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The topic of odd perfect numbers likely needs no introduction.

I would like to revisit these two questions:

Is it possible to improve on the bound $D(q^k) < \varphi(q^k)$ if $k>1$?

On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$

Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, $\varphi(x)$ is the Euler totient function of $x$, and $I(x)=\sigma(x)/x$ is the abundancy index of $x$.

In the OP of the second question, the following lower bound is derived: $$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2).$$

In the accepted answer to the first question, MSE user mathlove gives the following improved lower bound, under the assumption that $k>1$: $$3 - \bigg(\frac{q-1}{q^2}\bigg) + \bigg(\frac{q^4 - 1}{q^{k+1} (q - 1)}\bigg) = 3 - \frac{1}{q^{k+1}}\bigg(\varphi(q^k) - \frac{q^4 - 1}{q - 1}\bigg) < I(q^k) + I(n^2).$$

Here are my two questions:

(1) How does one prove the improved lower bound given by mathlove?

(2) Is the improved lower bound given by mathlove already "best-possible", in the sense of being obtainable via the $D(q^k)$-route?

(Note that I am already aware of the lower bound $$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2),$$ which does not answer Question (2), because while this latter lower bound improves on mathlove's result, to the best of my knowledge it is not obtainable via the $D(q^k)$-route.)

Jose Arnaldo Bebita Dris
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1 Answers1

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This is an answer to the question (1).

Let $$f(k):=\varphi(q^k)-D(q^k)=q^{k-1}(q-1)-\bigg(2q^k-\frac{q^{k+1}-1}{q-1}\bigg)=\frac{q^{k-1}-1}{q-1}$$ Then, we see that $f(k)$ is increasing.

So, if $k\gt 1$ and $k\equiv 1\pmod 4$, then we have $f(k)\ge f(5)=\dfrac{q^{4}-1}{q-1}$ which is equivalent to $$D(q^k)\le\varphi(q^k)-\frac{q^4-1}{q-1}$$ Using $$D(n^2) < \frac{2n^2}{q}$$ we get $$D(q^k)D(n^2) < \frac{2n^2}{q}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$ Dividing the both sides by $2q^k n^2$, we obtain $$\frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$ which can be written as $$3 - \bigg(I(q^k) + I(n^2)\bigg) < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$

We therefore finally have $$3 - \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg) < I(q^k) + I(n^2) $$

mathlove
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