Can an odd perfect number be divisible by $825$?

I know that an odd perfect number cannot be divisible by $105$. I wonder if that's also the case for $825$.

Solution 1:

No it cannot.

Let $n=\prod p_i^{\alpha_i}=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots$ where $p_i$ is the $i$th prime number and let $S(n)$ be the sum of the divisors of $n$.

Suppose $n$ is an odd perfect number divisible by $825=3\cdot 5^2\cdot 11$, then $S(n)=2n$ and $\alpha_1=0,\alpha_2\ge 1,\alpha_3\ge 2,\alpha_5 \ge 1$.

Since $$ \begin{align} S(n)& = n\left(1+\frac{1}{3}+\cdots+\frac{1}{3^{\alpha_2}}\right) \left(1+\frac{1}{5}+\cdots+\frac{1}{5^{\alpha_3}}\right)\cdots \\ & = n \prod_i \left(\sum_{j=0}^{\alpha_i} p_i^{-j}\right) \end{align}$$

then we must have $\alpha_2\ge 2$, since $1+1/3 = 4/3$ but $S(n)=2n$ is not divisible by 4. Likewise $\alpha_4 \ge 2$ since $1+1/11=12/11$.

Then either $\alpha_2=2$, and since $1+1/3+1/9=13/9$ it must be that $13\mid n$. Or $\alpha_2>2$. Both cases lead to contradictions so there cannot be such an odd perfect number.

If $\alpha_2=2$ then $$ 2 = \frac{S(n)}{n} \ge \frac{13}{9} \left(1+\frac{1}{5}+\frac{1}{25}\right) \left(1+\frac{1}{11}+\frac{1}{121}\right)\left(1+\frac{1}{13}\right) > 2 $$ which cannot be, or if $\alpha_2>2$ $$ 2 = \frac{S(n)}{n} \ge \left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \left(1+\frac{1}{5}+\frac{1}{25}\right) \left(1+\frac{1}{11}+\frac{1}{121}\right) > 2 $$ which also cannot be.

This is adapted from this solution mentioned in the comments to the linked question.