# A Group Theory Approach towards Some Rational Difference Equations

Authors: Mensah Folly-Gbetoula; Nkosingiphile Mnguni; A. H. Kara
Journal: Journal of Mathematics
 (2019)

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## Abstract

A full Lie point symmetry analysis of rational difference equations is performed. Nontrivial symmetries are derived, and exact solutions using these symmetries are obtained.


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## Body

## 1. Introduction

Over a century ago, symmetries became a centre of interest of several authors after the work of Sophus Lie [1] on differential equations. He studied the continuous group of transformations that leaves the differential equations invariant. This concept of symmetries is strongly related to the existence of conservation laws, and the relationship between them has attracted great interest among researchers following the work of Noether [2]. The extension of this idea to difference equations is now well documented (see [3] and references herein). In [3], Hydon developed a symmetry-based algorithm enabling one to derive solutions of difference equations without making any special lucky guesses. Hydon emphasized on second-order difference equations, although his algorithm is valid for any order. When it comes to higher-order equations, computations can be cumbersome and extra ansatz may be needed to ease the calculations.We aim to extend the work by Elsayed [4] where the author investigated the dynamics and solutions of(1)xn+1=xn−5±1±xn−1xn−3xn−5,where the initial conditions x−5,x−4,x−3,x−2,x−1, and x0 are arbitrary nonzero real numbers. For related work, see [5, 6]. One can notice that equation (1) is just a special case of a more general form(2)xn+1=xn−5an+bnxn−1xn−3xn−5,where ann∈ℕ0 and bnn∈ℕ0 are arbitrary sequences. We will use a symmetry-based method to solve (2). Equivalently, we study(3)un+6=unAn+Bnunun+2un+4,instead, where Ann∈ℕ0 and Bnn∈ℕ0 are arbitrary sequences. This means we can only compare xi with ui+5. Furthermore, we use the same technique to obtain exact solutions of(4)xn+1=xn−4xn−3xn−2xn−1xnλ+μxn−4xn−3xn−2,where λ,μ∈ℝ, and again we study(5)un+5=unun+1un+2un+3un+4λ+μunun+1un+2,instead. Note that solutions of (4) are found in [7]; however, their method is completely different from ours.


## 2. Definitions and Notation

The definitions are taken from Hydon [3], and most of the notation follows from the same book.Definition 1.
A parameterized set of point transformations,(6)Γε:x⟼x^x;ε,where x=xi and i=1,…,p are continuous variables, is a one-parameter local Lie group of transformations if the following conditions are satisfied:(1)
Γ0 is the identity map if x^=x when ε=0(2)
ΓaΓb=Γa+b for every a and b sufficiently close to 0(3)
Eachx^i can be represented as a Taylor series (in a neighborhood of ε=0 that is determined by x), and therefore(7)x^ix;ε=xi+εξix+Oε2,i=1,…,p.
Consider thepth-order difference equation(8)un+p=Ωn,un,…,un+p−1,for some function Ω. Assume the point transformations are of the form(9)n^=n,u^n=un+εQn,un+Oε2,with the corresponding infinitesimal symmetry generator(10)X=Qn,un∂∂un+SQn,un∂∂un+1+⋯+Sp−1Qn,un∂∂un+p−1,where S is the shift operator, i.e., S:n⟼n+1. The symmetry condition is defined as(11)u^n+p=Ωn,u^n,u^n+1,…,u^n+p−1,whenever (8) is true. Substituting the Lie point symmetries (9) into the symmetry condition (11) leads to the linearized symmetry condition:(12)SpQ−XΩ=0,whenever (8) holds.Definition 2.
V
n is invariant under the Lie group of transformations (9) if and only if XVn=0.
We define the functions(13)Θnθj,s,l=∏j=slθ2j+n,n=0,1,and we adopt the standard conventions(14)∏j=slθj=1,whens>l,∑j=nl0θj=0,whenn>l0.
We refer the reader to [3] for a deeper understanding of the concept of symmetry analysis of difference equations.


