%0 Journal Article
%T 关于丢番图方程(an)x+(bn)y=(cn)z<br>On the Diophantine Equation (an)x
%A 孙翠芳
%A 汤　敏
%J 数学年刊（A辑）
%D 2018
%R 10.16205/j.cnki.cama.2018.0009
%X 设$n,a,b,c$是正整数, $\gcd(a,b,c)=1,\ a,b\geqslant 3$, 且丢番图方程$a^{x}+b^{y}=c^{z}$只有正整数解$(x,y,z)=(1,1,1)$. 证明了若$(x,y,z)$是丢番图方程$(an)^{x}+(bn)^{y}=(cn)^{z}$ 的正整数解且$(x,y,z)\neq (1,1,1)$, 则$y<br>Let $n,a,b,c$ be positive integers with $\gcd(a,b,c)=1,\ a,b\geqslant 3$ and the Diophantine equation $a^{x}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(1,1,1)$. In this paper, the authors prove that if $(x,y,z)$ is a positive integer solution of the Diophantine equation $(an)^{x}+(bn)^{y}=(cn)^{z}$ with $(x,y,z)\neq (1,1,1)$, then $y
%K Je'{s}manowicz猜想
%K 丢番图方程
%K Fibonacci序列&lt
%K br&gt
%K Je'{s}manowicz' conjecture
%K Diophantine equation
%K Fibonacci sequence
%U http://www.camath.fudan.edu.cn/camacn/ch/reader/view_abstract.aspx?file_no=39A109&flag=1