关于Diophantine方程Dx^2+2^(2m)=y^n

On the diophantine equation Dx^2+2~2m=y^n

设D是无平方主因子正整数,h是判别式等于-4D的二元二次原型的类数.本文解决了方程Dx2+22m=Yn,x,y,m,n ∈N,gcd(x,y)=1,n>2,gcd(n,h)=1求解问题.

Let D be a positive integer with squarefree. Let h denote the class number of primitive binary quadratic torms of discriminant - 4D. In this paper , the equation Dx^2 ＋ 2^m = y^n,, x, y, m, n ∈N, gcd （ x ,y ） =1 , n 〉 2 , gcd （ n , h ） = 1 is solved.