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

Question

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 $X = A/B = C/D$ ($B \neq 0$ and $D \neq 0$), 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 $n$: $$\dfrac{2n}{n + D(n)} < I(n) < \dfrac{2n + D(n)}{n + D(n)}.$$

Answer 1

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