# Does $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < 10$ hold in general for an odd perfect number $q^k n^2$ with special prime $q$?

This question is an offshoot of this MSE answer.

Let $$\sigma(x)$$ be the sum of the divisors of the positive integer $$x$$. (Denote the abundancy index of $$x$$ by $$I(x)=\sigma(x)/x$$.)

If $$\sigma(M) = 2M$$, then $$M$$ is said to be perfect.

Currently, as of December 2018, there are $$51$$ known examples of even perfect numbers -- on the other hand, we still do not know whether there are any odd perfect numbers.

Euler derived the general form that an odd perfect number $$N$$ must take:

$$N = {q^k}{n^2},$$

where $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n) = 1$$. We call $$q$$ the special or Euler prime of $$N$$.

Descartes, Frenicle and subsequently Sorli conjectured that $$k = 1$$.

In [Dris, 2012], it was shown that the implications

$$n < q \Longrightarrow k = 1$$ and $$n < q^2 \Longrightarrow k = 1$$

are true.

Now, note that, since $$q$$ and $$\sigma(q) = q + 1$$ are consecutive integers, then the following implications are true.

Case 1: $$q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$$

Case 2: $$n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$$

The remaining cases to be considered are:

Case 3: $$q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$$

Case 4: $$n < \sigma(n) < q^k < \sigma(q^k) \land k \geq 1$$

Case 5: $$n < q^k \leq \sigma(n) < \sigma(q^k) \land k \geq 1$$

Case 6: $$n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$$

mathlove (in the hyperlinked MSE answer) is of the opinion that we should separate the cases in the following way:

Case 1 : $$q^k < n < \sigma(q^k) < \sigma(n) \Longrightarrow k > 1$$

Case 2 : $$n < q^k < \sigma(n) < \sigma(q^k) \Longrightarrow k > 1$$

Case 3 : $$q^k\lt\sigma(q^k)\lt n\lt\sigma(n)\land k \geq 1$$

Case 6 : $$n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$$

Case 7 : $$n\lt \sigma(n)\le q^k\lt\sigma(q^k)\land k \geq 1$$

We want to rule out the following scenario:

Case 6 : $$n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$$

in order to show that the biconditional $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$ holds.

I noticed that $$\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \sqrt{I(n)} < \sqrt{I(q^k n)} < \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{2}$$

But $$\dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{2} < \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{I(q^k n)}$$ holds since $$q^k n$$ is deficient, being a proper factor of the perfect number $$q^k n^2$$.

However, we can rewrite this as $$\dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{I(q^k n)} = \dfrac{\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}}{\Bigg(\dfrac{\sigma(q^k)}{n}\Bigg)\cdot\Bigg(\dfrac{\sigma(n)}{q^k}\Bigg)} = \dfrac{q^k}{\sigma(n)} + \dfrac{n}{\sigma(q^k)}.$$

Hence, we have the simultaneous inequalities: $$\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \dfrac{q^k}{\sigma(n)} + \dfrac{n}{\sigma(q^k)}$$ and $$2\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}.$$

Under this scenario:

Case 6 : $$n\lt q^k\lt\sigma(q^k)\le\sigma(n)\land k \geq 1$$

we obtain the upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < \dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} < 10,$$ as $$\dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} \approx 9.909120785838094255.$$

Here are my:

QUESTIONS Using the ideas in this post, would it be possible to show that $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < 10$$ holds in all cases? If it is not possible, can you explain why?

• FYI, under Case 1,2 or 6, one has $\bigg(\dfrac{\sigma(q^k)}{n}-1\bigg)\bigg(\dfrac{\sigma(n)}{q^k}-1\bigg)\gt 0$ which implies $\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k}\lt I(nq^k)+1\lt 3$. Jan 26 at 10:51
• Thank you for your time and attention, @mathlove! Please write out your last comment as an actual answer so that I can upvote it. =) Jan 26 at 10:59

On OP's request, I am converting my comment into an answer.

