Show me the steps of determine green and inf is here The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-kXn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+dxn-k– exŋ-n= 0,1,2,.....(1)where the coefficients A, B, C, D, b, d, e E (0,0), whilek, 1 and o are positive integers. The initial conditionsX_g,..., X_1,.., X_k, ….., X_1, X are arbitrary positive realnumbers such that k <1< 0. Note that the special casesof Eq.(1) have been studied in [1] when B=C= D=0,and k= 0,1= 1, b is replaced by – b and in [27] whenB=C= D=0, and k= 0, b is replaced by[33] when B = C = D = 0, 1= 0 and in [32] whenA=C= D=0, 1=0, b is replaced by – b.b and in%3| Theorem 8. If o is even and k, 1 are odd positive integersand (A+D+ 1) + (B+ C), then Eq.(1) has no primeperiod two solution.Proof.Following the proof of Theorem 5, we deduce that ifo is even and k, 1 are odd positive integers, then x, = Xn-oand Xn+1 = xn-k= Xn=1. It follows from Eq.(1) thatbP=(A+D) Q+(B+C)P –(18)(e – d)andbQ = (A+D) P+ (B+C) Q –(19)(e – d)By subtracting (19) from (18), we get(P- Q) [(A+D+ 1) – (B+C)] = 0,Since (A+D+1) – (B+C) + 0, then P= Q. This is acontradiction. Thus, the proof is now completed.O

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Theorem 8. If o is even and k, 1 are odd positive integers and (A+D+1) + (B+C), then Eq.(1) has no prime period two solution. Proof.Following the proof of Theorem 5, we deduce that if o is even and k,1 are odd positive integers, then x, = Xn-o and Xn+1 = xn-k= Xn=1. It follows from Eq.(1) that b P=(A+D) Q+(B+C)P – (18) (e – d) and b Q = (A+D) P+ (B+C) Q – (19) (e – d) - By subtracting (19) from (18), we get (P- Q) [(A+D+ 1) – (B+C)] = 0, Since (A+D+1) – (B+C) + 0, then P= Q. This is a contradiction. Thus, the proof is now completed.O