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(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 $k$?

Jose Arnaldo Bebita Dris
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    You used an inequality $$\sigma\bigg({q^{(k+2)/2}\bigg(q^{(k-1)/2}{n^2}\bigg)}\bigg)\le \sigma\bigg({q^{(k+1)/2}}\bigg)\sigma\bigg({q^{(k-1)/2}}{n^2}\bigg)$$This inequality is true, but I would like to point out that we can have $$\sigma\bigg({q^{(k+1)/2}\bigg(q^{(k-1)/2}{n^2}\bigg)}\bigg)=\sigma\bigg({q^{(k+1)/2}}\bigg)\sigma\bigg({q^{(k-1)/2}}{n^2}\bigg)-\frac{q(q^{\frac{k-1}{2}}-1)(q^{\frac{k+1}{2}}-1)}{(q-1)^2}\sigma(n^2)$$ though I'm not sure if this helps. – mathlove Oct 17 '20 at 06:21
  • Thank you for your comment, @mathlove, but I do not follow. What identity/property are you using to derive the last equation in your comment? – Jose Arnaldo Bebita Dris Oct 17 '20 at 06:27
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    I've just added an explanation. – mathlove Oct 17 '20 at 06:37
  • I am accepting @mathlove's answer for now, since it introduces a new attack/approach to the problem. – Jose Arnaldo Bebita Dris Dec 01 '20 at 07:11

1 Answers1

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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 $$\begin{align}&\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg) \\\\&=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)\sigma({n^2})-\sigma(q^k)\sigma(n^2) \\\\&=\Bigg(\frac{q^{\dfrac{k+3}{2}}-1}{q-1}\cdot\frac{q^{\dfrac{k+1}{2}}-1}{q-1}-\frac{q^{k+1}-1}{q-1}\Bigg)\sigma(n^2) \\\\&=\frac{\Bigg(q^{\dfrac{k+3}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)-(q^{k+1}-1)(q-1)}{(q-1)^2}\sigma(n^2) \\\\&=\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)\end{align}$$

mathlove
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