# Two inequalities for proving that there are no odd perfect numbers?

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). 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

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

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

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 $$k \ge 1$$