## 3. Main Results

### 3.1. On the Difference Equation (3)

Consider the sixth-order difference equations of the form (3), that is,(15)un+6=unAn+Bnunun+2un+4.We impose the symmetry condition (12) and simplify the resulting equation to get(16)S6Q+Bnun2un+2S4Q+Bnun2un+4S2Q−AnQBnunun+2un+4+An2=0.To solve forQ, we first differentiate (16) with respect to un (keeping Ω fixed and viewing un+2 as a function of un,un+4, and Ω). This leads, after simplification, to(17)S4Q+un+4S2Q′−un+4Q′+2un+4unQ=0,where ′ denotes the derivative with respect to the independent variable. We then differentiate (17) with respect to un to get(18)Q″n,un−2unQ′n,un+2un2Qn,un=0.The solutions of (18) are given by(19)Qn,un=αnun2+βnun,for some functions αn and βn of n. To obtain more information on αn and βn, we substitute (19) in (16) and we split the resulting equation to get the following:(20a)unun+2un+42:Bnαn+4=0,(20b)unun+22un+4:Bnαn+2=0,(20c)unun+2un+4:Bnβn+2+βn+4+βn+6=0,(20d)un:−Anαn+αn+6=0,(20e)1:Anβn−βn+6=0.These equations ((20a)–(20e)) reduce to(21)αn=0,βn+4+βn+2+βn=0.The expression ofβn in (21) is merely obtained by solving the corresponding characteristic equation r4+r2+1=0 for r. Assuming that ri,i=1,2,3,4, are the solutions of this characteristic equation, then βn is a linear combination of the rin’s. In other words, the solutions of (21) are(22)αn=0,βn=−1nβnc1+−1nβ¯nc2+β¯nc3+βnc4,for some arbitrary constants ci,i=1,…,4, where β=expπi/3. Thus, we obtain four characteristics given by(23)Q1=−1nβnun,Q2=−1nβ¯nun,Q3=β¯nun,Q4=βnun.The four corresponding symmetry generatorsX1,X2,X3, and X4 are given by(24a)X1=−1nβnun∂un−−1nβn+1un+1∂un+1+−1nβn+2un+2∂un+2−−1nβn+3un+3∂un+3+−1nβn+4un+4∂un+4−−1nβn+5un+5∂un+5,(24b)X2=−1nβ¯nun∂un−−1nβ¯n+1un+1∂un+1+−1nβ¯n+2un+2∂un+2−−1nβ¯n+3un+3∂un+3+−1nβ¯n+4un+4∂un+4−−1nβ¯n+5un+5∂un+5,(24c)X3=β¯nun∂un+β¯n+1un+1∂un+1+β¯n+2un+2∂un+2+β¯n+3un+3∂un+3+β¯n+4un+4∂un+4+β¯n+5un+5∂un+5,(24d)X4=βnun∂un+βn+1un+1∂un+1+βn+2un+2∂un+2+βn+3un+3∂un+3+βn+4un+4∂un+4+βn+5un+5∂un+5.Here, usingQ4, we introduce the canonical coordinate [8](25)sn=∫dunβnun=1βnlnun.In view of (21), we introduce the variable(26)rn=βn+4sn+4+βn+2sn+2+βnsn.It is easy to check that(27)Xirn=βn+4+βn+2+βn=0,i=1,…,4.Therefore,rn is invariant under X4. It is advantageous to use(28)r˜n=exp−rn,that is, r˜n=±1/unun+2un+4. Here, we choose to use the plus sign and we show, using (3), that(29)r˜n+2=Anr˜n+Bn.Therefore,(30)r˜2n+k=r˜k∏k1=0n−1A2k1+k+∑l=0n−1B2l+k∏k2=l+1n−1A2k2+k,k=0,1.It is worthwhile to mention that equation (30) gives the solution of (29) for all n. By reversing all the change of variables, we have(31)un=exp−βnc5+βnc6+β¯nc7+−β¯nc8+−βn+1∑k1=0n−1−i36−β¯k1rk1+βn+1∑k2=0n−1i36β¯k2rk2+β¯n+1∑k3=0n−1−i36βk3rk3+−β¯n+1∑k4=0n−1i36−βk4rk4=Γnexp∑k=0n−133−1n+k+1Imγn+1,klnr˜k,where r˜k is given in (30), Γn=exp−βnc5+βnc6+β¯nc7+−β¯nc8, and γl,k=βlβ¯k.Note.
Equation (31) gives the solution of (3) in a unified manner.
We can simplify (31) further by splitting it into six categories. We have(32)u6n=Γ6nexp∑k=06n−133−16n+k+1Imγ6n+1,klnr˜k,u6n=u0∏s=0n−1r˜6sr˜6s+2.
Similarly, after a straightforward but lengthy computation, we get(33)u6n+i=ui∏s=0n−1r˜6s+ir˜6s+2+i,i=0,…,5.Note.
We can obtain (33) using (28) without considering the absolute value function.
Now, using (30) and (33), we obtain the solutions of (3) as follows:(34a)u6n=u0∏s=0n−1∏k1=03s−1A2k1+u0u2u4∑l=03s−1B2l∏k2=l+13s−1A2k2∏k1=03sA2k1+u0u2u4∑l=03sB2l∏k2=l+13sA2k2,(34b)u6n+1=u1∏k1=03s−1A2k1+1+u1u3u5∑l=03s−1B2l+1∏k2=l+13s−1A2k2+1∏k1=03sA2k1+1+u1u3u5∑l=03sB2l+1∏k2=l+13sA2k2+1,(34c)u6n+2=u2∏s=0n−1∏k1=03sA2k1+u0u2u4∑l=03sB2l∏k2=l+13sA2k2∏k1=03s+1A2k1+u0u2u4∑l=03s+1B2l∏k2=l+13s+1A2k2,(34d)u6n+3=u3∏s=0n−1∏k1=03sA2k1+1+u1u3u5∑l=03sB2l+1∏k2=l+13sA2k2+1∏k1=03s+1A2k1+1+u1u3u5∑l=03s+1B2l+1∏k2=l+13s+1A2k2+1,(34e)u6n+4=u4∏s=0n−1∏k1=03s+1A2k1+u0u2u4∑l=03s+1B2l∏k2=l+13s+1A2k2∏k1=03s+2A2k1+u0u2u4∑l=03s+2B2l∏k2=l+13s+2A2k2,(34f)u6n+5=u5∏s=0n−1∏k1=03s+1A2k1+1+u1u3u5∑l=03s+1B2l+1∏k2=l+13s+1A2k2+1∏k1=03s+2A2k1+1+u1u3u5∑l=03s+2B2l+1∏k2=l+13s+2A2k2+1,whenever the denominators do not vanish, i.e., u0u2u4∑l=0sB2lΘ0Ak2,l+1,s≠−Θ0Ak1,0,s and u1u3u5∑l=0sB2l+1Θ1Ak2,l+1,s≠−Θ1Ak1,0,s.
The implication is that solutions of (2) are given by(35a)x6n−5=x−5∏s=0n−1∏k1=03s−1a2k1+x−5x−3x−1∑l=03s−1b2l∏k2=l+13s−1a2k2∏k1=03sa2k1+x−5x−3x−1∑l=03sb2l∏k2=l+13sa2k2,(35b)x6n−4=x−4∏s=0n−1∏k1=03s−1a2k1+1+x−4x−2x0∑l=03s−1b2l+1∏k2=l+13s−1a2k2+1∏k1=03sa2k1+1+x−4x−2x0∑l=03sb2l+1∏k2=l+13sa2k2+1,(35c)x6n−3=x−3∏s=0n−1∏k1=03sa2k1+x−5x−3x−1∑l=03sb2l∏k2=l+13sa2k2∏k1=03s+1a2k1+x−5x−3x−1∑l=03s+1b2l∏k2=l+13s+1a2k2,(35d)x6n−2=x−2∏s=0n−1∏k1=03sa2k1+1+x−4x−2x0∑l=03sb2l+1∏k2=l+13sa2k2+1∏k1=03s+1a2k1+1+x−4x−2x0∑l=03s+1b2l+1∏k2=l+13s+1a2k2+1,(35e)x6n−1=x−1∏s=0n−1∏k1=03s+1a2k1+x−5x−3x−1∑l=03s+1b2l∏k2=l+13s+1a2k2∏k1=03s+2a2k1+x−5x−3x−1∑l=03s+2b2l∏k2=l+13s+2a2k2,(35f)x6n=x0∏s=0n−1∏k1=03s+1a2k1+1+x−4x−2x0∑l=03s+1b2l+1∏k2=l+13s+1a2k2+1∏k1=03s+2a2k1+1+x−4x−2x0∑l=03s+2b2l+1∏k2=l+13s+2a2k2+1,whenever the denominators do not vanish, i.e., x−5x−3x−1∑l=0sb2lΘ0ak2,l+1,s≠−Θ0ak1,0,s and x−4x−2x−0∑l=0sb2l+1Θ1ak2,l+1,s≠−Θ1ak1,0,s.


