Question

x,y,z和n都是正整数,n^x+n^y=n^z.

Answer 1

显然z>x,z>y,x和y是对称的,不防设x>=y

首先显然n=1时等式不成立

n^x+n^y=n^z=>n^(x-y)+1=n^(z-y) ----(1)

n^(z-y)=n^(x-y)+1>n^(x-y)

=>z-y>x-y又x,y,z均为正整数=>z-y>=x-y+1=>z-x>=1

=>n^(x-y)+1=n^(z-y)=>n^(x-y)*(n^(z-x)-1)=1=>n^(x-y)=1=n^(z-x)-1

=>x=y n^(z-x)=2=>z-x=1 n=2

故最后解为（x,x,x+1,2)x取任何正整数