FYI, under Case 1,2 or 6, one has $$\bigg(\dfrac{\sigma(q^k)}{n}-1\bigg)\bigg(\dfrac{\sigma(n)}{q^k}-1\bigg)\gt 0$$ which implies $$\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k}\lt I(nq^k)+1\lt 3$$

• Thank you for your answer, @mathlove. Using your idea, I can prove that $$\dfrac{2n}{q} + \dfrac{q}{n} > \Bigg(\dfrac{n}{q}\Bigg) I(q^k n) + \Bigg(\dfrac{q}{n}\Bigg) > \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}$$ holds under Case 3, since $$\dfrac{q}{n} < \dfrac{\sigma(q^k)}{n} < 1 < \dfrac{\sigma(n)}{q^k}$$ is true under this scenario, which implies that $$\Bigg(\dfrac{\sigma(q^k)}{n} - \dfrac{q}{n}\Bigg)\Bigg(\dfrac{\sigma(n)}{q^k} - \dfrac{q}{n}\Bigg) > 0,$$ from which the first inequality above follows. Jan 26 at 11:26
• Is it possible to show that $$\dfrac{2n}{q} + \dfrac{q}{n} < 3$$ if the inequality $q < n$ holds, @mathlove? Jan 26 at 11:30
• @Arnie Bebita-Dris : I don't think that it is possible. When $n$ is large, LHS will be large. Jan 26 at 11:37
• From the inequality $q < n$, we obtain $$\dfrac{q}{n} < 1 < 2 < \dfrac{2n}{q}$$ so that $$\Bigg(\bigg(\dfrac{q}{n}\bigg) - 1\Bigg)\Bigg(\bigg(\dfrac{2n}{q}\bigg) - 1\Bigg) < 0$$ from which it follows that $$3 = 2 + 1 < \Bigg(\dfrac{q}{n}\Bigg) + \Bigg(\dfrac{2n}{q}\Bigg),$$ so that my conjecture for Case 3 does not hold. Jan 26 at 11:38

(This is not a direct answer to the original question, just some remarks that are too long to fit in the Comments section.)

This answer improves on the upper bound for $$\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k}$$ under Case 6:

Case 6: $$n < q^k < \sigma(q^k) \leq \sigma(n) \land k \geq 1$$

Since the biconditionals $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n}<\dfrac{\sigma(n)}{q^k}$$ and $$n < q^k \iff \sigma(n) < \sigma(q^k) \iff \dfrac{\sigma(n)}{q^k}<\dfrac{\sigma(q^k)}{n}$$ do not hold if $$n < q^k < \sigma(q^k) < \sigma(n)$$, and since the equation $$\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k}=I(q^k)+I(n)$$ holds if and only if $$\sigma(q^k)=\sigma(n)$$, and since the two biconditionals above are equivalent to the truth of the inequality $$I(q^k)+I(n)<\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k},$$ it follows that, under Case 6, we actually have the upper bound $$\dfrac{\sigma(q^k)}{n}+\dfrac{\sigma(n)}{q^k} \leq I(q^k)+I(n) < I(q^k)+I(n^2) < 3.$$

(Note: This is only a partial answer.)

We consider the other cases separately below.

Case 1: $$q^k < n < \sigma(q^k) < \sigma(n) \implies k > 1$$

Similar to the proof for Case 6 in the original post, it follows that $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < \dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} < 10,$$ under Case 1.

Case 2: $$n < q^k < \sigma(n) < \sigma(q^k) \implies k > 1$$

Similar to the proof for Case 6 in the original post, it follows that $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < \dfrac{2}{\sqrt{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} - 1} < 10,$$ under Case 2.

Case 3: $$q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$$

Note that, under this case, we have that the biconditional $$q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}$$ holds. This is equivalent to the truth of the inequality $$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k},$$ whence it appears that we cannot obtain an upper bound for $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}$$ using the ideas in the original post. (This is because we are unable to obtain a nonzero (constant) lower bound for $$\dfrac{\sigma(q^k)}{n}$$.)

Case 7: $$n < \sigma(n) \leq q^k < \sigma(q^k) \land k \geq 1$$

Note that, under this case, we have that the biconditional $$n < q^k \iff \sigma(n) < \sigma(q^k) \iff \dfrac{\sigma(n)}{q^k} < \dfrac{\sigma(q^k)}{n}$$ holds. This is equivalent to the truth of the inequality $$I(q^k) + I(n) < \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k},$$ whence it appears that we cannot obtain an upper bound for $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}$$ using the ideas in the original post. (This is because we are unable to obtain a nonzero (constant) lower bound for $$\dfrac{\sigma(n)}{q^k}$$.)

We therefore conclude that it remains to consider the following scenarios:

• Case 3: $$q^k < \sigma(q^k) < n < \sigma(n) \land k \geq 1$$
• Case 7: $$n < \sigma(n) \leq q^k < \sigma(q^k) \land k \geq 1$$
• Blimey, how could I have missed it! Posting another answer in a bit. Feb 17 at 5:22

From this other answer, we know that it remains to consider the following cases:

• Case 3: $$q^k \lt \sigma(q^k) \lt n \lt \sigma(n)$$

Under this scenario, we get the bounds $$\dfrac{q^k + 1}{n} \leqslant \dfrac{\sigma(q^k)}{n} \lt 1 \lt \Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \lt \dfrac{\sigma(n)}{n} \lt \dfrac{\sigma(n)}{q^k} \lt \dfrac{2n}{q^k},$$ where we have used the bounds $$\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \lt \dfrac{\sigma(n)}{n} \lt \dfrac{\sigma(q^k)}{n}\cdot\dfrac{\sigma(n)}{q^k} \lt 2.$$

Consider the product $$\Bigg(\dfrac{\sigma(q^k)}{n} - \dfrac{2n}{q^k}\Bigg)\Bigg(\dfrac{\sigma(n)}{q^k} - \dfrac{2n}{q^k}\Bigg).$$ This product is positive. Hence, we derive the bound $$2 + \Bigg(\dfrac{2n}{q^k}\Bigg)^2 > \dfrac{\sigma(q^k)}{n}\cdot\dfrac{\sigma(n)}{q^k} + \Bigg(\dfrac{2n}{q^k}\Bigg)^2 \gt \dfrac{2n}{q^k}\cdot\Bigg(\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}\Bigg).$$ It follows that we have the upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \lt \dfrac{q^k}{n} + \dfrac{2n}{q^k}.$$ Similarly, by considering $$\dfrac{q^k + 1}{n} \leqslant \dfrac{\sigma(q^k)}{n} \lt \dfrac{\sigma(n)}{q^k},$$ we get the marginally improved upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \leqslant \dfrac{q^k + 1}{n} + \dfrac{2n}{q^k + 1}.$$

• Case 7: $$n \lt \sigma(n) \leq q^k \lt \sigma(q^k)$$

Under this scenario, we get the bounds $$\Bigg(\dfrac{n}{q^k}\Bigg)\cdot{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} \lt \dfrac{\sigma(n)}{q^k} \leqslant 1 \lt \Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \lt \dfrac{\sigma(n)}{n} \lt \dfrac{\sigma(q^k)}{n} \lt \dfrac{5q^k}{4n},$$ where we have used the bounds $$1 \lt \dfrac{\sigma(q^k)}{q^k} \lt \dfrac{5}{4} \lt \Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \lt \dfrac{\sigma(n)}{n} \lt \dfrac{\sigma(q^k)}{n}\cdot\dfrac{\sigma(n)}{q^k} \lt 2.$$

Consider the product $$\Bigg(\dfrac{\sigma(q^k)}{n} - \dfrac{5q^k}{4n}\Bigg)\Bigg(\dfrac{\sigma(n)}{q^k} - \dfrac{5q^k}{4n}\Bigg).$$ This product is positive. Hence, we derive the bound $$2 + \Bigg(\dfrac{5q^k}{4n}\Bigg)^2 \gt \dfrac{\sigma(q^k)}{n}\cdot\dfrac{\sigma(n)}{q^k} + \Bigg(\dfrac{5q^k}{4n}\Bigg)^2 \gt \dfrac{5q^k}{4n}\cdot\Bigg(\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}\Bigg).$$ It follows that we have the upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \lt \dfrac{8n}{5q^k} + \dfrac{5q^k}{4n}.$$ Similarly, by considering $$\Bigg(\dfrac{n}{q^k}\Bigg)\cdot{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} \lt \dfrac{\sigma(n)}{q^k} \lt \dfrac{\sigma(q^k)}{n},$$ we get the trivial upper bound $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} \lt \Bigg(\dfrac{n}{q^k}\Bigg)\cdot{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}} + \dfrac{2q^k}{n\cdot{\Bigg(\dfrac{8}{5}\Bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}}}.$$

CONCLUSION: So ultimately, the problem of determining if $$\dfrac{q^k}{n} + \dfrac{n}{q^k} < \mathscr{S}(q,k,n) := \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} < 2\cdot\Bigg(\dfrac{q^k}{n}+\dfrac{n}{q^k}\Bigg)$$ is bounded would depend on at least one of the following things:

• if an equation or a relationship exists between $$q^k$$ and $$n$$; or
• if there is no region (or domain) over which $$\mathscr{S}(q,k,n)$$ is unbounded.

Note that we only need to consider the rational quantity $$\mathscr{R}(q,k,n) = \dfrac{q^k}{n}$$ for the whole analysis of this problem.

Alas, this is where I am currently stuck!