### 3.2. The Case Wherean and bn Are Two Periodic Sequences

Letan=λ,μ,λ,μ,λ,…, bn=η,ζ,η,ζ,…, Φ=x−5x−3x−1, and Ψ=x−4x−2x0. The solution in this case is given by the following equations:(36a)x6n−5=x−5∏s=0n−1λ3s+ηΦ∑l=03s−1λlλ3s+1+ηΦ∑l=03sλl,x6n−4=x−4∏s=0n−1μ3s+ζΨ∑l=03s−1μlμ3s+1+ζΨ∑l=03sμl,(36b)x6n−3=x−3∏s=0n−1λ3s+1+ηΦ∑l=03sλlλ3s+2+ηΦ∑l=03s+1λl,x6n−2=x−2∏s=0n−1μ3s+1+ζΨ∑l=03sμlμ3s+2+ζΨ∑l=03s+1μl,(36c)x6n−1=x−1∏s=0n−1λ3s+2+ηΦ∑l=03s+1λlλ3s+3+ηΦ∑l=03s+2λl,x6n=x0∏s=0n−1μ3s+2+ζΨ∑l=03s+1μlμ3s+3+ζΨ∑l=03s+2μl,where ∑l=0s−1λlηΦ≠−λs and ∑l=0s−1λlηΨ≠−λs,s≤3n.


### 3.3. The Case Wherean and bn Are Constants

Here,an=μ=λ, bn=η=ζ, Φ=x−5x−3x−1, and Ψ=x−4x−2x0. Equation (2) becomes xn+1=xn−5/λ+ηxn−1xn−3xn−5.


#### 3.3.1.λ=1

The solution given in (35a)–(35f) simplifies to(37a)x6n−5=x−5∏s=0n−11+3sηΦ1+3s+1ηΦ,x6n−4=x−4∏s=0n−11+3sηΨ1+3s+1ηΨ,(37b)x6n−3=x−3∏s=0n−11+3s+1ηΦ1+3s+2ηΦ,x6n−2=x−2∏s=0n−11+3s+1ηΨ1+3s+2ηΨ,(37c)x6n−1=x−1∏s=0n−11+3s+2ηΦ1+3s+3ηΦ,x6n=x0∏s=0n−11+3s+2ηΨ1+3s+3ηΨ,where sηΦ≠−1 and sηΨ≠−1,s≤3n.(i)
Ifη=1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 2.1 in [4] for(38)xn+1=xn−51+xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero positive real numbers) is a special case of our restriction (the initial conditions are arbitrary nonzero real numbers andsΦ≠−1,sΨ≠−1,and0≤s≤3n).(ii)
Ifη=−1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 3.1 in [4] for(39)xn+1=xn−51−xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero real numbers andjbdf≠1andjace≠1) coincides with our restriction (the initial conditions are arbitrary nonzero real numbers and sΦ≠1,sΨ≠1,and0≤s≤3n).


#### 3.3.2. The Case Whereλ≠1

In this case, the solution given in (35a)–(35f) simplifies to(40)x6n−5=x−5∏s=0n−1λ3s+ηΦ1−λ3s/1−λλ3s+1+ηΦ1−λ3s+1/1−λ,x6n−4=x−4∏s=0n−1λ3s+ζΨ1−λ3s/1−λλ3s+1+ζΨ1−λ3s+1/1−λ,x6n−3=x−3∏s=0n−1λ3s+1+ηΦ1−λ3s+1/1−λλ3s+2+ηΦ1−λ3s+2/1−λ,x6n−2=x−2∏s=0n−1λ3s+1+ζΨ1−λ3s+1/1−λλ3s+2+ζΨ1−λ3s+2/1−λ,x6n−1=x−1∏s=0n−1λ3s+2+ηΦ1−λ3s+2/1−λλ3s+3+ηΦ1−λ3s+3/1−λ,x6n=x0∏s=0n−1λ3s+2+ζΨ1−λ3s+2/1−λλ3s+3+ζΨ1−λ3s+3/1−λ,where 1−λsηΦ≠−λs1−λ and 1−λsζΨ≠−λs1−λ,s≤3n.Note.
Ifλ=−1, then the solution given in Section 3.3.2 simplifies to(41)x12n−5=x62n−5=x−5,x12n−4=x−4,x12n−3=x−3,x12n−2=x−2,x12n−1=x−1,x12n=x0,x12n+1=x−5−1+ηΦ,x12n+2=x−4−1+ζΨ,x12n+3=x−3−1+ηΦ,x12n+4=x−2−1+ζΨ,x12n+5=x−1−1+ηΦ,x12n+6=x0−1+ζΨ,x12n+7=x62n+2−5=x−5.(i)
When settingη=1 in Section 3.3.2, we get the result obtained in Theorem 4.1 in [4] for(42)xn+1=xn−5−1+xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠1 and Ψ≠1).(ii)
When settingη=−1 in Section 3.3.2, we get the result obtained in Theorem 5.1 in [4] for(43)xn+1=xn−5−1−xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠−1 and Ψ≠−1).


