On a class of Lebesgue-Ramanujan-Nagell equations

Pub Date : 2023-12-14 DOI:10.1007/s10998-023-00564-z
Azizul Hoque
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引用次数: 0

Abstract

We deeply investigate the Diophantine equation \(cx^2+d^{2m+1}=2y^n\) in integers \(x, y\ge 1, m\ge 0\) and \(n\ge 3\), where c and d are coprime positive integers satisfying \(cd\not \equiv 3 \pmod 4\). We first solve this equation for prime n under the condition \(\gcd (n, h(-cd))=1\), where \(h(-cd)\) denotes the class number of the imaginary quadratic field \({\mathbb {Q}}(\sqrt{-cd})\). We then completely solve this equation for both c and d primes under the assumption \(\gcd (n, h(-cd))=1\). We also completely solve this equation for \(c=1\) and \(d\equiv 1 \pmod 4\) 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.

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关于一类 Lebesgue-Ramanujan-Nagell 方程
我们深入研究了整数 \(x, y\ge 1, m\ge 0\) 和 \(n\ge 3\) 中的二叉方程 \(cx^2+d^{2m+1}=2y^n\) ,其中 c 和 d 是满足 \(cd\not \equiv 3 \pmod 4\) 的共正整数。我们首先在质数 n 的条件下求解这个方程(\(gcd (n, h(-cd))=1),其中\(h(-cd)\)表示虚二次域\({\mathbb {Q}}(\sqrt{-cd})\) 的类数。然后,我们在假设 \(\gcd (n, h(-cd))=1\) 的条件下对 c 和 d 素数完全求解这个方程。在(\gcd (n, h(-d))=1)的条件下,我们还可以完全求解这个方程的(c=1)和(d\equiv 1 \pmod 4\ )。对于 c 和 d 的一些固定值,我们得出了一些关于这个方程可解性的结果。
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