18
$\begingroup$

Let $\sigma(x)$ denote the sum of the divisors of $x$, and denote the abundancy index of $x$ as $$I(x) = \dfrac{\sigma(x)}{x},$$ and the deficiency of $x$ as $$D(x) = 2x - \sigma(x).$$ If the equation $I(a)=b/c$ has no solution $a \in \mathbb{N}$, then $b/c$ is said to be an abundancy outlaw.

Statement of the Problem

When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index?

Preliminary Results

The following lemmas are easy to show:

Lemma 1. If $p$ is an odd prime, and $(p+2)/p$ is the abundancy index of some integer $n$, then $n$ is deficient.

Lemma 2. If $p$ is odd, then $\gcd(p,p+2)=1$.

Lemma 3. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $D(n) = 2n - \sigma(n) \neq 1$.

Lemma 4. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $p < n$.

Lemma 5. If $p$ is an odd prime and $I(n) = (p+2)/p$, then $\gcd(n,\sigma(n)) \neq 1$.

Remarks

In fact, one can show that, if $p$ is an odd prime and $I(n) = (p+2)/p$, then $$\sigma(n) = \bigg(\dfrac{n}{p}\bigg)\cdot(p+2)$$ and $$n = \bigg(\dfrac{\sigma(n)}{p+2}\bigg)\cdot(p).$$ (Note that $n/p$ and $\sigma(n)/(p+2)$ are (equal) integers because of Lemma 2.) Consequently, we obtain $$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2}.$$ (Note further that both $\gcd(n,\sigma(n)) \leq n/3$ and $\gcd(n,\sigma(n)) \leq \sigma(n)/5$ hold.)

Added September 16 2017

Given that $X = A/B = C/D$ ($B \neq 0$, $D \neq 0$, and $B \neq D$), we can make use of the algebraic identity $$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}$$ to get another expression for $$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2}.$$

Indeed, $$\gcd(n,\sigma(n)) = \dfrac{n}{p} = \dfrac{\sigma(n)}{p+2} = \frac{\sigma(n) - n}{2}.$$ This last finding implies that $$\bigg(\frac{\sigma(n) - n}{2}\bigg) \mid n \iff (\sigma(n) - n) \mid (2n) \iff 2n = (\sigma(n) - n){d_1}$$ and $$\bigg(\frac{\sigma(n) - n}{2}\bigg) \mid \sigma(n) \iff (\sigma(n) - n) \mid (2\sigma(n)) \iff 2\sigma(n) = (\sigma(n) - n){d_2}.$$ Note that $2 \mid (\sigma(n) - n)$. Additionally, notice that $$2\gcd\left(n,\sigma(n)\right) = \gcd\left(2n, 2\sigma(n)\right) = \gcd\left((\sigma(n) - n){d_1},(\sigma(n) - n){d_2}\right)$$ $$= \left(\sigma(n) - n\right)\gcd({d_1},{d_2}) \iff \frac{2\gcd\left(n,\sigma(n)\right)}{\left(\sigma(n) - n\right)}=1=\gcd({d_1},{d_2}).$$ In fact, $$d_1 = \frac{2n}{\sigma(n) - n} = p$$ and $$d_2 = \frac{2\sigma(n)}{\sigma(n) - n} = p+2.$$ Double-checking if it is indeed the case that ${d_1}+2={d_2}$: $$d_1 = \frac{2n}{\sigma(n) - n} + 2 = \frac{2n + 2(\sigma(n) - n)}{\sigma(n) - n} = \frac{2\sigma(n)}{\sigma(n) - n} = d_2.$$ So far so good!

More is actually true. One can also show that $$p(2n - \sigma(n)) = (p - 2)n$$ so that $$D(n) = (p - 2)\cdot\bigg(\dfrac{n}{p}\bigg) = (p - 2)\cdot\bigg(\dfrac{\sigma(n)}{p + 2}\bigg) = (p - 2)\cdot\gcd(n,\sigma(n)).$$

We therefore conclude that $$\dfrac{D(n)}{n} = \dfrac{p - 2}{p} = \bigg(\dfrac{p - 2}{p + 2}\bigg)\cdot{I(n)}.$$

Added October 8 2017

We deduce that $$\dfrac{p-2}{p}=\dfrac{D(n)}{n}<\dfrac{\phi(n)}{n}<\dfrac{n}{\sigma(n)}=\dfrac{p}{p+2},$$ whence there is still no contradiction.

Motivation

It is conjectured that $(p+2)/p$ is an outlaw, since if it were an index, then we would be able to produce an odd perfect number for $p=3$.

Here is my question:

To what extent can the following theorem be improved to hopefully produce some results towards proving the aforementioned conjecture?

Theorem If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy of $n$ in terms of the deficiency of $n$: $$\dfrac{2n}{n + D(n)} < I(n) < \dfrac{2n + D(n)}{n + D(n)}.$$

$\endgroup$
  • $\begingroup$ Commented on September 16 2017: Note that $\sigma(n) - n$ is called the sum of the aliquot parts of $n$. (This is tabulated in OEIS sequence A001065.) $\endgroup$ – Arnie Bebita-Dris Sep 15 '17 at 18:14
  • $\begingroup$ Note that, since $p$ and $p+2$ are both odd, then $I(n)=(p+2)/p$ implies that $n$ is an odd square, from which it follows that $p^2$ divides $n$, since $p$ is a prime. $\endgroup$ – Arnie Bebita-Dris Feb 6 '18 at 2:32
  • $\begingroup$ The proof for the assertion in the preceding comment (which turned out to be nontrivial) is in this answer to a related MSE question. $\endgroup$ – Arnie Bebita-Dris Mar 5 '18 at 9:43
  • $\begingroup$ Plugging in the value $D(n)/n = (p-2)/p$ into the inequality $$\frac{2}{1 + (D(n)/n)} < I(n) = \frac{p+2}{p} < \frac{2 + (D(n)/n)}{1 + (D(n)/n)}$$ we obtain $$\frac{2p}{p + (p - 2)} = \frac{2p}{2p - 2} = \frac{p}{p - 1} < \frac{p+2}{p} < \frac{3p - 2}{2(p - 1)} = \frac{2p + (p - 2)}{p + (p - 2)}$$ which implies that $$p^2 < (p - 1)(p + 2) = p^2 + p - 2 \implies 2 < p$$ and $$2(p^2 + p - 2) = 2(p - 1)(p + 2) < p(3p - 2) = 3p^2 - 2p \implies 0 < p^2 - 4p + 4 = (p - 2)^2 \implies 2 < p.$$ $\endgroup$ – Arnie Bebita-Dris Apr 11 at 12:04
  • 1
    $\begingroup$ One can prove that $\frac{2n}{n+D(n)}\lt\frac{(2a+2)n-aD(n)}{(a+1)n+D(n)}\lt I(n)$ holds for any $a\gt 0$, but it seems that we cannot prove the conjecture using this inequality. $\endgroup$ – mathlove Apr 12 at 5:44
3
+250
$\begingroup$

Too long to comment.

One can prove that

$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n)\tag1$$

$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag2$$

hold for any $a>0,b\gt -1$. Note here that $a,b$ are not necessarily integers.

However, it seems that we cannot prove the conjecture using $(1)(2)$.


About $(1)$ :

Let $x:=\frac{\sigma(n)}{n}$. Then, we get $$1\lt x\lt 2\tag3$$

Trying to find $a,b,c$ such that $-x+c\gt 0$ and $$\frac{2}{3-x}\lt\frac{ax+b}{-x+c}\lt x$$ which is equivalent to $$ax^2+(b-3a-2)x+2c-3b\lt 0\quad\text{and}\quad x^2+(a-c)x+b\lt 0\tag4$$

every $x$ such that $(4)$ has to satisfy $(3)$.

So, trying to find $a,b,c$ such that

$$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(a-c)^2-4b}\le \min(4+a-c,-a+c-2)\\\sqrt{(3a+2-b)^2-4a(2c-3b)}\le \min(a+2-b,a-2+b)\end{cases}$$ and choosing $b=2$ give $$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(c-a)^2-8}\le \min(4-(c-a),(c-a)-2)\\\sqrt{9a^2-8a(c-3)}\le a\end{cases}$$

So, we see that choosing $c=a+3$ works, and that $$\frac{2}{3-x}\lt\frac{2+ax}{a+3-x}\lt x,$$ i.e. $$\frac{2n}{n+D}\lt\frac{(2a+2)n-aD}{(a+1)n+D}\lt \frac{\sigma(n)}{n}$$ holds for any $a\gt 0$.


About $(2)$ :

Trying to find $a,b,c$ such that $-x+c\gt 0$ and $$x\lt\frac{a+bx}{-x+c}\lt\frac{4-x}{3-x}$$ which is equivalent to $$x^2+(b-c)x+a\gt 0\quad\text{and}\quad (b+1)x^2+(a-3b-c-4)x+4c-3a\gt 0\tag6$$ every $x$ such that $(6)$ has to satisfy $x\not=2$.

So, trying to find $a,b,c$ such that

$$\begin{cases}b+1\gt 0\\ (b-c)^2-4a\le 0\\ (a-3b-c-4)^2-4(b+1)(4c-3a)\le 0\\ 2^2+(b-c)\times 2+a=0\\ (b+1)\times 2^2+(a-3b-c-4)\times 2+4c-3a=0\end{cases}$$

and choosing $a=4$ give $$\begin{cases}b\gt -1\\ c=b+4\end{cases}$$

So, we see that $$x\lt\frac{4+bx}{-x+b+4}\lt\frac{4-x}{3-x},$$ i.e. $$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}$$ holds for any $b\gt -1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your sharing the proof of your inequalities for the improved lower and upper bounds for $I(n)$ (in terms of $n$ and $D(n)$). On closer look, I have realized that I am getting different lower bounds for $I(n)$ (in terms of $n$ and $D(n)$) when I used the Mediant Inequality. An upvote for now, @mathlove! =) $\endgroup$ – Arnie Bebita-Dris Apr 12 at 8:03
  • $\begingroup$ I will be reviewing your proof in the next couple of days. Thank you for your time and attention, @mathlove! =) $\endgroup$ – Arnie Bebita-Dris Apr 12 at 8:04
  • $\begingroup$ Perhaps these same inequalities that you have obtained in this answer could be used to tackle my (related) problem in this other MSE question, @mathlove? $\endgroup$ – Arnie Bebita-Dris May 6 at 13:28
3
$\begingroup$

Not an answer, just some remarks that are too long to fit in the Comments section.

Notice that, if $I(n)=(p+2)/p$, then $$\frac{n}{D(n)}=\frac{p}{p-2}.$$

Since $p$ is an odd prime, then $\gcd(p,p-2)=1$. Thus, $D(n) \nmid n$, unless $p=3$.

This implies that if $I(n)=(p+2)/p=5/3$ (when $p=3$), then $n$ is deficient-perfect.

Otherwise, if $p>3$, then since $p$ is an odd prime, $p \geq 5$, so that $$\frac{n}{D(n)}=\frac{p}{p-2}=\frac{1}{1-\frac{2}{p}} \leq \frac{1}{1-\frac{2}{5}}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$$ and $$\frac{n}{D(n)}=\frac{(p-2)+2}{p-2}=1+\frac{2}{p-2}>1,$$ from which we obtain $$1 < \frac{n}{D(n)} \leq \frac{5}{3},$$ which implies that $D(n) \nmid n$ for $p>3$.

Since $$1 < \frac{n}{D(n)} \leq \frac{5}{3}$$ implies that $$1 < I(n) \leq \frac{7}{5},$$ and since we have $n$ is a square if $I(n)=(p+2)/p$ and $p$ is an odd prime, and because $$\frac{8}{5} < I(m^2) < 2$$ if $q^k m^2$ is an odd perfect number with Euler prime $q$, then we have that $$I(m^2)=\frac{p+2}{p} \iff p=3.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.