Let A∈N,B∈Z with gcd(A,B)=1,B{-1,0,1}. For the binary recurrence (Lucas sequence) of the form u 0=0, u 1=1, u n+2 =Au n+1 +Bu n, let N 1(A,B,k) be the number of the terms n of |u n|=k, where k∈N. In this paper, usi...Let A∈N,B∈Z with gcd(A,B)=1,B{-1,0,1}. For the binary recurrence (Lucas sequence) of the form u 0=0, u 1=1, u n+2 =Au n+1 +Bu n, let N 1(A,B,k) be the number of the terms n of |u n|=k, where k∈N. In this paper, using a new result of Bilu, Hanrot and Voutier on primitive divisors, we proved that N 1(A,B,k)≤1 except N 1(1,-2,1)=5[n=1,2,3,5,13], N 1(1,-3,1)=3, N 1(1,-5,1)=3,N 1(1,B,1)=2(B{-2,-3,-5}), N 1(12,-55,1)=2, N 1(12,-377,1)=2, N 1(A,B,1)=2(A 2+B=±1, A>1), N 1(1,-2,3)=2, N 1(A,B,A)=2(A 2+2B=±1,A>1. For Lehmer sequence, we got a similar result. In addition, we also obtained some applications of the above results to some Diophantime equations.展开更多
We apply a new, deep theorem of Bilu, Hanrot ＆ Voutier and some fine results on the representation of the solutions of quadratic Diophantine equations to solve completely the exponential Diophantine equation x^2＋（3...We apply a new, deep theorem of Bilu, Hanrot ＆ Voutier and some fine results on the representation of the solutions of quadratic Diophantine equations to solve completely the exponential Diophantine equation x^2＋（3a^2-1）^m = （4a^2-1）^n when 3a^2-1 is a prime or a prime power.展开更多
Let a and b be positive integers, with a not perfect square and b > 1. Recently, He, Togband Walsh proved that the Diophantine equation x2-a((bk-1)/(b-1))2=1 has at most three solutions in positive integers. Moreov...Let a and b be positive integers, with a not perfect square and b > 1. Recently, He, Togband Walsh proved that the Diophantine equation x2-a((bk-1)/(b-1))2=1 has at most three solutions in positive integers. Moreover, they showed that if max{a,b} > 4.76·1051, then there are at most two positive integer solutions (x,k). In this paper, we sharpen their result by proving that this equation always has at most two solutions.展开更多
文摘Let A∈N,B∈Z with gcd(A,B)=1,B{-1,0,1}. For the binary recurrence (Lucas sequence) of the form u 0=0, u 1=1, u n+2 =Au n+1 +Bu n, let N 1(A,B,k) be the number of the terms n of |u n|=k, where k∈N. In this paper, using a new result of Bilu, Hanrot and Voutier on primitive divisors, we proved that N 1(A,B,k)≤1 except N 1(1,-2,1)=5[n=1,2,3,5,13], N 1(1,-3,1)=3, N 1(1,-5,1)=3,N 1(1,B,1)=2(B{-2,-3,-5}), N 1(12,-55,1)=2, N 1(12,-377,1)=2, N 1(A,B,1)=2(A 2+B=±1, A>1), N 1(1,-2,3)=2, N 1(A,B,A)=2(A 2+2B=±1,A>1. For Lehmer sequence, we got a similar result. In addition, we also obtained some applications of the above results to some Diophantime equations.
基金the Natural Science Foundation of Guangdong Province (04009801)the Important Science Research Foundation of Foshan University.
文摘We apply a new, deep theorem of Bilu, Hanrot ＆ Voutier and some fine results on the representation of the solutions of quadratic Diophantine equations to solve completely the exponential Diophantine equation x^2＋（3a^2-1）^m = （4a^2-1）^n when 3a^2-1 is a prime or a prime power.
基金the first two authors has been partially supported by a LEA Franco-Roumain Math-Mode projectPurdue University North Central for the support
文摘Let a and b be positive integers, with a not perfect square and b > 1. Recently, He, Togband Walsh proved that the Diophantine equation x2-a((bk-1)/(b-1))2=1 has at most three solutions in positive integers. Moreover, they showed that if max{a,b} > 4.76·1051, then there are at most two positive integer solutions (x,k). In this paper, we sharpen their result by proving that this equation always has at most two solutions.
