THEOREM 4.1. Let l, k and t are both odd positive integers then for all B, a, a, b and c are positive real numbers, then Eq. (1) has a prime period two solution if a +B <1 and c(1- a - 3) < a. (7) Proof: First, suppose that there exists distinct nonnegative solution P and Q, such that ...Р, Q, Р, Q, .., is a prime period two solution of Eq.(1). We see from Eq. (1) when l, k and t are both odd, then xn+1 = xn=1 = Xn-k = xn-t = P. It follows Eq. (1) that P = BP+ aP + , and Q = BQ +aQ + Therefore, b (1— а - B) Р? + (с (1 —а - 3) — а) Р %3D0, (8) b(1 - a - B) Q + (c (1 - a - 3) - a) Q = 0, (9) Subtracting (9) from (8) gives P+Q= a-c(1-a-8) b(1-a-8) (10)

THEOREM 4.1. Let l, k and t are both odd positive integers then for all B, a, a, b and c are positive real numbers, then Eq. (1) has a prime period two solution if a +B <1 and c(1- a - 3) < a. (7) Proof: First, suppose that there exists distinct nonnegative solution P and Q, such that ...Р, Q, Р, Q, .., is a prime period two solution of Eq.(1). We see from Eq. (1) when l, k and t are both odd, then xn+1 = xn=1 = Xn-k = xn-t = P. It follows Eq. (1) that P = BP+ aP + , and Q = BQ +aQ + Therefore, b (1— а - B) Р? + (с (1 —а - 3) — а) Р %3D0, (8) b(1 - a - B) Q + (c (1 - a - 3) - a) Q = 0, (9) Subtracting (9) from (8) gives P+Q= a-c(1-a-8) b(1-a-8) (10)

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.2: Linear Independence, Basis, And Dimension

Problem 61EQ

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