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 $a(n)<n$. 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!

*

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


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