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arXiv 2109.13652math.NT
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关于量$m^2-p^k$(其中$p^k m^2$是奇完满数)--第II部分

On the quantity $m^2-p^k$ where $p^k m^2$ is an odd perfect number -- Part II

Jose Arnaldo Bebita Dris, Immanuel Tobias San Diego

摘要:

设$p^k m^2$是具有特殊素数$p$的奇完满数.推广了作者以前的工作,我们证明了不等式$m < p^k$是由$m^2 - p^k = 2^r t$导出的,其中$r \geq 2$和$\gcd(2,t)=1$在下列假设下成立:(a)$m > t > 2^r$,或(b)$m > 2^r > t$ .我们还证明了估计$m^2 - p^k > 2m$成立。我们还可以将该无条件估计改进为$m^2 - p^k > {313m^2}/315$。

英文摘要:

Let $p^k m^2$ be an odd perfect number with special prime $p$. Extending previous work of the authors, we prove that the inequality $m < p^k$ follows from $m^2 - p^k = 2^r t$, where $r \geq 2$ and $\gcd(2,t)=1$, under the following hypotheses: (a) $m > t > 2^r$, or (b) $m > 2^r > t$. We also prove that the estimate $m^2 - p^k > 2m$ holds. We can also improve this unconditional estimate to $m^2 - p^k > {313m^2}/315$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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