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Let $\sigma(x)$ denote the sum of the divisors of $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.

If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number. Euler showed that an odd perfect number, if one exists, must necessarily have the so-called Eulerian form $N=q^k n^2$, where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $k \equiv 1 \pmod 4$ and $q$ is prime, we obtain $$\sigma(q^k) = \frac{q^{k+1}-1}{q-1} = \bigg(\frac{q^{\frac{k+1}{2}}-1}{q-1}\bigg)\bigg(q^{\frac{k+1}{2}}+1\bigg).$$

But we can rewrite the first factor as $$\frac{q^{\frac{k+1}{2}}-1}{q-1} = \frac{q^{\frac{k-1}{2}+1}-1}{q-1} = \sigma(q^{\frac{k-1}{2}}),$$ so that we have $$\sigma(q^k) = \sigma(q^{\frac{k-1}{2}})\bigg(q^{\frac{k+1}{2}}+1\bigg).$$

Note that the divisor-sum function $\sigma$ satisfies $\sigma(yz) \leq \sigma(y)\sigma(z)$, where equality holds if and only if $\gcd(y,z)=1$. Setting $y=z=q^{\frac{k-1}{2}}$ and noting that $\gcd(y,z)=q^{\frac{k-1}{2}}>1$ (unless $k=1$), we obtain $$\sqrt{\sigma(q^{k-1})} < \sigma(q^{\frac{k-1}{2}}).$$

Now, it is known that $$\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}$$ from which it follows that $$\frac{D(n^2)}{\sigma(q^{k-1})}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k}=\frac{2n^2}{\sigma(q^k)},$$ whereby we obtain $$\frac{2n^2}{D(n^2)}=\frac{\sigma(q^k)}{\sigma(q^{k-1})}=\frac{\sigma(q^{\frac{k-1}{2}})\bigg(q^{\frac{k+1}{2}}+1\bigg)}{\sigma(q^{k-1})}>\frac{\sqrt{\sigma(q^{k-1})}\bigg(q^{\frac{k+1}{2}}+1\bigg)}{\sigma(q^{k-1})}=\frac{q^{\frac{k+1}{2}}+1}{\sqrt{\sigma(q^{k-1})}}.$$ This implies that $$\frac{n^2}{D(n^2)}>\frac{q^{\frac{k+1}{2}}+1}{2\sqrt{\sigma(q^{k-1})}}.$$

We now claim the following proposition.

If $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \leq \frac{2q}{q+1}.$$ Equality holds if and only if $k=1$.

Proof

$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=2q^k n^2$$ $$\frac{q+1}{q}=\frac{\sigma(q)}{q} \leq \frac{\sigma(q^k)}{q^k}=\frac{2n^2}{\sigma(n^2)}$$ implying $$\frac{\sigma(n^2)}{n^2} \leq \frac{2q}{q+1}.$$ Equality holds if and only if $k=1$.

From the inequality $$\frac{\sigma(n^2)}{n^2} \leq \frac{2q}{q+1}$$ we obtain $$(q+1)\sigma(n^2) \leq 2qn^2$$ which implies that $$2n^2 \leq 2qn^2 + 2n^2 - q\sigma(n^2) - \sigma(n^2) = {2n^2}(q+1) - (q+1)\sigma(n^2) = (q+1)(2n^2 - \sigma(n^2)) = (q+1)D(n^2)$$ from which it follows that $$\frac{n^2}{D(n^2)} \leq \frac{q+1}{2}.$$ Equality holds if and only if $k=1$.

Now assume that $k>1$. It follows from our previous considerations in this post that $$\frac{q+1}{2}>\frac{n^2}{D(n^2)}>\frac{q^{\frac{k+1}{2}}+1}{2\sqrt{\sigma(q^{k-1})}}$$ which implies that $$\sqrt{\sigma(q^{k-1})}>\frac{q^{\frac{k+1}{2}}+1}{q+1}$$ from which it follows that $$\frac{q^k - 1}{q - 1}=\sigma(q^{k-1})>\bigg(\frac{q^{\frac{k+1}{2}}+1}{q+1}\bigg)^2.$$ Since $k>1$ and $k \equiv 1 \pmod 4$, then the lowest possible value of $k$ is $5$. Plugging this as a test value, we obtain $$\frac{q^5 - 1}{q - 1}=q^4 + q^3 + q^2 + q + 1$$ and $$\bigg(\frac{q^3 + 1}{q + 1}\bigg)^2 = (q^2 - q + 1)^2 = q^4 - 2q^3 + 3q^2 - 2q + 1$$ so that we get $$q^4 + q^3 + q^2 + q + 1 > q^4 - 2q^3 + 3q^2 - 2q + 1$$ which simplifies to $$3q^3 - 2q^2 + 3q = q(q(3q - 2) + 3) > 0$$ which is trivially true as it is known that $q \geq 5$ (since $q$ is a prime satisfying $q \equiv 1 \pmod 4$). Indeed, by a lengthy (but doable) computation, it is possible to show that the inequality $$\frac{q^k - 1}{q - 1}=\sigma(q^{k-1})>\bigg(\frac{q^{\frac{k+1}{2}}+1}{q+1}\bigg)^2$$ is trivial.

Here is my question:

The crux of the inequalities in this post is $$\sqrt{\sigma(q^{k-1})} < \sigma(q^{\frac{k-1}{2}}).$$ Will it be possible to improve on this inequality, by a similar reasoning for the derivation of the inequality $$\bigg(\frac{\sigma(n^2)}{n^2}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} < \frac{\sigma(n)}{n},$$ as detailed in this MSE question?

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  • $\begingroup$ I conjecture that $$\bigg(\sigma(q^{k-1})\bigg)^{\frac{\ln(6/5)}{\ln(31/25)}} < \sigma(q^{\frac{k-1}{2}})$$ but I have no proof. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 9 at 12:34
  • $\begingroup$ If this conjecture is true, then the resulting inequality is $$\frac{q^k - 1}{q - 1} > \bigg(\frac{q^{\frac{k+1}{2}+1}}{q+1}\bigg)^{\frac{\ln(31/25)}{\ln(31/30)}}$$ where the exponent $$\frac{\ln(31/25)}{\ln(31/30)} \approx 6.5603093$$ I am not 100% sure, but I think this last inequality will lead to a contradiction, thereby yielding a proof of the Descartes-Frenicle-Sorli Conjecture that $k=1$, if $q^k n^2$ is an odd perfect number with special/Euler prime $q$. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 9 at 13:08
  • $\begingroup$ The conjecture is false. The following counterexample was communicated to me by Pascal Ochem. (He adds that the inequality cannot be improved much.) With $q = 40009$ and $k = 5$: $$(q^4+q^3+q^2+q+1)^{1/2} = 1600740085.875$$ $$q^2+q+1 = 1600760091$$ $$(q^4+q^3+q^2+q+1)^{\log(6/5)/\log(31/25)} = 4004779171150154.163$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 10 at 1:35
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The conjecture in the comments is false. The following counterexample was communicated to me by Pascal Ochem. (He adds that the inequality $$\sqrt{\sigma(q^{k-1})} < \sigma(q^{\frac{k-1}{2}})$$ cannot be improved much.)

With $q = 40009$ and $k = 5$: $$(q^4+q^3+q^2+q+1)^{1/2} = 1600740085.875$$ $$q^2+q+1 = 1600760091$$ $$(q^4+q^3+q^2+q+1)^{\log(6/5)/\log(31/25)} = 4004779171150154.163$$

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