Can an odd perfect number be divisible by $165$?

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 $165$.