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 $\gcd(n,\sigma(n)) \neq 1$.


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.


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

  • $\begingroup$ 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.) $\endgroup$ Sep 15, 2017 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. $\endgroup$ Feb 6, 2018 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. $\endgroup$ Mar 5, 2018 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 $$2(p^2 + p - 2) = 2(p - 1)(p + 2) < p(3p - 2) = 3p^2 - 2p \implies 0 < p^2 - 4p + 4 = (p - 2)^2 \implies 2 < p.$$ $\endgroup$ Apr 11, 2020 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, 2020 at 5:44

3 Answers 3


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


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

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 $b\gt -1$.

  • $\begingroup$ 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! =) $\endgroup$ Apr 12, 2020 at 8:03
  • $\begingroup$ I will be reviewing your proof in the next couple of days. Thank you for your time and attention, @mathlove! =) $\endgroup$ Apr 12, 2020 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? $\endgroup$ 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 $s=3<p$, 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.

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

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