Similar reasoning gives the necessary and sufficient conditions for the existence of a period-two solution to be 𝐵=𝑎₁−𝑎₀+𝑎₂,𝐵≠𝑎₁. (39) In this case there is an infinite number of period-two solutions of the form {𝑥₋₁,−𝑥₋₁}, 𝑥₋₁≠0.

Corollary 17. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Assume that 𝐴₀/(1−𝐵₀)=𝐴₁/(1−𝐵₁) and either 𝐵_𝑛≠1 or 𝐴_𝑛≠0 for 𝑛≥0. Then (1) has a minimal period-two solution if and only if 𝐵_2𝑛+1−𝐴_2𝑛+1=1𝑎𝑛𝑑𝐵_2𝑛−𝐴_2𝑛=1 for 𝑛≥0.

Proof. The necessary part follows from Corollary 14 part (1b).
For the proof of sufficient part choose the initial conditions 𝑥₀=𝑥₋₂=⋯,𝑥₋₁=𝑥₋₃=⋯, and 𝑥₀=−𝑥₋₁. Setting 𝑛=0,1,… we obtain 𝑥₁=𝐴₀𝑥₀+𝐵₀𝑥₋₁=−𝐴₀𝑥₋₁+𝐵₀𝑥₋₁=𝑥₋₁,𝑥₂=𝐴₁𝑥₁+𝐵₁𝑥₀=𝐴₁𝑥₋₁−𝐵₁𝑥₋₁=−𝑥₋₁,𝑥₃=𝐴₂𝑥₂+𝐵₂𝑥₁=−𝐴₂𝑥₋₁+𝐵₂𝑥₋₁=𝑥₋₁,𝑥₄=𝐴₃𝑥₃+𝐵₃𝑥₂=𝐴₃𝑥₋₁−𝐵₃𝑥₋₁=−𝑥₋₁. (40) Simple induction completes the proof.

So far we have considered the cases when 𝑙∈{1,3,…} and 𝐵_𝑛 is either equal or not equal to one for all 𝑛≥0. But what happens when 𝑙∈{1,3,…} and 𝐵_𝑛=1 for some 𝑛’s and 𝐵_𝑛≠1 for other 𝑛’s or when 𝐴_𝑛=0 for some 𝑛’s and 𝐴_𝑛≠0 for other 𝑛’s? We will now investigate these cases along with the cases when 𝑙∈{2,4,…} and 𝐵_𝑛=0 for some 𝑛’s and 𝐵_𝑛≠0 for other 𝑛’s or when 𝐴_𝑛=1 for some 𝑛’s and 𝐴_𝑛≠1 for other 𝑛’s.

Theorem 18. Let 𝑙∈{1,3,…}. Suppose that (1) has the linearization (2) and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (1)Let ΦΨ≠0. Then(a)for some 𝑁∈{0,1,…}𝐵_𝑁=1 if and only if 𝐴_𝑁=0;(b)if some 𝑁 are even and 𝐵_𝑁≠1, then 𝐴_𝑁=(Φ/Ψ)(1−𝐵_𝑁);(c)if some 𝑁 are odd and 𝐵_𝑁≠1, then 𝐴_𝑁=(Ψ/Φ)(1−𝐵_𝑁).(2)If  Ψ=0, then for all 𝑛≥0𝐵_2𝑛=1 and 𝐴_2𝑛+1=0.(3)If  Φ=0, then for all 𝑛≥0𝐵_2𝑛+1=1 and 𝐴_2𝑛=0.

Proof. The proof is as follows.(1)The result follows from Lemma 9 part (a).(2)Since Ψ=0, then Φ≠0 and the result follows from Lemma 9 part (a).(3)Since Φ=0, then Ψ≠0 and the result follows from Lemma 9 part (a).

Remark 19. Note that in part (2) of Theorem 18  𝐵_2𝑛+1∈ℝ and 𝐴_2𝑛∈ℝ for all 𝑛≥0. Similarly in part (3) of Theorem 18  𝐵_2𝑛∈ℝ and 𝐴_2𝑛+1∈ℝ for all 𝑛≥0.

Theorem 20. Let 𝑙∈{2,4,…}. Suppose that (1) has the linearization (2) and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Assume that (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0. (1)Let ΦΨ≠0. Then(a)for some 𝑁∈{0,1,…}𝐴_𝑁=1 if and only if 𝐵_𝑁=0;(b)if some 𝑁 are even and 𝐴_𝑁≠1, then 𝐵_𝑁=(Ψ/Φ)(1−𝐴_𝑁);(c)if some 𝑁 are odd and 𝐴_𝑁≠1, then 𝐵_𝑁=(Φ/Ψ)(1−𝐴_𝑁). (2)If  Ψ=0, then for all 𝑛≥0𝐵_2𝑛=0 and 𝐴_2𝑛+1=1.(3)If  Φ=0, then for all 𝑛≥0𝐵_2𝑛+1=0 and 𝐴_2𝑛=1.

Proof. The results follow from Lemma 9 part (b).

Theorem 21. Suppose that (1) has the linearization (2) with 𝑙=1 and that 𝐴_𝑛,𝐵_𝑛 are as in (23). Then (1) has a minimal period-two solution …,Φ,Ψ,…, where Φ≠Ψ,Ψ=𝑥_2𝑛,𝑎𝑛𝑑Φ=𝑥_2𝑛−1 for 𝑛≥0 if and only if one of the following holds: (1)ΦΨ≠0 and for each 𝑛≥0 either (a)𝐵_𝑛=1 and 𝐴_𝑛=0;(b)𝑛 is even, 𝐵_𝑛≠1, and 𝐴_𝑛=(Φ/Ψ)(1−𝐵_𝑛);(c)𝑛 is odd, 𝐵_𝑛≠1, and 𝐴_𝑛=(Ψ/Φ)(1−𝐵_𝑛). (2) If  Ψ=0, then for all 𝑛≥0𝐵_2𝑛=1 and 𝐴_2𝑛+1=0.(3) If  Φ=0, then for all 𝑛≥0𝐵_2𝑛+1=1 and 𝐴_2𝑛=0.

Proof. The necessary part follows from Theorem 18. The sufficient part follows by direct calculation of the period-two solution which satisfies specific initial conditions in a similar way as in the proof of Theorem 15.

An illustration of Theorem 20 is the following example.

Example 22. The difference equation 𝑥_𝑛+1=sin𝑥_𝑛−1𝑥𝑛+1^𝑛+𝑥_𝑛−1𝑒^𝑥𝑛,𝑛=0,1,…, (41) has an infinite number of period-two solutions of the form {0,𝑥₀},𝑥₀≠0, which can be seen by immediate checking. In view of Theorem 20 part (3)  𝐴_2𝑛=sin(𝑥_2𝑛−1)/(2𝑛+1) and 𝐵_2𝑛+1=𝑒^𝑥2𝑛+1=1 for 𝑛=0,1,… when Φ=0.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.