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
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.
