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(Note: This post is an offshoot of this earlier MSE question.)

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Define the abundancy index $$I(x)=\frac{\sigma(x)}{x}$$ where $\sigma(x)$ is the classical sum of divisors of $x$.

Since $q$ is prime, we have the bounds $$1 < \frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1} \leq \frac{5}{4},$$ which implies, since $N$ is perfect, that $$\frac{8}{5} \leq \frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1} < 2.$$

By considering the nonnegative product $$\bigg(I(q^k) - \frac{q+1}{q}\bigg)\bigg(I(n^2) - \frac{q+1}{q}\bigg) \geq 0,$$ then after some routine algebraic manipulations, we arrive at the upper bound $$I(q^k) + I(n^2) \leq 3 - \Bigg(\frac{q-1}{q(q+1)}\Bigg) = \frac{3q^2 + 2q + 1}{q(q + 1)}.$$

Here, I shall derive the following slightly different bound:

If $q^k n^2$ is an odd perfect number with special prime $q$, then $$I(q^k) + I(n^2) \leq \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}.$$

PROOF: Suppose that $q^k n^2$ is an odd perfect number with special prime $q$. Then we have $$\sigma(q^k) \geq q^k + 1$$ so that $$\frac{1}{4} \geq \frac{1}{q-1} > I(q^k) - 1 \geq \frac{1}{q^k}$$ and $$1 > \frac{q-1}{q+1} \geq I(n^2) - 1 > \frac{q-2}{q} \geq \frac{3}{5}$$ so that we obtain $$\Bigg(\bigg(I(q^k) - 1\bigg) - \frac{1}{q^k}\Bigg)\Bigg(\bigg(I(n^2) - 1\bigg) - \frac{1}{q^k}\Bigg)$$ is nonnegative. Thus, we have $$\Bigg(\bigg(I(q^k) - 1\bigg) - \frac{1}{q^k}\Bigg)\Bigg(\bigg(I(n^2) - 1\bigg) - \frac{1}{q^k}\Bigg) \geq 0$$ which is equivalent to $$\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) + \frac{1}{q^{2k}} \geq \frac{1}{q^k}\Bigg(I(q^k) + I(n^2) - 2\Bigg)$$ $$\Bigg(3 - \bigg(I(q^k) + I(n^2)\bigg)\Bigg) + \frac{1}{q^{2k}} \geq \frac{1}{q^k}\Bigg(I(q^k) + I(n^2) - 3\Bigg) + \frac{1}{q^k}$$ $$\Bigg(\frac{q^k + 1}{q^k}\Bigg)\Bigg(3 - \bigg(I(q^k) + I(n^2)\bigg)\Bigg) = \Bigg(1 + \frac{1}{q^k}\Bigg)\Bigg(3 - \bigg(I(q^k) + I(n^2)\bigg)\Bigg) \geq \frac{1}{q^k} - \frac{1}{q^{2k}} = \frac{q^k - 1}{q^{2k}}$$ $$3 - \bigg(I(q^k) + I(n^2)\bigg) \geq \frac{q^k - 1}{q^k (q^k + 1)}.$$ Hence, we finally get $$I(q^k) + I(n^2) \leq 3 - \Bigg(\frac{q^k - 1}{q^k (q^k + 1)}\Bigg) = \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}.$$ QED.

Notice the striking similarity between the upper bound $$I(q^k) + I(n^2) \leq 3 - \Bigg(\frac{q-1}{q(q+1)}\Bigg) = \frac{3q^2 + 2q + 1}{q(q + 1)},$$ and the upper bound $$I(q^k) + I(n^2) \leq 3 - \Bigg(\frac{q^k - 1}{q^k (q^k + 1)}\Bigg) = \frac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}.$$

Here are my:

QUESTIONS:

(1) Does this argument imply that $k=1$? (i.e. in the sense that $k$ is taken to be a placeholder for $1$?)

(2) Similar to the train of thought in this post, can you think of a way to show that $$I(q^k) + I(n^2) > 3 - \Bigg(\frac{q^k - 2}{q^k (q^k - 1)}\Bigg) = \frac{3q^{2k} - 4q^k + 2}{q^k (q^k - 1)}?$$

(3) If it is not possible to prove the inequality in (2), can you explain why?

Jose Arnaldo Bebita Dris
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    I think that the answer to the question (1) is no. Let $f(k):=\dfrac{3q^{2k} + 2q^k + 1}{q^k (q^k + 1)}$. It follows from $I(q^k)+I(n^2)\leqslant f(1)$ that $I(q^k)+I(n^2)\leqslant f(k)$ since $f'(k)$ is positive. – mathlove Dec 08 '21 at 06:26
  • Thank you for your time and attention, @mathlove. Do you have some comments regarding my other questions? – Jose Arnaldo Bebita Dris Dec 08 '21 at 06:47
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    I don't know how to prove the inequality in (2). Let $g(k):=\dfrac{3q^{2k} -4q^k + 2}{q^k (q^k -1)}$. Then, since $g'(k)$ is positive, one gets $g(1)\leqslant g(k)$. So, it does not follow from $I(q^k)+I(n^2)\gt g(1)$ that $I(q^k)+I(n^2)\gt g(k)$. This does not mean that it is not possible to prove $I(q^k)+I(n^2)\gt g(k)$. – mathlove Dec 08 '21 at 07:15
  • If you would write out your last two comments as an answer, I'd be more than willing to upvote then accept it, @mathlove. – Jose Arnaldo Bebita Dris Dec 08 '21 at 08:05

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I think that the answer to the question (1) is no. Let $f(k):=\dfrac{3q^{2k}+2q^k+1}{q^k(q^k+1)}$. It follows from $I(q^k)+I(n^2)\leqslant f(1)$ that $I(q^k)+I(n^2)\leqslant f(k)$ since $f′(k)$ is positive.

I don't know how to prove the inequality in (2). Let $g(k):=\dfrac{3q^{2k} -4q^k + 2}{q^k (q^k -1)}$. Then, since $g'(k)$ is positive, one gets $g(1)\leqslant g(k)$. So, it does not follow from $I(q^k)+I(n^2)\gt g(1)$ that $I(q^k)+I(n^2)\gt g(k)$. This does not mean that it is not possible to prove $I(q^k)+I(n^2)\gt g(k)$.

mathlove
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