I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $165$ (this is actually a stronger statement).

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    $\begingroup$ Related $\endgroup$ – MJD May 23 '13 at 20:11
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – user77400 May 23 '13 at 20:48
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    $\begingroup$ Are you going to continue by this fashion and ask infinitely many questions? $\endgroup$ – Asaf Karagila May 25 '13 at 23:06
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    $\begingroup$ @AsafKaragila Don't ask me, I am only an editor. Besides, is there a problem with posting generalizations of previous questions? $\endgroup$ – Kortlek May 25 '13 at 23:25

An odd perfect number, if one exists, could in principle be divisible by 165.

Let $\sigma(n)$ denote the sum of the positive divisors of $n$, and let the "abundancy" be $a(n)=\sigma(n)/n$. A perfect number is then defined by $\sigma(n)=2n$, or equivalently $a(n)=2$.

Let $$n = \prod_{i=1}^k q_i^{\alpha_i}$$ be the prime factorization of $n$, where the $q_i$ are distinct primes. Then \begin{align} \sigma(n) &= \sum_{d|n} d\\ &= \prod_{i=1}^k \sum_{\alpha=0}^{ \alpha_i } q_i^{\alpha}\\ &= \prod_{i=1}^k \frac{q_i^{\alpha_i+1}-1}{q_i-1} \end{align} So \begin{align} a(n) &= \prod_{i=1}^k \frac{q_i^{\alpha_i+1}-1}{ q_i^{\alpha_i}(q_i-1)}\\ &= \prod_{i=1}^k h(q_i, \alpha_i), \end{align} where $$ h(q, \alpha) = \frac{q^{\alpha+1}-1}{ q^{\alpha}(q-1)}. $$

We note that $h(q,\alpha)>1\ \forall\ q\geq2\textrm{ and }\alpha>0$, and that $h(q,\alpha)$ is an increasing function of $\alpha$ for fixed $q$.

Suppose $n$ is an odd perfect number. Then $n = p^e m^2$ for some prime $p$ with $p\equiv e \equiv 1\ (\operatorname{mod} 4)$ and $\operatorname{gcd}(p,m)=1$. (This is a well-known theorem due to Euler.) Thus, if $105 | n$ then $3^2\cdot5\cdot7^2\ |\ n$, which would imply $a(n) > h(3,2)h(5,1)h(7,2) = 2.01\ldots$, a contradiciton. This is the reason that no odd perfect number can be divisible by $105$: any odd perfect number $n$ divisible by $105$ would, by Euler, be divisible by $3^2\cdot5\cdot7^2$ and would therefore be "abundant", i.e. $a(n)>2$ for such $n$, a contradiction. The same argument gives $a(n)>1.905$ for $n$ divisible by $165$. There is nothing wrong with this statement; it does not lead to a contradiction.

It's difficult to find a reference that something has not been proven to be impossible. However, in the normal machinations to show that a prospective odd perfect number can't have a certain divisibility property, there's nothing obvious to say that one couldn't be divisible by $165$.

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    $\begingroup$ Since $3^2\,||\,n \implies (1+3+3^2)\,|\,n$, we can conclude that $3^4\,|\,n$; this improves the lower bound to $1.97$. However, we can only go so far, as $3^\infty\cdot5\cdot11^\infty\,|\,n \implies \frac{\sigma(n)}{n}\ge\frac{3}{2}\cdot\frac{6}{5}\cdot\frac{11}{10}=1.98$, which is not enough. Nevertheless, we still have not used the fact that the Eulerian factor $p^e$ of $n$ must be $5^1$, which is a strong restriction, so we might be able to further improve the bound. But how? $\endgroup$ – Librecoin May 29 '13 at 21:56
  • $\begingroup$ @Tharsis That is an interesting line of reasoning. Right away this gives $3^4 | n$, $5 || n$, and $11^4 | n$, so $a(n) \geq 6442/3267 = 1.97\ldots$. I don't immediately see how to push the exponents on $3$ and $11$ higher than $4$; but regardless, as you point out, using $3$ and $11$ only we can't get higher than $1.98$. $\endgroup$ – Douglas B. Staple May 30 '13 at 2:23
  • $\begingroup$ I would be pleasantly surprised if someone does in fact manage to prove that $165\nmid n$, but my instinct is that this isn't going to happen. $\endgroup$ – Douglas B. Staple May 30 '13 at 2:27

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