论文标题
通过GCD的奇数完美数字的新方法
A new approach to odd perfect numbers via GCDs
论文作者
论文摘要
让$ q^k n^2 $是一个奇怪的完美数字，具有特殊的Prime $ Q $。 Define the GCDs $$G = \gcd\bigg(σ(q^k),σ(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,σ(n^2)\bigg)$$ and $$I = \gcd\bigg(n,σ(n^2)\bigg).$$ We prove that $G \times H = i^2 $。 （请注意，表明$ g \ mid i $和$ i \ mid h $都持有。之后，我们证明，如果$ g = h = i $，则$σ（q^k）/2 $不是平方的。其他自然和相关的结果将进一步得出。最后，我们猜想$$ \ mathscr {a} = \ {m：\ gcd（m，σ（m^2））= \ gcd（m^2，σ（m^2））\} $ $渐近密度零。
Let $q^k n^2$ be an odd perfect number with special prime $q$. Define the GCDs $$G = \gcd\bigg(σ(q^k),σ(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,σ(n^2)\bigg)$$ and $$I = \gcd\bigg(n,σ(n^2)\bigg).$$ We prove that $G \times H = I^2$. (Note that it is trivial to show that $G \mid I$ and $I \mid H$ both hold.) We then compute expressions for $G, H,$ and $I$ in terms of $σ(q^k)/2, n,$ and $\gcd\bigg(σ(q^k)/2,n\bigg)$. Afterwards, we prove that if $G = H = I$, then $σ(q^k)/2$ is not squarefree. Other natural and related results are derived further. Lastly, we conjecture that the set $$\mathscr{A} = \{m : \gcd(m,σ(m^2))=\gcd(m^2,σ(m^2))\}$$ has asymptotic density zero.
