# 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 $825$.

• On the contrary, every odd perfect number is divisible by 825. – MJD May 11 '13 at 20:54
• Those who vote to reopen should probably bother with editing the question into something which is an actual question, and not a political reference by the former user[s] involved in this thread (which are now deleted). – Asaf Karagila May 12 '13 at 0:15
• The edit was "minor" – Stahl May 12 '13 at 23:17
• Is this a coincidence, or is it bannable to ask about odd perfect numbers? – user77400 May 16 '13 at 19:52
• I do not think that it is bannable to ask about odd perfect numbers here at MSE. See here for some of my own questions and some of my own answers to other users' questions on odd perfect numbers. – Jose Arnaldo Bebita-Dris Nov 7 '18 at 7:11

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.