Let $n$ be a natural number. Let $U_n = \{d \in \mathbb{N}\mid d\mid n \text{ and } \gcd(d,n/d)=1 \}$ be the set of unitary divisors, $D_n$ be the set of divisors and $S_n=\{d \in \mathbb{N}\mid d^2 \mid n\}$ be the set of square divisors of $n$.

The set $U_n$ is a group with $a\oplus b := \frac{ab}{\gcd(a,b)^2}$. It operates on $D_n$ via:

$$ u \oplus d := \frac{ud}{\gcd(u,d)^2}$$

The orbits of this operation "seem" to be

$$ U_n \oplus d = d \cdot U_{\frac{n}{d^2}} \text{ for each } d \in S_n$$

From this conjecture it follows (also one can prove this directly since both sides are multiplicative and equal on prime powers):

$$\sigma(n) = \sum_{d\in S_n} d\sigma^*(\frac{n}{d^2})$$

where $\sigma^*$ denotes the sum of unitary divisors.

Since $\sigma^*(k)$ is divisible by $2^{\omega(k)}$ if $k$ is odd, where $\omega=$ counts the number of distinct prime divisors of $k$, for an odd perfect number $n$ we get (Let now $n$ be an odd perfect number):

$$2n = \sigma(n) = \sum_{d \in S_n} d \sigma^*(\frac{n}{d^2}) = \sum_{d \in S_n} d 2^{\omega(n/d^2)} k_d $$

where $k_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$ are natural numbers. Let $\hat{d}$ be the largest square divisor of $n$. Then: $\omega(n/d^2)\ge \omega(n/\hat{d}^2)$.

Hence we get:

$$2n = 2^{\omega(n/\hat{d}^2)} \sum_{d \in S_n} d l_d$$ for some natural numbers $l_d$.

If the prime $2$ divides not the prime power $2^{\omega(n/\hat{d}^2})$, we must have $\omega(n/\hat{d}^2)=0$ hence $n=\hat{d}^2$ is a square number, which is in contradiction to Eulers theorem on odd perfect numbers.

So the prime $2$ must divide the prime power $2^{\omega(n/\hat{d}^2})$ and we get:

$$n = 2^{\omega(n/\hat{d}^2)-1} \sum_{d \in S_n} d l_d$$

with $l_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$. Hence the odd perfect number, satisifies:

$$n = \sum_{d^2\mid n} d \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}=:a(n)$$

Hence an odd perfect number satisifies:

$$n = a(n)$$

So my idea was to study the function $a(n)$, which is multiplicative on odd numbers, on the right hand side and what properties it has to maybe derive insights into odd perfect numbers.

The question is if it ever can happen that an odd number $n$ satisfies: $n=a(n)$? (checked for $n=2k+1$ and $1 \le k \le 10^7$)

Edit: Conjecture: For all odd $n \ge 3$ we have $a(n)<n$. This would prove that there exists no odd perfect number.

This conjecture could be proved as follows: Since $a(n)$ is multiplicative, it is enough to show that for an odd prime power $p^k$ we have

$$a(p^k) < p^k$$

The values of $a$ at prime powers are not difficult to compute and they are:

$$a(p^{2k+1})= \frac{p^{2(k+1)}-1}{2(p-1)}$$


$$a(p^{2k}) = \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}$$

However, I am not very good at proving inequalities, so:

If someone has an idea how to prove the following inequalities for odd primes $p$ that would be very nice:

$$p^{2k+1} > \frac{p^{2(k+1)}-1}{2(p-1)}, \text{ for all } k \ge 0$$


$$p^{2k} > \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}, \text{ for all } k \ge 1$$

Thanks for your help!

The inequalities have been proved here: https://math.stackexchange.com/questions/3807399/two-inequalities-for-proving-that-there-are-no-odd-perfect-numbers

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    $\begingroup$ I think this question should be broken up into separate questions. The final one is not appropriate for MO, which is not for proof verification (MSE would be better) - I'm also not sure what statement 'the proof' is trying to prove? $\endgroup$ Commented Aug 27, 2020 at 13:43
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    $\begingroup$ Yes, I know, I meant what relevance do they have to the question. Do you mean to take $n$ to be an odd perfect number throughout the question, even the very first part? if so you should state this. $\endgroup$ Commented Aug 27, 2020 at 14:04
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    $\begingroup$ In the equation defining $a(n)$, shouldn't the denominator be $2^{\omega(n/\hat{d}^2)}$? $\endgroup$ Commented Aug 29, 2020 at 16:49
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    $\begingroup$ It does matter. As written, that specific equation is false even for odd perfect numbers. $\endgroup$ Commented Aug 29, 2020 at 16:59
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    $\begingroup$ @GjergjiZaimi gave a good answer below. But I want to note something more general: The group operation in this case corresponds directly to the symmetric difference operation on sets where we are thinking of a number n corresponding to the set S of prime powers $p^a$ such that $p^a ||n$ . So there's little actual number theoretic content here, and we shouldn't expect that thinking about this group operation will help us specifically with the problem of understanding odd perfect numbers. $\endgroup$
    – JoshuaZ
    Commented Aug 29, 2020 at 18:35

1 Answer 1


Here are some general comments:

  1. You don't need to bring these actions of abelian groups on various sets of divisors. The identity $$\sigma(n)=\sum_{d^2|n}d\sigma^{*}(\frac{n}{d^2})$$ is easy to check directly, without appeal to anything fancy.

  2. Let's call $\alpha(n)$ the number of prime divisors of $n$ which appear with an odd exponent in the factorization of $n$. This is what you call $\omega(n/\hat{d}^2)$. You are right in observing that $2^{\alpha(n)}$ divides $\sigma(n)$. This is where Euler's result comes from: If $n$ is an odd perfect number then $\alpha(n)=1$.

  3. It seems you want to define a new function $a(n)=\frac{\sigma(n)}{2^{\alpha(n)}}$, and you conjecture that $$a(n)<n$$ for all odd numbers $n$. If true this conjecture would imply that there are no odd perfect numbers. Unfortunately it is false. For example the inequality is reversed at $n=3^35^2 7^2$.

  • $\begingroup$ Faleminderit Gjergji. You are right, one could prove the first equality without the group, but how do you come up with such an equality without the group structure in the first place? $\endgroup$
    – user6671
    Commented Aug 29, 2020 at 17:57
  • $\begingroup$ However, it seems that $a(n)< n$ whenever is not of the form $m^2$ or $pm^2$ for a prime $p$ not necessary $\gcd(p,m)=1$. If $n$ is of the latter form it seems that $a(n)>n$. $\endgroup$
    – user6671
    Commented Aug 29, 2020 at 19:29
  • $\begingroup$ It's false in that case too, but the counterexamples are larger. $\endgroup$ Commented Aug 29, 2020 at 19:34
  • $\begingroup$ Can you give a counterexample? $\endgroup$
    – user6671
    Commented Aug 29, 2020 at 19:37
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    $\begingroup$ There exist numbers $n$ with $\sigma(n)/n$ arbitrarily large. Moreover $\sigma(n)/n$ increases when you increase the exponent of any of the prime divisors of $n$. Together these two facts imply that $a(n)<n$ is violated infinitely often for any value of $\alpha(n)$. $\endgroup$ Commented Aug 29, 2020 at 19:41

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