On the Exponential Diophantine Equation $(a^n-1)(b^n-1)=x^2$

Let $a$ and $b$ be two distinct fixed positive integers such that $\min {a, b} > 1.$ First, we correct an oversight from \cite{X-Z}. Then, we show that the equation in the title with $b \equiv 3 (m o d 8)$ , $b$ prime and $a$ even has no solution in positive integers $n, x$ . This generalizes a result of Szalay \cite{L}.

On the Exponential Diophantine Equation $(a^{n} - 1) (b^{n} - 1) = x^{2}$

文献摘要

讨论

讲解博客

AI智能校验

On the Exponential Diophantine Equation $(a^{n} - 1) (b^{n} - 1) = x^{2}$

文献摘要

讨论

讲解博客

AI智能校验

On the Exponential Diophantine Equation (an−1)(bn−1)=x2(an−1)(bn−1)=x2

文献摘要

讨论

|讲解博客

|AI智能校验

On the Exponential Diophantine Equation (an−1)(bn−1)=x2(an−1)(bn−1)=x2

文献摘要

讨论

|讲解博客

|AI智能校验

On the Exponential Diophantine Equation $(a^{n} - 1) (b^{n} - 1) = x^{2}$

讲解博客

AI智能校验

On the Exponential Diophantine Equation $(a^{n} - 1) (b^{n} - 1) = x^{2}$

讲解博客

AI智能校验