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

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 $$n$$: $$\dfrac{2n}{n + D(n)} < I(n) < \dfrac{2n + D(n)}{n + D(n)}.$$

• 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.) Sep 15, 2017 at 18:14
• 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. Feb 6, 2018 at 2:32
• The proof for the assertion in the preceding comment (which turned out to be nontrivial) is in this answer to a related MSE question. Mar 5, 2018 at 9:43
• 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 $$2(p^2 + p - 2) = 2(p - 1)(p + 2) < p(3p - 2) = 3p^2 - 2p \implies 0 < p^2 - 4p + 4 = (p - 2)^2 \implies 2 < p.$$ Apr 11, 2020 at 12:04
• 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. Apr 12, 2020 at 5:44

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 $$I(m^2)=\frac{p+2}{p} \iff p=3.$$

Too long to comment.

One can prove that

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

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

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 $$b\gt -1$$.

• 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! =) Apr 12, 2020 at 8:03
• I will be reviewing your proof in the next couple of days. Thank you for your time and attention, @mathlove! =) Apr 12, 2020 at 8:04
• 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? May 6, 2020 at 13:28

This is far from a complete answer, but rather a collection of necessary conditions.

If $$I(n)=\frac{p+2}{p}$$, then you already know that $$n$$ is of the form $$p^2s^2$$ where $$s$$ is odd.

We can say the followings :

• $$s\geqslant 5$$

• $$2n-\sigma(n)\geqslant 25p(p-2)$$

• $$p\leqslant \frac{n}{75}$$

• If $$p\gt 3$$, then $$3\not\mid n$$.

• If $$p\not=5$$ and $$5\mid n$$, then $$p=3$$

• $$s$$ is not of the form $$p^m$$

• $$s$$ is not of the form $$q^m$$ where $$q\not=p$$ is a prime number.

• $$\gcd(s,p)\times\gcd(s^2,\sigma(s^2))\gt 1$$.

We have $$I(n)=\frac{p+2}{p}\iff p(p+2)s^2=\sigma(p^2s^2)$$

• $$s\geqslant 5$$
Proof : If $$s=1$$, then $$p(p+2)=1+p+p^2$$ implies $$p=1$$, a contradiction. If $$s=p=3$$, then $$(1)$$ does not hold. If $$s=3, then there is no prime $$p$$ such that $$9p(p+2)=13(1+p+p^2)$$.

• $$2n-\sigma(n)\geqslant 25p(p-2)$$
Proof : $$2n-\sigma(n)=2p^2s^2-p(p+2)s^2=p(p-2)s^2\geqslant 25p(p-2)$$.

• $$p\leqslant \frac{n}{75}$$
Proof : $$\frac{p}{n}=\frac{1}{ps^2}\leqslant \frac{1}{3\times 5^2}=\frac{1}{75}$$.

• If $$p\gt 3$$, then $$3\not\mid n$$.
Proof : Suppose that $$3\mid n$$. Then, since $$3p\mid n$$, we have $$\frac{4(p+1)}{3p}=I(3)I(p)=I(3p)\leqslant I(n)=\frac{p+2}{p}\implies p\leqslant 2$$, a contradiction.

• If $$p\not=5$$ and $$5\mid n$$, then $$p=3$$
Proof : $$5p\mid n\implies \frac{6(p+1)}{5p}=I(5)I(p)=I(5p)\leqslant I(n)=\frac{p+2}{p}\implies p=3$$.

• $$s$$ is not of the form $$p^m$$
Proof : Suppose that $$s=p^m$$. Then, $$p^{2m+1}(p+2)=\sigma(p^{2m+2})$$. LHS is divisible by $$p$$ while RHS isn't, a contradiction.

• $$s$$ is not of the form $$q^m$$ where $$q\not=p$$ is a prime number.
Proof : Suppose that $$s=q^m$$. Then, $$p(p+2)q^{2m}=(1+p+p^2)(1+q+\cdots +q^{2m})$$. There is a positive integer $$t$$ such that $$1+p+p^2=q^{2m}t$$ and $$(1+q+\cdots+ q^{2m})t=p^2+2p$$. We have $$p=(1+q+\cdots +q^{2m-1})t+1\gt q^{2m-1}t$$, so $$q^{2m}t\gt p^2\gt q^{4m-2}t^2$$ implies $$1\gt q^{2m-2}t$$, a contradiction.

• $$\gcd(s,p)\times\gcd(s^2,\sigma(s^2))\gt 1$$.
Proof : Suppose that $$\gcd(s,p)=\gcd(s^2,\sigma(s^2))=1$$. Then, from $$p(p+2)s^2=(1+p+p^2)\sigma(s^2)$$, there is an integer $$t$$ such that $$p(p+2)=t\sigma(s^2)$$ and $$ts^2=1+p+p^2$$. Since $$p=p^2+2p-(1+p+p^2)+1=t\sigma(s^2)-ts^2+1\gt t(1+s)$$, we get $$s^2t\gt p^2\gt t^2(1+s)^2$$ which implies $$s^2\gt t(s^2+2s+1)$$, a contradiction.

• An upvote for now, @mathlove. I will be reviewing your answer in a few. May 20, 2023 at 12:14