On the Difference Equation xn 1 xnxn − 2 − 1

and Applied Analysis 3 From 2.3 we obtain c 1 b b 1 2b 1 b 1 ⇒ b 1 b 1 2b 1 3b 2 2b 1 , 2.4 which implies b2 − b − 1 0. Hence b x1 or b x2. From this and 2.3 it follows that a b c x1 or a b c x2, from which the result follows. Theorem 2.3. Equation 1.1 has no prime period-four solutions. Proof. Let xn n −2 be a prime period-four solution of 1.1 and x−2 a, x−1 b, x0 c. Then we have x1 x0x−2 − 1 ac − 1, 2.5 x2 x1x−1 − 1 ac − 1 b − 1 a, 2.6 x3 x2x0 − 1 ac − 1 b, 2.7 x4 x3x1 − 1 b ac − 1 − 1 c. 2.8 Thus, from 2.6 and 2.8 , we have that a c. This along with 2.7 gives a2 − 1 b, 2.9 while from 2.8 we get b a2 − 1 − 1 a or equivalently a 1 b a − 1 − 1 0. 2.10 Case 1. Suppose a −1. Then b a2 − 1 0 and c a −1, which, by Theorem 2.1, yields a period-two solution. Suppose a/ − 1. If a 1, then from 2.10 we get a contradiction. If a/ 1, then a2 − 1 b 1/ a − 1 , so that a a2 − a − 1 0. Hence a 0, a x1, or a x2. Case 2. Suppose a 0. Then b a2 − 1 −1 and c a 0 which results in a period-two solution as proved in Theorem 2.1. Case 3. Suppose a x1. Then b a2 − 1 x1 and c x1 which is an equilibrium solution. Case 4. Suppose a x2. Then b a2 − 1 x2 and c x2 which is the second equilibrium solution. Proof is complete. 3. Local Stability Here we study the local stability at the equilibrium points x1 and x2. Theorem 3.1. The negative equilibrium of 1.1 , x1, is unstable. Moreover, it is a hyperbolic equilibrium. 4 Abstract and Applied Analysis Proof. The linearized equation associated with the equilibrium x1 1 − √ 5 /2 ∈ −1, 0 is xn 1 − x1xn − x1xn−2 0. 3.1 Its characteristic polynomial is Px1 λ λ 3 − x1λ − x1. 3.2 Hence P ′ x1 λ 3λ 2 − 2x1λ λ 3λ − 2x1 . 3.3 Since Px1 −2 −8 − 5x1 < 0, Px1 −1 −1 − 2x1 √ 5 − 2 > 0, 3.4 there is a zero λ1 ∈ −2,−1 of Px1 . On the other hand, from Px1 0 −x1 > 0 and 3.3 , it follows that λ1 is a unique real zero of Px1 . Hence, the other two roots λ2,3 are conjugate complex. Since λ1|λ2| x1, 3.5 we obtain |λ2| |λ3| < 1. From this, the theorem follows. Theorem 3.2. The positive equilibrium of 1.1 , x2, is unstable. Moreover, it is also a hyperbolic equilibrium. Proof. The linearized equation associated with the equilibrium x2 1 √ 5 /2 ∈ 1, 2 is xn 1 − x2xn − x2xn−2 0. 3.6 Its characteristic polynomial is Px2 λ λ 3 − x2λ − x2. 3.7 We have P ′ x2 λ 3λ 2 − 2x2λ λ 3λ − 2x2 . 3.8