Theorem 6 If (b+ f) > (c+r) and (d+ g) > (e + s), then the necessary and 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 bxn-1 + cxn-2+ fan-3+ rXn-4 Xn+1 = axn + %3D dxn-1 + exn-2+ gan-3 + sxn-4 (b+ 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) Q² + (b + f)P+(c+r)Q, (14) %3D