# An interesting identity involving the abundancy index of divisors of odd perfect numbers

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:

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

 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

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