Theorem 6 If (b+ f) > (c+r) und (d+g) > (e+ s), then the necessary und sufficient condition for Eq. (1) to have positive solutions of prime period two is that the inequality [(a + 1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]² +4 [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e + s) (b+ f)] > 0. (13) is valid. Proof: Suppose that there exist positive distinctive solutions of prime period two ..., P, Q, P,Q,.. of Eq.(1). From Eq.(1) we have bæn-1+ can-2+ fæn-3 + rxn-4 In+1 = axn+ dæn-1 + exn-2+ gæn-3 + sæn-4 (6+ f)P+(c+r) Q (d + g) P+ (e+ s) Q' (b+f)Q+(c+r) P (d+ g) Q + (e + s) P' P = aQ+ Q = aP+ Consequently, we obtain (d+ g) P² + (e+s) PQ = a (d+ g) PQ+a(e+s)Q?+(b+ f) P+(c+r)Q, (14) and (d + g) Q? + (e+ s) PQ = a (d+ g) PQ+a(e+s) P2 + (b6+ f)Q+(c+r) P. (15) By subtracting (14) from (15), we have (a (e + s) + (d+ g)I (P² – Q²) = [(b+ f) - (c+r)] (P - Q) - %3D Since P+ Q, it follows that [(b+f)- (c+r)] [a (e + s) + (d + g)] P+Q = (16) while, by adding (14) and (15) and by using the relation p² + Q? = (P+ Q)² – 2PQ for all P,Q ER, %3D we obtain [(b+ f) – (c+r)] [(c+r) (d+ g) + a (e +s) (b+ f)] (17) PQ = [a (e + s) + (d+ g)] [((e+ s) - (d+ g)) (a + 1)] Assume that P and Q are two distinct real roots of the quadratic equation t2 - (P+Q)t+ PQ = 0. (a (e + s) + (d+ g)] t² – [(b+ f) – (c+r)] t [(b+ f) - (c+r)] [(c+r) (d+ g) + a (e + s) (b+f)] [a (e+ s) + (d+ g)] [(e+ s) - (d+ g)) (a + 1)] 0, (18) and so - (c+r))? – 4 (b +f)- (c+r)][(c+r) (d + g) + a (e + s) (b+f)] [(a + 1) ((e + s) - (d+g))] >0, or + S) – (c+r))² + 4 [(b+ f) – (c+r)[(c+r) (d + g) + a (e + s) (b+ f)] [(a +1) ((d + g) –- (e+s))] >0. (19) From (19), we obtain [(a + 1) ((d + g) – (e + s))] [(b + f) – (c + r)]? +4 [(b + f) – (c+r)] [(c+ r) (d + g) + a(e + s) (b+ f)] > 0. Thus, the condition (13) is valid. Conversely, suppose that the condition (13) is valid where (b+ f) > (c+r) and (d+g) > (e+ s). Then, we deduce