关于丢番图方程Dx^2+1=ca^n(英文)

On the Diophantine equation Dx^2+1 = ca^n

设c和a为正整数,D为与ca互素的正整数.记N(D;c,a)为方程Dx2+1=can的解(x,n)的个数,其中x及n是正整数.利用Nagell和Ljunggren的一个结果和Walker的一个结果,证明了:除N(2;1,3)=3,N(6;1,7)=N(7;1,2)=2和N(D;1,b2-1)=2,其中b>1为正整数且Ds2=b2-2,s为整数,均有N(D;1,a)≤1;除N(2;1,3):3,均有N(D;c,a)≤2.

Let c and a be positive integers, and let D be a positive integer coprime with ca. Denote by N(D;c,a) the number of solutions (x,n) of the equation Dx2 + 1 =can in positive integers x and n. By using a result of Nagell and Ljunggren and a result of Walker, it is shown that; N(D;1 ,a) ≤1 except for N(2;1,3) =3,N(6;1,7) = N(7;1,2) =2, and N(D;1,b2 -1) =2, where A > 1 is a positive integer and Ds2 =b2-2 for some integer s; N(D;c,a) ≤2 except for N(2; 1 ,3) =3.