Mathematics Asked on January 7, 2022

Let $sigma(x)$ denote the sum of divisors of the positive integer $x$.

A number $y$ is said to be perfect if $sigma(y)=2y$.

Denote the abundancy index of $z$ by $I(z)=sigma(z)/z$.

Euler proved that an *odd perfect number* $N$, if one exists, must necessarily have the form $N = q^k n^2$ where $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.

While considering the difference

$$I(n^2) – I(q^k)$$

for $k=1$, I came across the interesting identity

$$frac{d}{dq}bigg(I(n^2)-I(q)bigg)=frac{d}{dq}bigg(frac{q^2 – 2q – 1}{q(q+1)}bigg)=frac{3q^2 + 2q + 1}{q^2 (q+1)^2}.$$

This *is interesting* because of

$$I(n^2)+I(q)=frac{2q}{q+1}+frac{q+1}{q}=frac{3q^2 + 2q + 1}{q(q+1)}=q(q+1)bigg(frac{3q^2 + 2q + 1}{q^2 (q+1)^2}bigg)$$

so that we have the *identity* (or *differential equation* (?))

$$q(q+1)frac{d}{dq}bigg(I(n^2)-I(q)bigg)=I(n^2)+I(q).$$

Two questions:

[1]Is there a simple explanation for why the identity (or differential equation (?)) holds?

[2]Are there any other identities that could be derived in a similar fashion?

**Update (July 25, 2020 – 10:15 AM Manila time)**

I tried computing the derivative

$$frac{d}{dq}bigg(I(n^2)+I(q)bigg)$$

and I got

$$q(q+1)frac{d}{dq}bigg(I(n^2)+I(q)bigg)=q(q+1)frac{d}{dq}bigg(frac{3q^2 + 2q + 1}{q(q+1)}bigg)=q(q+1)bigg(frac{q^2 – 2q – 1}{q^2 (q+1)^2}bigg)=I(n^2)-I(q).$$

**There is nothing too special about the identities, as we shall soon see.**

Hereinafter, we let $q^k n^2$ be an odd perfect number with special prime $q$, and we assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds true. Also, we denote the *abundancy index* of the positive integer $x$ by $I(x)=sigma(x)/x$, where $sigma(x)$ is the classical *sum of divisors* of $x$.

We compute:

$$frac{d}{dq}(I(n^2) + I(q))=frac{d}{dq}bigg(frac{3q^2 + 2q + 1}{q(q+1)}bigg)=frac{d}{dq}bigg(3 - frac{q - 1}{q(q + 1)}bigg)=frac{d}{dq}bigg(frac{-1}{q+1}+frac{1}{q(q+1)}bigg)$$ $$=frac{d}{dq}bigg(frac{-2}{q+1}+frac{1}{q}bigg)=frac{2}{(q+1)^2}-frac{1}{q^2}= frac{2q^2 - (q+1)^2}{(q(q+1))^2}$$

But $$I(n^2) - I(q) = frac{2q}{q+1} - frac{q+1}{q} = frac{2q^2 - (q+1)^2}{q(q+1)}.$$

Similarly, we obtain:

$$frac{d}{dq}(I(n^2) - I(q))=frac{d}{dq}bigg(frac{q^2 - 2q - 1}{q(q+1)}bigg)=frac{d}{dq}(I(n^2) + I(q) - 2I(q))$$ $$=frac{d}{dq}bigg(3 - frac{q - 1}{q(q + 1)} - frac{2(q+1)}{q}bigg)=frac{d}{dq}bigg(frac{-1}{q+1}+frac{1}{q(q+1)}+frac{-2}{q}-2bigg)$$ $$=frac{d}{dq}bigg(frac{-2}{q+1}+frac{1}{q}+frac{-2}{q}-2bigg)=frac{d}{dq}bigg(frac{-2}{q+1}-frac{1}{q}-2bigg)=frac{2}{(q+1)^2}+frac{1}{q^2}$$ $$=frac{2q^2 + (q+1)^2}{(q(q+1))^2}.$$

But $$I(n^2)+I(q)=frac{2q}{q+1}+frac{q+1}{q}=frac{2q^2 + (q+1)^2}{q(q+1)}.$$

In the above computations, we have used the identity $$frac{1}{q(q+1)}=frac{1}{q}-frac{1}{q+1}.$$

Answered by Arnie Bebita-Dris on January 7, 2022

3 Asked on December 9, 2020

1 Asked on December 9, 2020 by buyanov-igor

1 Asked on December 9, 2020 by pazu

image processing inverse problems reference request regularization total variation

1 Asked on December 8, 2020

1 Asked on December 8, 2020 by carlos-andres-henao-acevedo

1 Asked on December 8, 2020 by kevin_h

4 Asked on December 8, 2020 by sandro-gakharia

0 Asked on December 8, 2020 by motherboard

approximate integration combinatorial proofs combinatorics contour integration

2 Asked on December 8, 2020 by hachiman-hikigaya

hypothesis testing normal distribution statistical inference

1 Asked on December 8, 2020 by tchappy-ha

0 Asked on December 7, 2020 by dev-dhruv

1 Asked on December 7, 2020 by julian-ven

1 Asked on December 7, 2020 by confusion

1 Asked on December 7, 2020 by james-black

0 Asked on December 7, 2020 by francesco-totti

2 Asked on December 7, 2020 by cryo

2 Asked on December 6, 2020 by ivan-bravo

bilinear form elementary set theory linear algebra orthogonality vector spaces

Get help from others!

Recent Answers

- Peter Machado on Why fry rice before boiling?
- Jon Church on Why fry rice before boiling?
- haakon.io on Why fry rice before boiling?
- Lex on Does Google Analytics track 404 page responses as valid page views?
- Joshua Engel on Why fry rice before boiling?

Recent Questions

- How Do I Get The Ifruit App Off Of Gta 5 / Grand Theft Auto 5
- Iv’e designed a space elevator using a series of lasers. do you know anybody i could submit the designs too that could manufacture the concept and put it to use
- Need help finding a book. Female OP protagonist, magic
- Why is the WWF pending games (“Your turn”) area replaced w/ a column of “Bonus & Reward”gift boxes?
- Does Google Analytics track 404 page responses as valid page views?

© 2023 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP