(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?
Best Answer
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 $I(q^k)+I(n^2)\gt g(k)$.