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

 A: 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
$$I(m^2)=\frac{p+2}{p} \iff p=3.$$
A: 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 $b\gt -1$.
