论文标题
在Lebesgue Ramanujan-Nagell方程式上
On a class of Lebesgue-Ramanujan-Nagell equations
论文作者
论文摘要
我们深入研究了diophantine方程$ cx^2+d^{2m+1} = 2y^n $ in Integers $ x，y \ geq 1，m \ geq 0 $和$ n \ geq 3 $，其中$ c $和$ c $和$ d $均给予coprime正整数，以使$ cd \ cd \ cd \ cd \ equiv equiv equiv equiv equiv 3 \ pmod 4 $ 4 $ pmod 4 $ pmod 4 $ 4。我们首先在条件$ n \ nmid h（-cd）$的情况下为prime $ n $求解此方程，其中$ h（-cd）$表示二次字段$ \ mathbb {q}（\ sqrt {-cd}}）$的二次字段$ \ mathbb {q}）$。然后，在假设$ \ gcd（n，h（-cd））= 1 $的假设下，我们将此方程式完全求解了$ c $和$ d $ primes。在条件$ \ gcd（n，h（-d））= 1 $的情况下，我们还以$ c = 1 $和$ d \ equiv1 \ pmod 4 $的方式完全求解了此方程。对于$ c $和$ d $的某些固定值，我们获得了有关该方程的解决性的一些结果。
We deeply investigate the Diophantine equation $cx^2+d^{2m+1}=2y^n$ in integers $x, y\geq 1, m\geq 0$ and $n\geq 3$, where $c$ and $d$ are given coprime positive integers such that $cd\not\equiv 3 \pmod 4$. We first solve this equation for prime $n$, under the condition $n\nmid h(-cd)$, where $h(-cd)$ denotes the class number of the quadratic field $\mathbb{Q}(\sqrt{-cd})$. We then completely solve this equation for both $c$ and $d$ primes under the assumption that $\gcd(n, h(-cd))=1$. We also completely solve this equation for $c=1$ and $d\equiv1 \pmod 4$, under the condition $\gcd(n, h(-d))=1$. For some fixed values of $c$ and $d$, we derive some results concerning the solvability of this equation.
