(This question is an offshoot of this earlier post.)
Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x)=\sigma(x)-x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $\sigma(m)=2m$ and $m$ is odd, then $m$ is called an odd perfect number. It is currently unknown whether there are any odd perfect numbers, although it is widely believed that there is none.
Euler proved that an odd perfect number $m$, if one exists, must have the so-called Eulerian form
$$m = q^k n^2$$
where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since the divisor sum function $\sigma$ is a multiplicative function and $m = q^k n^2$ is perfect, we obtain
$$2 q^k n^2 = 2m=\sigma(m)=\sigma(q^k n^2)=\sigma(q^k)\sigma(n^2)$$
which implies that
$$2 = \dfrac{\sigma(m)}{m} = \dfrac{\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)}{{q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}} \leq \dfrac{\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)}{q^{\dfrac{k+1}{2}}}\cdot\dfrac{\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)}{q^{\dfrac{k-1}{2}}{n^2}} = I\bigg(q^{\dfrac{k+1}{2}}\bigg)I(d),$$
where
$$d = q^{\dfrac{k-1}{2}}{n^2}.$$
We now compute
$$\dfrac{d}{D(d)} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} – \dfrac{\bigg(q^{\dfrac{k+1}{2}} – 1\bigg)\sigma(n^2)}{q – 1}} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} – \dfrac{\sigma(q^k)\sigma(n^2)}{q^{\dfrac{k+1}{2}} + 1}} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} – \dfrac{2q^k n^2}{q^{\dfrac{k+1}{2}} + 1}},$$
which simplifies to
$$\dfrac{d}{D(d)} = \dfrac{q^{\dfrac{k+1}{2}} + 1}{2\Bigg(q^{\dfrac{k+1}{2}} + 1\Bigg) – 2\Bigg(q^{\dfrac{k+1}{2}}\Bigg)} = \dfrac{q^{\dfrac{k+1}{2}} + 1}{2}.$$
This confirms a theorem of Holdener and Rachfal. (That is, the proper divisor
$$d = q^{\dfrac{k-1}{2}}{n^2}$$
of an odd perfect number
$$m = q^k n^2$$
must be deficient-perfect.)
So now we have the inequality
$$\dfrac{2}{I\bigg(q^{\dfrac{k+1}{2}}\bigg)} \leq I(d),$$
which implies that
$$\dfrac{D(d)}{d} = 2 – I(d) \leq 2 – \dfrac{2}{I\bigg(q^{\dfrac{k+1}{2}}\bigg)},$$
from which it follows that
$$\dfrac{q^{\dfrac{k+1}{2}} + 1}{2} = \dfrac{d}{D(d)} \geq \dfrac{I\bigg(q^{\dfrac{k+1}{2}}\bigg)}{2\Bigg(I\bigg(q^{\dfrac{k+1}{2}}\bigg) – 1\Bigg)}.$$
We therefore have to solve the resulting inequality
$$q^{\dfrac{k+1}{2}} + 1 \geq \dfrac{I\bigg(q^{\dfrac{k+1}{2}}\bigg)}{I\bigg(q^{\dfrac{k+1}{2}}\bigg) – 1} = \dfrac{\dfrac{q^{\dfrac{k+3}{2}} – 1}{q^{\dfrac{k+1}{2}} \bigg(q – 1\bigg)}}{\Bigg(\dfrac{q^{\dfrac{k+3}{2}} – 1}{q^{\dfrac{k+1}{2}} \bigg(q – 1\bigg)}\Bigg) – 1}$$
which implies that
$$\Bigg(q^{\dfrac{k+1}{2}} + 1\Bigg)\cdot\Bigg(q^{\dfrac{k+1}{2}} – 1\Bigg) \geq q^{\dfrac{k+3}{2}} – 1$$
from which it follows that
$$q^{k+1} – 1 \geq q^{\dfrac{k+3}{2}} – 1 \implies q^{k+1} \geq q^{\dfrac{k+3}{2}} \implies q^{2k+2} \geq q^{k+3} \implies 2k+2 \geq k+3 \implies k \geq 1.$$
INQUIRY
Is it possible to improve on the approach/attack as outlined in this post to hopefully produce a nontrivial lower bound for $k$?
Best Answer
Too long to comment :
You used an inequality $$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)\le \sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)$$
This inequality is true, but we can have $$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)$$ $$=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)$$ since we have $$\begin{align}&\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg) \\\\&=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)\sigma({n^2})-\sigma(q^k)\sigma(n^2) \\\\&=\Bigg(\frac{q^{\dfrac{k+3}{2}}-1}{q-1}\cdot\frac{q^{\dfrac{k+1}{2}}-1}{q-1}-\frac{q^{k+1}-1}{q-1}\Bigg)\sigma(n^2) \\\\&=\frac{\Bigg(q^{\dfrac{k+3}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)-(q^{k+1}-1)(q-1)}{(q-1)^2}\sigma(n^2) \\\\&=\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)\end{align}$$