## 3.1. On the Difference Equation (3)

Consider the sixth-order difference equations of the form (3), that is,(15)un+6=unAn+Bnunun+2un+4.We impose the symmetry condition (12) and simplify the resulting equation to get(16)S6Q+Bnun2un+2S4Q+Bnun2un+4S2Q−AnQBnunun+2un+4+An2=0.To solve forQ, we first differentiate (16) with respect to un (keeping Ω fixed and viewing un+2 as a function of un,un+4, and Ω). This leads, after simplification, to(17)S4Q+un+4S2Q′−un+4Q′+2un+4unQ=0,where ′ denotes the derivative with respect to the independent variable. We then differentiate (17) with respect to un to get(18)Q″n,un−2unQ′n,un+2un2Qn,un=0.The solutions of (18) are given by(19)Qn,un=αnun2+βnun,for some functions αn and βn of n. To obtain more information on αn and βn, we substitute (19) in (16) and we split the resulting equation to get the following:(20a)unun+2un+42:Bnαn+4=0,(20b)unun+22un+4:Bnαn+2=0,(20c)unun+2un+4:Bnβn+2+βn+4+βn+6=0,(20d)un:−Anαn+αn+6=0,(20e)1:Anβn−βn+6=0.These equations ((20a)–(20e)) reduce to(21)αn=0,βn+4+βn+2+βn=0.The expression ofβn in (21) is merely obtained by solving the corresponding characteristic equation r4+r2+1=0 for r. Assuming that ri,i=1,2,3,4, are the solutions of this characteristic equation, then βn is a linear combination of the rin’s. In other words, the solutions of (21) are(22)αn=0,βn=−1nβnc1+−1nβ¯nc2+β¯nc3+βnc4,for some arbitrary constants ci,i=1,…,4, where β=expπi/3. Thus, we obtain four characteristics given by(23)Q1=−1nβnun,Q2=−1nβ¯nun,Q3=β¯nun,Q4=βnun.The four corresponding symmetry generatorsX1,X2,X3, and X4 are given by(24a)X1=−1nβnun∂un−−1nβn+1un+1∂un+1+−1nβn+2un+2∂un+2−−1nβn+3un+3∂un+3+−1nβn+4un+4∂un+4−−1nβn+5un+5∂un+5,(24b)X2=−1nβ¯nun∂un−−1nβ¯n+1un+1∂un+1+−1nβ¯n+2un+2∂un+2−−1nβ¯n+3un+3∂un+3+−1nβ¯n+4un+4∂un+4−−1nβ¯n+5un+5∂un+5,(24c)X3=β¯nun∂un+β¯n+1un+1∂un+1+β¯n+2un+2∂un+2+β¯n+3un+3∂un+3+β¯n+4un+4∂un+4+β¯n+5un+5∂un+5,(24d)X4=βnun∂un+βn+1un+1∂un+1+βn+2un+2∂un+2+βn+3un+3∂un+3+βn+4un+4∂un+4+βn+5un+5∂un+5.Here, usingQ4, we introduce the canonical coordinate [8](25)sn=∫dunβnun=1βnlnun.In view of (21), we introduce the variable(26)rn=βn+4sn+4+βn+2sn+2+βnsn.It is easy to check that(27)Xirn=βn+4+βn+2+βn=0,i=1,…,4.Therefore,rn is invariant under X4. It is advantageous to use(28)r˜n=exp−rn,that is, r˜n=±1/unun+2un+4. Here, we choose to use the plus sign and we show, using (3), that(29)r˜n+2=Anr˜n+Bn.Therefore,(30)r˜2n+k=r˜k∏k1=0n−1A2k1+k+∑l=0n−1B2l+k∏k2=l+1n−1A2k2+k,k=0,1.It is worthwhile to mention that equation (30) gives the solution of (29) for all n. By reversing all the change of variables, we have(31)un=exp−βnc5+βnc6+β¯nc7+−β¯nc8+−βn+1∑k1=0n−1−i36−β¯k1rk1+βn+1∑k2=0n−1i36β¯k2rk2+β¯n+1∑k3=0n−1−i36βk3rk3+−β¯n+1∑k4=0n−1i36−βk4rk4=Γnexp∑k=0n−133−1n+k+1Imγn+1,klnr˜k,where r˜k is given in (30), Γn=exp−βnc5+βnc6+β¯nc7+−β¯nc8, and γl,k=βlβ¯k.Note.
Equation (31) gives the solution of (3) in a unified manner.
We can simplify (31) further by splitting it into six categories. We have(32)u6n=Γ6nexp∑k=06n−133−16n+k+1Imγ6n+1,klnr˜k,u6n=u0∏s=0n−1r˜6sr˜6s+2.
Similarly, after a straightforward but lengthy computation, we get(33)u6n+i=ui∏s=0n−1r˜6s+ir˜6s+2+i,i=0,…,5.Note.
We can obtain (33) using (28) without considering the absolute value function.
Now, using (30) and (33), we obtain the solutions of (3) as follows:(34a)u6n=u0∏s=0n−1∏k1=03s−1A2k1+u0u2u4∑l=03s−1B2l∏k2=l+13s−1A2k2∏k1=03sA2k1+u0u2u4∑l=03sB2l∏k2=l+13sA2k2,(34b)u6n+1=u1∏k1=03s−1A2k1+1+u1u3u5∑l=03s−1B2l+1∏k2=l+13s−1A2k2+1∏k1=03sA2k1+1+u1u3u5∑l=03sB2l+1∏k2=l+13sA2k2+1,(34c)u6n+2=u2∏s=0n−1∏k1=03sA2k1+u0u2u4∑l=03sB2l∏k2=l+13sA2k2∏k1=03s+1A2k1+u0u2u4∑l=03s+1B2l∏k2=l+13s+1A2k2,(34d)u6n+3=u3∏s=0n−1∏k1=03sA2k1+1+u1u3u5∑l=03sB2l+1∏k2=l+13sA2k2+1∏k1=03s+1A2k1+1+u1u3u5∑l=03s+1B2l+1∏k2=l+13s+1A2k2+1,(34e)u6n+4=u4∏s=0n−1∏k1=03s+1A2k1+u0u2u4∑l=03s+1B2l∏k2=l+13s+1A2k2∏k1=03s+2A2k1+u0u2u4∑l=03s+2B2l∏k2=l+13s+2A2k2,(34f)u6n+5=u5∏s=0n−1∏k1=03s+1A2k1+1+u1u3u5∑l=03s+1B2l+1∏k2=l+13s+1A2k2+1∏k1=03s+2A2k1+1+u1u3u5∑l=03s+2B2l+1∏k2=l+13s+2A2k2+1,whenever the denominators do not vanish, i.e., u0u2u4∑l=0sB2lΘ0Ak2,l+1,s≠−Θ0Ak1,0,s and u1u3u5∑l=0sB2l+1Θ1Ak2,l+1,s≠−Θ1Ak1,0,s.
The implication is that solutions of (2) are given by(35a)x6n−5=x−5∏s=0n−1∏k1=03s−1a2k1+x−5x−3x−1∑l=03s−1b2l∏k2=l+13s−1a2k2∏k1=03sa2k1+x−5x−3x−1∑l=03sb2l∏k2=l+13sa2k2,(35b)x6n−4=x−4∏s=0n−1∏k1=03s−1a2k1+1+x−4x−2x0∑l=03s−1b2l+1∏k2=l+13s−1a2k2+1∏k1=03sa2k1+1+x−4x−2x0∑l=03sb2l+1∏k2=l+13sa2k2+1,(35c)x6n−3=x−3∏s=0n−1∏k1=03sa2k1+x−5x−3x−1∑l=03sb2l∏k2=l+13sa2k2∏k1=03s+1a2k1+x−5x−3x−1∑l=03s+1b2l∏k2=l+13s+1a2k2,(35d)x6n−2=x−2∏s=0n−1∏k1=03sa2k1+1+x−4x−2x0∑l=03sb2l+1∏k2=l+13sa2k2+1∏k1=03s+1a2k1+1+x−4x−2x0∑l=03s+1b2l+1∏k2=l+13s+1a2k2+1,(35e)x6n−1=x−1∏s=0n−1∏k1=03s+1a2k1+x−5x−3x−1∑l=03s+1b2l∏k2=l+13s+1a2k2∏k1=03s+2a2k1+x−5x−3x−1∑l=03s+2b2l∏k2=l+13s+2a2k2,(35f)x6n=x0∏s=0n−1∏k1=03s+1a2k1+1+x−4x−2x0∑l=03s+1b2l+1∏k2=l+13s+1a2k2+1∏k1=03s+2a2k1+1+x−4x−2x0∑l=03s+2b2l+1∏k2=l+13s+2a2k2+1,whenever the denominators do not vanish, i.e., x−5x−3x−1∑l=0sb2lΘ0ak2,l+1,s≠−Θ0ak1,0,s and x−4x−2x−0∑l=0sb2l+1Θ1ak2,l+1,s≠−Θ1ak1,0,s.


