关于丢番图方程x^2+b^y=c^z的一个注记

A Note on the Diophantine Equation x^2-b^y=c^z

设正整数a=|m(m^4-10m^2+5)|,b=5m^4-10m^2+1,c=m^2+1,其中m是偶数,b是素数的方幂。利用Bilu,Hanrot and Voutiers关于本原素除子的深刻结果证明了:如果二次数域Q((-b)^(1/2))的类数是2的方幂,则丢番图方程x^2+b^y=c^z仅有正整数解(z,y,z)=(a,2,5)适合min(x,y,z)>1。

Let a=｜m（m^4-10m^2＋5）｜,b=5m^4-10m^2＋1,c=m^2＋1, where rn is a positive even integer and b is a prime power. We apply a deep result of Bilu, Hanrot and Voutier on primitive divisors to show that if the class number of quadratic field Q（√-b） is a power of 2, then, the Diophantine equation x^2＋b^y=c^z has only the positive integes solution （x, y, z） = （a, 2, 5） with min（x, y, z） 〉 1.