In this post we denote the sum of positive divisors of a natural $n\geq 1$ as $$\sigma(n)=\sum_{d\mid n}d.$$

We consider integers $m\geq 1$ and $k\geq 1$ where $m\equiv 1\text{ mod }2$ (thus $m$ is an odd integer) that are solutions $(m,k)$ of the equation $$\sigma(m)=105\cdot k.\tag{1}$$

Example. The first few solutions of $(1)$ satisfying previous requirements are $(m,k)=(377,4),(419,4),(695,8),(767,8),(779,8)$ and $(839,8)$.

Definition. We denote the set of all $m$'s being $(m,k)$ a solution as $\mathcal{M}$. That is

$$\mathcal{M}=\{\text{odd integers }m\geq 1\text{ s.t. }\sigma(m)=105k,\text{ for some integer }k\geq 1\}.$$

Question 1. Prove or refute these statements

1) If $m\in\mathcal{M}$ then $m$ is deficient (this last condition means that $\sigma(m)<2m$).

2) Let $$\mathcal{D}=\{\text{integers }m\in\mathcal{M}\text{ s.t. }\sigma(m)<2m\},$$ thus it denotes the set of odd integers $m\geq 1$ satisfying that $(m,k)$ is a solution of $(1)$ and that $m$ is a deficient number. Then $\mathcal{D}$ has infinitely many elements.

Many thanks.

I was inspired in my experiments to state the conjectures 1 and 2. As aside remark we can to prove that on assumption that $M$ is an odd perfect number, then doesn't safisfy $(1)$, that is $\sigma(M)=105k$ is an absurd (when we assume that there exists an odd perfect number $M$).

  • $\begingroup$ I add that there is an article from Wikipedia for Deficient number. $\endgroup$ – user243301 Jun 5 '18 at 15:11

$1)$ The smallest counterexample is $12\ 285$. There are many more.

$2)$ Every prime of the form $105k-1$ does the job, and there are infinite many such primes (Dirichlet's theorem)

| cite | improve this answer | |
  • 1
    $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 Jun 6 '18 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy