# Can this inequality involving the deficiency and sum of aliquot divisors be improved? - Part II

This MSE question (from April 2020) asked whether the inequality $$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}$$ could be improved, where $$D(x)=2x-\sigma(x)$$ is the deficiency of the positive integer $$x$$, $$s(x)=\sigma(x)-x$$ is the aliquot sum of $$x$$, and $$\sigma(x)=\sigma_1(x)$$ is the classical sum of divisors of $$x$$.

In the accepted answer, MSE user Adam Ledger asserts that:

I am yet to find a counterexample for the following, I have not spent much time on your problem so will appreciate if this is not considered an improvement on your original inequality, but none the less I hope that it helps in some small way:

Denoting the Kronecker delta as follows: $$\delta \left( x,y \right) =\cases{1&x=y\cr 0&x\neq y \cr}\tag{ 0}$$

up to $$n \leq 2 \cdot 10^7$$ I have found the following to be satisfied: $${\frac {D \left( n \right) }{s \left( n \right) }}-{\frac { D \left( {n}^{2} \right) }{s \left( {n}^{2} \right) }}-\frac{1}{4} \delta \left( n-2\,\left\lfloor \frac{n}{2}\right\rfloor,1 \right) \lt \frac{3}{4} \tag{1}$$

Adam's result implies that we have the bounds $$0 < \frac{D(n)}{s(n)} - \frac{D(n^2)}{s(n^2)} < 1.$$

Here is my inquiry:

QUESTION: Does anybody here know how to prove Adam's Inequality $$(1)$$?

MOTIVATION FOR THE INQUIRY

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

This answer shows that we have $$\frac{D(m^2)}{s(m^2)}=\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)},$$ which means that, conjecturally, we should have $$\frac{D(p^k m)}{D(p^k)s(m)}=\frac{D(m)}{s(m)}-\frac{D(m^2)}{s(m^2)} < 1$$ by Adam's result.

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

Note that we have the numerical bounds $$1 < I(p^k) < \frac{5}{4} < \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < I(m) < 2,$$ from which we get $$0 < \frac{D(p^k m)}{D(p^k)s(m)}=\frac{2 - I(p^k)I(m)}{(2 - I(p^k))(I(m) - 1)} < \dfrac{2-\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}}{\dfrac{3}{4}\bigg(\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} - 1\bigg)} \approx 1.666929067.$$

Alas, this is where I get stuck!

This answer proves that for every $$n\gt 1$$, $$\dfrac{D(n)}{s(n)} - \dfrac{D(n^2)}{s(n^2)} < 1\tag2$$ holds.

To prove $$(2)$$, it is sufficient to prove the following claims.

Let $$f(n):=(\sigma(n)-2n)\sigma(n^2)+n^3$$.

Claim 1 : $$(2)$$ is equivalent to $$f(n)\gt 0$$.

Claim 2 : If $$p$$ is a prime number and $$k$$ is a positive integer, then $$f(p^k)\gt 0$$.

Claim 3 : If $$f(n)\gt 0$$ and $$p$$ is a prime number satisfying $$\gcd(p,n)=1$$ and $$k$$ is a positive integer, then $$f(p^kn)\gt 0$$.

Example : $$f(2^4)\gt 0$$ (claim 2) $$\implies f(2^43^2)\gt 0$$ (claim 3) $$\implies f(2^43^25^3)\gt 0$$ (claim 3).

Claim 1 : $$(2)$$ is equivalent to $$f(n)\gt 0$$.

Proof :

$$\frac{D(n)}{s(n)}-\frac{D(n^2)}{s(n^2)}=\bigg(\frac{n}{\sigma(n)-n}-1\bigg)-\bigg(\frac{n^2}{\sigma(n^2)-n^2}-1\bigg)$$ $$=\frac{n}{\sigma(n)-n}-\frac{n^2}{\sigma(n^2)-n^2}=\frac{n\sigma(n^2)-n^2\sigma(n)}{(\sigma(n)-n)(\sigma(n^2)-n^2)}$$

So, we have \begin{align}&\frac{D(n)}{s(n)}-\frac{D(n^2)}{s(n^2)}\lt 1\iff \frac{n\sigma(n^2)-n^2\sigma(n)}{(\sigma(n)-n)(\sigma(n^2)-n^2)}\lt 1 \\\\&\iff n\sigma(n^2)-n^2\sigma(n)\lt (\sigma(n)-n)(\sigma(n^2)-n^2) \\\\&\iff n\sigma(n^2)-n^2\sigma(n)\lt \sigma(n)\sigma(n^2)-n^2\sigma(n)-n\sigma(n^2)+n^3 \\\\&\iff (\sigma(n)-2n)\sigma(n^2)+n^3\gt 0\qquad\blacksquare\end{align}

Claim 2 : If $$p$$ is a prime number and $$k$$ is a positive integer, then $$f(p^k)\gt 0$$.

Proof :

\begin{align}f(p^k)&=(\sigma(p^k)-2p^k)\sigma(p^{2k})+p^{3k} \\\\&=\bigg(\frac{p^{k+1}-1}{p-1}-2p^k\bigg)\frac{p^{2k+1}-1}{p-1}+p^{3k} \\\\&=\frac{p^k-1}{(p-1)^2}\bigg(\underbrace{p^k(p^{k}-p +1) - 1}_{\text{positive}}\bigg) \\\\&\gt 0\qquad\blacksquare\end{align}

Claim 3 : If $$f(n)\gt 0$$ and $$p$$ is a prime number satisfying $$\gcd(p,n)=1$$ and $$k$$ is a positive integer, then $$f(p^kn)\gt 0$$.

Proof :

\begin{align}f(p^kn)&=(\sigma(p^kn)-2p^kn)\sigma(p^{2k}n^2)+p^{3k}n^3 \\\\&=(\sigma(p^k)\sigma(n)-2p^kn)\sigma(p^{2k})\sigma(n^2)+p^{3k}n^3 \\\\&=\bigg(\frac{p^{k+1}-1}{p-1}\sigma(n)-2p^kn\bigg)\frac{p^{2k+1}-1}{p-1}\sigma(n^2)+p^{3k}n^3\end{align}

So, we have \small\begin{align}&(p-1)^2f(p^kn) \\\\&=\bigg((p^{k+1}-1)\sigma(n)-2p^kn(p-1)\bigg)(p^{2k+1}-1)\sigma(n^2)+p^{3k}n^3(p-1)^2 \\\\&=\bigg((p^{k+1}-1)\sigma(n)-p^k(p-1)\sigma(n)+p^k(p-1)\sigma(n)-2p^k(p-1)n\bigg)(p^{2k+1}-1)\sigma(n^2) \\&\qquad +p^{3k}n^3(p-1)^2 \\\\&=\bigg((p^{k+1}-1)\sigma(n)-p^k(p-1)\sigma(n)\bigg)(p^{2k+1}-1)\sigma(n^2) \\&\qquad +\bigg(p^k(p-1)\sigma(n)-2p^k(p-1)n\bigg)(p^{2k+1}-1)\sigma(n^2)+p^{3k}n^3(p-1)^2 \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) \\&\qquad +p^k(p-1)\underbrace{(\sigma(n)-2n)\sigma(n^2)}_{\gt -n^3}(p^{2k+1}-1)+p^{3k}n^3(p-1)^2 \\\\&\gt(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) \\&\qquad +p^k(p-1)(-n^3)(p^{2k+1}-1)+p^{3k}n^3(p-1)^2 \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) -n^3 (p - 1) p^k (p^{2 k} - 1) \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2)-(p - 1) p^k (p^{2 k} - 1)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)\sigma(n)\sigma(n^2) - (p - 1) p^k (p^{2 k} - 1)n^3 \\\\&=\bigg((p^k-1)(p^{2k+1}-1)-(p - 1) p^k (p^{2 k} - 1)\bigg)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)(\sigma(n)\sigma(n^2) - n^3) \\\\&=(p^k - 1)\bigg( \underbrace{p^k(p^{k}-p +1) - 1}_{\text{positive}}\bigg)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)\bigg(\underbrace{\sigma(n)\sigma(n^2) - n^3}_{\text{positive}}\bigg) \\\\&\gt 0\end{align} Hence, $$f(p^kn)\gt 0$$ follows.$$\quad\blacksquare$$

• Allow me to review your answer in the next couple of days, @mathlove! An upvote for now. =) Commented Dec 29, 2022 at 22:06
• Great answer. I was wondering about a justification of the $(σ(n)−2n)σ(n^2)>-n^3$ inequality and I think one is that it holds for all primes and is equivalent to $(I(n)-2)I(n^2)>-1$. Therefore the latter inequality holds for all primes, and as if $p$ divides $n$ then $I(p)\leq I(n)$, $(I(n)-2)I(n^2)\geq (I(p)-2)I(p^2)>-1$.
– SFA
Commented Dec 31, 2022 at 22:29
• I agree that it is a great answer. I am hereby accepting your answer, @mathlove! =) Commented Jan 1, 2023 at 4:30