## 3.2. The Case Wherean and bn Are Two Periodic Sequences

Letan=λ,μ,λ,μ,λ,…, bn=η,ζ,η,ζ,…, Φ=x−5x−3x−1, and Ψ=x−4x−2x0. The solution in this case is given by the following equations:(36a)x6n−5=x−5∏s=0n−1λ3s+ηΦ∑l=03s−1λlλ3s+1+ηΦ∑l=03sλl,x6n−4=x−4∏s=0n−1μ3s+ζΨ∑l=03s−1μlμ3s+1+ζΨ∑l=03sμl,(36b)x6n−3=x−3∏s=0n−1λ3s+1+ηΦ∑l=03sλlλ3s+2+ηΦ∑l=03s+1λl,x6n−2=x−2∏s=0n−1μ3s+1+ζΨ∑l=03sμlμ3s+2+ζΨ∑l=03s+1μl,(36c)x6n−1=x−1∏s=0n−1λ3s+2+ηΦ∑l=03s+1λlλ3s+3+ηΦ∑l=03s+2λl,x6n=x0∏s=0n−1μ3s+2+ζΨ∑l=03s+1μlμ3s+3+ζΨ∑l=03s+2μl,where ∑l=0s−1λlηΦ≠−λs and ∑l=0s−1λlηΨ≠−λs,s≤3n.


## 3.3. The Case Wherean and bn Are Constants

Here,an=μ=λ, bn=η=ζ, Φ=x−5x−3x−1, and Ψ=x−4x−2x0. Equation (2) becomes xn+1=xn−5/λ+ηxn−1xn−3xn−5.


### 3.3.1.λ=1

The solution given in (35a)–(35f) simplifies to(37a)x6n−5=x−5∏s=0n−11+3sηΦ1+3s+1ηΦ,x6n−4=x−4∏s=0n−11+3sηΨ1+3s+1ηΨ,(37b)x6n−3=x−3∏s=0n−11+3s+1ηΦ1+3s+2ηΦ,x6n−2=x−2∏s=0n−11+3s+1ηΨ1+3s+2ηΨ,(37c)x6n−1=x−1∏s=0n−11+3s+2ηΦ1+3s+3ηΦ,x6n=x0∏s=0n−11+3s+2ηΨ1+3s+3ηΨ,where sηΦ≠−1 and sηΨ≠−1,s≤3n.(i)
Ifη=1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 2.1 in [4] for(38)xn+1=xn−51+xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero positive real numbers) is a special case of our restriction (the initial conditions are arbitrary nonzero real numbers andsΦ≠−1,sΨ≠−1,and0≤s≤3n).(ii)
Ifη=−1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 3.1 in [4] for(39)xn+1=xn−51−xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero real numbers andjbdf≠1andjace≠1) coincides with our restriction (the initial conditions are arbitrary nonzero real numbers and sΦ≠1,sΨ≠1,and0≤s≤3n).


### 3.3.2. The Case Whereλ≠1

In this case, the solution given in (35a)–(35f) simplifies to(40)x6n−5=x−5∏s=0n−1λ3s+ηΦ1−λ3s/1−λλ3s+1+ηΦ1−λ3s+1/1−λ,x6n−4=x−4∏s=0n−1λ3s+ζΨ1−λ3s/1−λλ3s+1+ζΨ1−λ3s+1/1−λ,x6n−3=x−3∏s=0n−1λ3s+1+ηΦ1−λ3s+1/1−λλ3s+2+ηΦ1−λ3s+2/1−λ,x6n−2=x−2∏s=0n−1λ3s+1+ζΨ1−λ3s+1/1−λλ3s+2+ζΨ1−λ3s+2/1−λ,x6n−1=x−1∏s=0n−1λ3s+2+ηΦ1−λ3s+2/1−λλ3s+3+ηΦ1−λ3s+3/1−λ,x6n=x0∏s=0n−1λ3s+2+ζΨ1−λ3s+2/1−λλ3s+3+ζΨ1−λ3s+3/1−λ,where 1−λsηΦ≠−λs1−λ and 1−λsζΨ≠−λs1−λ,s≤3n.Note.
Ifλ=−1, then the solution given in Section 3.3.2 simplifies to(41)x12n−5=x62n−5=x−5,x12n−4=x−4,x12n−3=x−3,x12n−2=x−2,x12n−1=x−1,x12n=x0,x12n+1=x−5−1+ηΦ,x12n+2=x−4−1+ζΨ,x12n+3=x−3−1+ηΦ,x12n+4=x−2−1+ζΨ,x12n+5=x−1−1+ηΦ,x12n+6=x0−1+ζΨ,x12n+7=x62n+2−5=x−5.(i)
When settingη=1 in Section 3.3.2, we get the result obtained in Theorem 4.1 in [4] for(42)xn+1=xn−5−1+xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠1 and Ψ≠1).(ii)
When settingη=−1 in Section 3.3.2, we get the result obtained in Theorem 5.1 in [4] for(43)xn+1=xn−5−1−xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠−1 and Ψ≠−1).


