In what follows, I will let $\sigma$ denote the classical sum-of-divisors function.
Example: $\sigma(6)=1+2+3+6=12$
Let us call a number $x \in \mathbb{N}$ perfect if $\sigma(x)=2x$.
Let $M=(2^p - 1)\cdot{2^{p-1}}$ be an even perfect number.
Notice that $\sigma(2^p - 1) = 2^p$ and $\sigma(2^{p-1}) = 2^p - 1$ are both deficient numbers.
Now, let $N={q^k}{n^2}$ be an odd perfect number.
Here is my question:
Is it possible to prove that both $\sigma(q^k)$ and $\sigma(n^2)$ are deficient?
Here is my attempt:
Since $$\sigma(N)=\sigma(q^k)\sigma(n^2)=2N=2{q^k}{n^2}$$ and $N$ is perfect, then it follows that $\frac{1}{2}\sigma(q^k)\sigma(n^2)$ is perfect. Since $2 \nmid \sigma(n^2)$, this implies that $\sigma(n^2)$ is deficient since it is a factor of the (odd) perfect number $N=\frac{1}{2}\sigma(q^k)\sigma(n^2)$.
Now how about $\sigma(q^k)$? The most that I could prove is that $\frac{1}{2}\sigma(q^k)$ is deficient. (I am currently unable to prove, for example, that $\sigma(\sigma(q^k))<2\sigma(q^k)$.)
I could only prove though, that $$3/2 < \sigma(\sigma(q^k))/\sigma(q^k) \leq 3,$$ using $$a\sigma(b) \leq \sigma(ab) \leq \sigma(a)\sigma(b)$$ which holds for all $a, b \in \mathbb{N}$.