论文标题
在Diophantine方程式上$（5pn^{2} -1）^{X}+（p（p-5）n^{2} +1）^{y} =（pn）^{z} $
On the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$
论文作者
论文摘要
令$ p $为$ p> 3 $，$ p \ equiv 3 \ pmod {4} $，让$ n $为正整数。 In this paper, we prove that the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $pn \equiv \pm1 \pmod 5$. As an another result, we show that the Diophantine equation $(35n^{2}-1)^{x}+(14n^{2}+1)^{y}=(7n)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $n\equiv \pm 3% \pmod{5}$ or $ 5 \中n $。在证据上，我们使用雅各比符号和贝克方法的属性。
Let $p$ be a prime number with $p>3$, $p\equiv 3\pmod{4}$ and let $n$ be a positive integer. In this paper, we prove that the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $pn \equiv \pm1 \pmod 5$. As an another result, we show that the Diophantine equation $(35n^{2}-1)^{x}+(14n^{2}+1)^{y}=(7n)^{z}$ has only the positive integer solution $(x,y,z)=(1,1,2)$ where $n\equiv \pm 3% \pmod{5}$ or $5\mid n$. On the proofs, we use the properties of Jacobi symbol and Baker's method.