## 3.3.1.λ=1

The solution given in (35a)–(35f) simplifies to(37a)x6n−5=x−5∏s=0n−11+3sηΦ1+3s+1ηΦ,x6n−4=x−4∏s=0n−11+3sηΨ1+3s+1ηΨ,(37b)x6n−3=x−3∏s=0n−11+3s+1ηΦ1+3s+2ηΦ,x6n−2=x−2∏s=0n−11+3s+1ηΨ1+3s+2ηΨ,(37c)x6n−1=x−1∏s=0n−11+3s+2ηΦ1+3s+3ηΦ,x6n=x0∏s=0n−11+3s+2ηΨ1+3s+3ηΨ,where sηΦ≠−1 and sηΨ≠−1,s≤3n.(i)
Ifη=1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 2.1 in [4] for(38)xn+1=xn−51+xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero positive real numbers) is a special case of our restriction (the initial conditions are arbitrary nonzero real numbers andsΦ≠−1,sΨ≠−1,and0≤s≤3n).(ii)
Ifη=−1, then equations in (37a)–(37c) are exactly the ones obtained in Theorem 3.1 in [4] for(39)xn+1=xn−51−xn−1xn−3xn−5,and their restriction (the initial conditions are arbitrary nonzero real numbers andjbdf≠1andjace≠1) coincides with our restriction (the initial conditions are arbitrary nonzero real numbers and sΦ≠1,sΨ≠1,and0≤s≤3n).


## 3.3.2. The Case Whereλ≠1

In this case, the solution given in (35a)–(35f) simplifies to(40)x6n−5=x−5∏s=0n−1λ3s+ηΦ1−λ3s/1−λλ3s+1+ηΦ1−λ3s+1/1−λ,x6n−4=x−4∏s=0n−1λ3s+ζΨ1−λ3s/1−λλ3s+1+ζΨ1−λ3s+1/1−λ,x6n−3=x−3∏s=0n−1λ3s+1+ηΦ1−λ3s+1/1−λλ3s+2+ηΦ1−λ3s+2/1−λ,x6n−2=x−2∏s=0n−1λ3s+1+ζΨ1−λ3s+1/1−λλ3s+2+ζΨ1−λ3s+2/1−λ,x6n−1=x−1∏s=0n−1λ3s+2+ηΦ1−λ3s+2/1−λλ3s+3+ηΦ1−λ3s+3/1−λ,x6n=x0∏s=0n−1λ3s+2+ζΨ1−λ3s+2/1−λλ3s+3+ζΨ1−λ3s+3/1−λ,where 1−λsηΦ≠−λs1−λ and 1−λsζΨ≠−λs1−λ,s≤3n.Note.
Ifλ=−1, then the solution given in Section 3.3.2 simplifies to(41)x12n−5=x62n−5=x−5,x12n−4=x−4,x12n−3=x−3,x12n−2=x−2,x12n−1=x−1,x12n=x0,x12n+1=x−5−1+ηΦ,x12n+2=x−4−1+ζΨ,x12n+3=x−3−1+ηΦ,x12n+4=x−2−1+ζΨ,x12n+5=x−1−1+ηΦ,x12n+6=x0−1+ζΨ,x12n+7=x62n+2−5=x−5.(i)
When settingη=1 in Section 3.3.2, we get the result obtained in Theorem 4.1 in [4] for(42)xn+1=xn−5−1+xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠1 and Ψ≠1).(ii)
When settingη=−1 in Section 3.3.2, we get the result obtained in Theorem 5.1 in [4] for(43)xn+1=xn−5−1−xn−1xn−3xn−5,and their restriction coincides with our restriction (the initial conditions are arbitrary nonzero real numbers,Φ≠−1 and Ψ≠−1).


## 4. On the Difference Equation (5)

Consider the fifth-order difference equations of the form (5), i.e.,(44)un+5=Ω=unun+1un+2un+3un+4λ+μunun+1un+2.Here, the procedure for finding the characteristics of (5) is similar as above and is as follows:(i)
Impose the symmetry condition (12) to (5)(ii)
Differentiate with respect toun (keeping Ω fixed) and viewing un+3 as a function of un,un+1,un+2, and Ω(iii)
Differentiate with respect toun twice (keeping un+1 fixed)(iv)
Use the method of separationAfter performing this series of operations, we obtain the following characteristics:(45)Q5n,un=βnun,Q6=β¯nun,where β=exp−2πi/3. Thus, we obtain two characteristics with corresponding generators given by(46a)X5=β¯nun∂un+β¯n+1un+1∂un+1+ββ¯nun+2∂un+2+β¯nun+3∂un+3+β¯n+1un+4∂un+4,(46b)X6=βnun∂un+βn+1un+1∂un+1+β¯βnun+2∂un+2+βnun+3∂un+3+βn+1un+4∂un+4.From the characteristic equations(47)dunβ¯nun=dun+1β¯n+1un+1=dun+2β¯n+2un+2=dun+3β¯n+3un+3=dun+4β¯n+4un+4=Vn0,we obtain the invariants c1=un+1β/un,c2=un+2β¯/un,c3=unun+3,c4=un+4β/un, and c5=Vn. We readily notice that(48)S3dunβ¯nun=dun+1β¯n+1un+1=dun+3β¯n+3un+3=dun+4β¯n+4un+4,and we choose Vn=c1β¯c2β, i.e.,(49)Vn=unun+1un+2.By shifting (49) thrice, we get(50)Vn+3=Vnλ+μVn,whose solution is given by(51)Vn=μ−12/3−1n−−11/3+1+β3−12/3−1+c6+β¯nc7+βnc8−1,ifλ=1,c10λn/3+−12/3λ1/3nc8+−−11/3λ1/3nc9+μ1−λ−1,ifλ≠1.The constantsci,i=6,…,10, can be obtained from the following equations:(52a)c6+c7+c8=1u0u1u2−μ−−11/3+1+β3−12/3−1,(52b)c6+β¯c7+βc8=1u1u2u3−μβ−−11/3+−12/33−12/3−1,(52c)c6+βc7+β¯c8=1u2u3u4−μ−1+2−12/3−−11/3+β3−12/3−1,(52d)c10+c8+c9=1u0u1u2−μ1−λ,(52e)λ1/3c10+−12/3λ1/3c8+−−11/3λ1/3c9=1u1u2u3−μ1−λ,(52f)λ2/3c10+−12/3λ1/32c8+−−11/3λ1/32c9=1u2u3u4−μ1−λ.Thanks to (49), we can express un in terms of Vn as follows:(53a)un=expβnc11+β¯nc12−23∑k=0n−1Imγn,klnVk,where Vk is given in (51) with γn,k=βnβ¯k+1. The constants c11 and c12 must satisfy(53b)c11+c12=lnu0,βc11+β¯c12=lnu1.Equations in (53a) and (53b) give the solutions of (5) in a unified manner.For the sake of clarification, we split solutions (53a) to realize the solutions in the existing literature. Using (53a) and (53b), we have(54)u6n=u0∏s=12nV3s−2V3s−3.Using the same approach, we show that(55)u6n+i=ui∏s=12nV3s−1+i+1V3s−1+i,i=0,…,5.


