4
$\begingroup$

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

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

Hence an odd perfect number satisifies:

$$n = a(n)$$

Edit: This equation is wrong for odd perfect numbers.

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.

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:

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

and

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

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:

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

and

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

Thanks for your help!

$\endgroup$
9
  • $\begingroup$ Are you sure about first inequality. That's true for all $p \ge 2$ $\endgroup$ – openspace Aug 29 '20 at 15:57
  • $\begingroup$ @openspace why is it true? $\endgroup$ – user276611 Aug 29 '20 at 15:59
  • $\begingroup$ Because $p^{2k+1}2(p-1) - p^{2k+2} + 1 = (p-2)p^{2k+1} + 1$ $\endgroup$ – openspace Aug 29 '20 at 16:01
  • $\begingroup$ And the second equals to $(p^k-1)((p-2)p^k-1)$ $\endgroup$ – openspace Aug 29 '20 at 16:06
  • $\begingroup$ @openspace ok that was easy. what about the second one? $\endgroup$ – user276611 Aug 29 '20 at 16:08
0
$\begingroup$
  1. $p^{2k+1} > \dfrac{p^{2(k+1)} - 1}{2(p-1)}$ equals to $(p-2)p^{2k+1} + 1 > 0$ for $p \ge 2$ and $k \ge 0$

  2. $p^{2k} > \dfrac{p^{2k+1} + p^{k+1} - p ^ k - 1}{2(p-1)}$ equals to $(p^k-1)((p-2)p^k-1) > 0$ for $p > 2$ and $k \ge 1$

$\endgroup$
1
  • $\begingroup$ thanks for your help $\endgroup$ – user276611 Aug 29 '20 at 16:16

Your Answer

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