### 4.1. The Case ofλ=1

Equation (5) becomes(56)un+5=unun+1un+2un+3un+41+μunun+1un+2,and we said earlier that the solution of (50), in this case, is (51), i.e.,(57)Vn=μ−12/3−1n−−11/3+1+β3−12/3−1+c6+β¯nc7+βnc8−1.We have(58)V3s=μ−12/3−13s−−11/3+1+β3−12/3−1+c6+c7+c8−1,and using (52a) in (58), we get(59)V3s=u0u1u21+μsu0u1u2.Using the same approach, we show that(60)V3s=u0u1u21+μsu0u1u2,V3s+1=u1u2u31+μsu1u2u3,V3s+2=u2u3u41+μsu2u3u4.Leta=x−4,b=x−3,c=x−2,andd=x−1 and A=u0,B=u1, C=u2, D=u3, and E=u4. Using (60) in (55), we obtain the solution of (56) as follows:(61)u6n=D2nA2n−1∏s=12n−11+μsABC1+μsBCD,u6n+1=E2nB2n−1∏s=12n−11+μsBCD1+μsCDE,u6n+2=CA2nB2nD2nE2n∏s=02n−11+μsCDE1+μs+1ABC,u6n+3=D2n+1A2n∏s=12n1+μsABC1+μsBCD,u6n+4=E2n+1B2n∏s=02n1+μsBCD1+μsCDE,u6n+5=CAB2n+1DE2n+1∏s=02n1+μsCDE1+μs+1ABC,whenever the denominators do not vanish.


### 4.2. The Case ofλ≠1

In this case, as we found earlier, the solution of (50) is given by (51), i.e.,(62)Vn=c10λn/3+−12/3λ1/3nc8+−−11/3λ1/3nc9+μ1−λ−1.Using this, we get(63)V3s=c10λs+c8λs+c9λs+μ1−λ−1,and using (52d) in (63), we find(64)V3s=ABCλs+μABC1−λs/1−λ.Using the same approach, we show that(65)V3s+i=uiui+1ui+2λs+μuiui+1ui+21−λs/1−λ,i=0,1,2.Using (65) in (55), we obtain the solution of (5) as follows:(66)u6n=D2nA2n−1∏s=22nλs−1+μΔ0∑j=0s−2λjλs−1+μΔ1∑j=0s−2λj,u6n+1=E2nB2n−1∏s=22nλs−1+μΔ1∑j=0s−2λjλs−1+μΔ2∑j=0s−2λj,u6n+2=CΔ02nΔ12n∏s=22nλs−1+μΔ2∑j=0s−2λj∏s=12nλs+μΔ0∑j=0s−1λj,u6n+3=D2n+1A2n∏s=22n+1λs−1+μΔ0∑j=0s−2λjλs−1+μΔ1∑j=0s−2λj,u6n+4=EΔ22nΔ12n∏s=22n+1λs−1+μΔ1∑j=0s−2λjλs−1+μΔ2∑j=0s−2λj,u6n+5=CΔ02n+1Δ12n+1∏s=22n+1λs−1+μΔ2∑j=0s−2λj∏s=12n+1λs+μΔ0∑j=0s−1λj,where Δi=uiui+1ui+2. Equations in Section 4.2 give the exact solution of (5) for any real values of λ and μ provided that the denominators do not vanish.Recall that we acted the shift operator on (4) to get (5). Hence, the solutions of (4) are obtained, using Section 4.2, as follows:(67)x6n−4=d2na2n−1∏s=22nλs−1+μabc∑j=0s−2λjλs−1+μbcd∑j=0s−2λj,x6n−3=e2nb2n−1∏s=22nλs−1+μbcd∑j=0s−2λjλs−1+μcde∑j=0s−2λj,x6n−2=cab2nde2n∏s=22nλs−1+μcde∑j=0s−2λj∏s=12nλs+μabc∑j=0s−1λj,x6n−1=d2n+1a2n∏s=22n+1λs−1+μabc∑j=0s−2λjλs−1+μbcd∑j=0s−2λj,x6n=e2n+1b2n∏s=22n+1λs−1+μbcd∑j=0s−2λjλs−1+μcde∑j=0s−2λj,x6n+1=cab2n+1de2n+1∏s=22n+1λs−1+μcde∑j=0s−2λj∏s=12n+1λs+μabc∑j=0s−1λj,for any real values of λ and μ as long as the denominators do not vanish.(i)
Whenλ=1 and μ=1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 5 in [9] for(68)xn+1=xn−2xn−3xn−4xnxn−11+xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are positive real numbers.(ii)
Whenλ=1 and μ=−1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 9 in [9] for(69)xn+1=xn−2xn−3xn−4xnxn−11−xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are positive real numbers with abc≠1 and cde≠1.Note.
There should not be a minus sign right after the expression ofx3n−2 in Theorem 9 in [9].(i)
Whenλ=−1 and μ=1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 7 in [9] for(70)xn+1=xn−2xn−3xn−4xnxn−1−1+xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,d, and e are nonzero real numbers with abc≠1, bcd≠1, and cde≠1.(ii)
Whenλ=−1 and μ=−1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 11 in [9] for(71)xn+1=xn−2xn−3xn−4xnxn−1−1−xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are nonzero real numbers with abc≠−1, bcd≠−1, and cde≠−1.Note.
There should be a minus sign right after the expression ofx6n+1 in Theorem 11 in [9].


## 4.1. The Case ofλ=1

Equation (5) becomes(56)un+5=unun+1un+2un+3un+41+μunun+1un+2,and we said earlier that the solution of (50), in this case, is (51), i.e.,(57)Vn=μ−12/3−1n−−11/3+1+β3−12/3−1+c6+β¯nc7+βnc8−1.We have(58)V3s=μ−12/3−13s−−11/3+1+β3−12/3−1+c6+c7+c8−1,and using (52a) in (58), we get(59)V3s=u0u1u21+μsu0u1u2.Using the same approach, we show that(60)V3s=u0u1u21+μsu0u1u2,V3s+1=u1u2u31+μsu1u2u3,V3s+2=u2u3u41+μsu2u3u4.Leta=x−4,b=x−3,c=x−2,andd=x−1 and A=u0,B=u1, C=u2, D=u3, and E=u4. Using (60) in (55), we obtain the solution of (56) as follows:(61)u6n=D2nA2n−1∏s=12n−11+μsABC1+μsBCD,u6n+1=E2nB2n−1∏s=12n−11+μsBCD1+μsCDE,u6n+2=CA2nB2nD2nE2n∏s=02n−11+μsCDE1+μs+1ABC,u6n+3=D2n+1A2n∏s=12n1+μsABC1+μsBCD,u6n+4=E2n+1B2n∏s=02n1+μsBCD1+μsCDE,u6n+5=CAB2n+1DE2n+1∏s=02n1+μsCDE1+μs+1ABC,whenever the denominators do not vanish.


## 4.2. The Case ofλ≠1

In this case, as we found earlier, the solution of (50) is given by (51), i.e.,(62)Vn=c10λn/3+−12/3λ1/3nc8+−−11/3λ1/3nc9+μ1−λ−1.Using this, we get(63)V3s=c10λs+c8λs+c9λs+μ1−λ−1,and using (52d) in (63), we find(64)V3s=ABCλs+μABC1−λs/1−λ.Using the same approach, we show that(65)V3s+i=uiui+1ui+2λs+μuiui+1ui+21−λs/1−λ,i=0,1,2.Using (65) in (55), we obtain the solution of (5) as follows:(66)u6n=D2nA2n−1∏s=22nλs−1+μΔ0∑j=0s−2λjλs−1+μΔ1∑j=0s−2λj,u6n+1=E2nB2n−1∏s=22nλs−1+μΔ1∑j=0s−2λjλs−1+μΔ2∑j=0s−2λj,u6n+2=CΔ02nΔ12n∏s=22nλs−1+μΔ2∑j=0s−2λj∏s=12nλs+μΔ0∑j=0s−1λj,u6n+3=D2n+1A2n∏s=22n+1λs−1+μΔ0∑j=0s−2λjλs−1+μΔ1∑j=0s−2λj,u6n+4=EΔ22nΔ12n∏s=22n+1λs−1+μΔ1∑j=0s−2λjλs−1+μΔ2∑j=0s−2λj,u6n+5=CΔ02n+1Δ12n+1∏s=22n+1λs−1+μΔ2∑j=0s−2λj∏s=12n+1λs+μΔ0∑j=0s−1λj,where Δi=uiui+1ui+2. Equations in Section 4.2 give the exact solution of (5) for any real values of λ and μ provided that the denominators do not vanish.Recall that we acted the shift operator on (4) to get (5). Hence, the solutions of (4) are obtained, using Section 4.2, as follows:(67)x6n−4=d2na2n−1∏s=22nλs−1+μabc∑j=0s−2λjλs−1+μbcd∑j=0s−2λj,x6n−3=e2nb2n−1∏s=22nλs−1+μbcd∑j=0s−2λjλs−1+μcde∑j=0s−2λj,x6n−2=cab2nde2n∏s=22nλs−1+μcde∑j=0s−2λj∏s=12nλs+μabc∑j=0s−1λj,x6n−1=d2n+1a2n∏s=22n+1λs−1+μabc∑j=0s−2λjλs−1+μbcd∑j=0s−2λj,x6n=e2n+1b2n∏s=22n+1λs−1+μbcd∑j=0s−2λjλs−1+μcde∑j=0s−2λj,x6n+1=cab2n+1de2n+1∏s=22n+1λs−1+μcde∑j=0s−2λj∏s=12n+1λs+μabc∑j=0s−1λj,for any real values of λ and μ as long as the denominators do not vanish.(i)
Whenλ=1 and μ=1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 5 in [9] for(68)xn+1=xn−2xn−3xn−4xnxn−11+xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are positive real numbers.(ii)
Whenλ=1 and μ=−1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 9 in [9] for(69)xn+1=xn−2xn−3xn−4xnxn−11−xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are positive real numbers with abc≠1 and cde≠1.Note.
There should not be a minus sign right after the expression ofx3n−2 in Theorem 9 in [9].(i)
Whenλ=−1 and μ=1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 7 in [9] for(70)xn+1=xn−2xn−3xn−4xnxn−1−1+xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,d, and e are nonzero real numbers with abc≠1, bcd≠1, and cde≠1.(ii)
Whenλ=−1 and μ=−1, equations in Section 4.2 yield the results obtained by Yasin Yazlik in Theorem 11 in [9] for(71)xn+1=xn−2xn−3xn−4xnxn−1−1−xn−2xn−3xn−4,n=0,1,2,…,wherea,b,c,c,d, and e are nonzero real numbers with abc≠−1, bcd≠−1, and cde≠−1.Note.
There should be a minus sign right after the expression ofx6n+1 in Theorem 11 in [9].


## 5. Conclusion

In this paper, we have obtained nontrivial symmetries of some rational ordinary difference equations and their exact solutions were obtained. Most importantly, we note that (21) gives a clear idea, without making any lucky guesses, for the most convenient choice of the invariant of (3).



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*Source: 1505619-2019-12-01.xml*