When $p$ is an odd prime, is $\frac{p+2}{p}$ an outlaw or an index?

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the abundancy index $I$ of $x$ by

$$I(x)=\frac{\sigma(x)}{x}$$

and the deficiency $D$ of $y$ by

$$D(y)=2y - \sigma(y).$$

We first show some preliminary lemmata.

Lemma 1.  If $p$ is an odd prime, and $\frac{p+2}{p}$ is the abundancy index of some integer $n$, then $n$ is deficient.

Proof.  Since $p$ is an odd prime, $p > 2$.  Furthermore, since

$$I(n) = \frac{p+2}{p} = 1 + \frac{2}{p} < 1 + 1 = 2,$$

then $n$ is deficient.          QED

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

$$\gcd(p,p+2)=\gcd(p,(p+2)-p)=\gcd(p,2)=1$$

where the last equality uses the fact that $p$ is odd.          QED

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

Proof.

Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Now, assume to the contrary that

$$D(n) = 2n - \sigma(n) = 1.$$

Note that this implies

$$\gcd(n,\sigma(n))=\gcd(n,2n-1)=1.$$

Additionally, since $p$ is odd, by Lemma 2 we obtain

$$\gcd(p,p+2)=1.$$

Consequently, since $I(n)=\frac{p+2}{p}$, we get

$$p\sigma(n)=(p+2)n.$$

But $\gcd(p,p+2)=1$ implies that $p \mid n$, and $\gcd(n,\sigma(n))=1$ implies that $n \mid p$.  Hence, it follows that $p=n$.

But since $p$ is an odd prime, $I(n)=I(p)=\frac{p+1}{p}$, which contradicts $I(n)=\frac{p+2}{p}$.

We conclude that $D(n) = 2n - \sigma(n) \neq 1$.          QED

The following lemmas show that $p < n$, if $I(n)=\frac{p+2}{p}$ is true.  In particular, $p < n$ follows from Lemma 4 and $p \mid n$.

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

Proof.

Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.

We have

$$I(p)=\frac{p+1}{p}<\frac{p+2}{p}=I(n).$$

In fact, we have

$$I(n) - I(p) > I(n) - \frac{p}{p-1} = \left(\frac{1}{p-1}\right)\cdot\left(\frac{D(n)}{n}\right).$$

QED

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

Proof.

Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Suppose to the contrary that we have $\gcd(n,\sigma(n))=1$.  Following the proof of Lemma 3, since $\gcd(p,p+2)=1$ (by Lemma 2), we have

$$n = p$$

and

$$\sigma(n) = p+2.$$

Thus,

$$p+1=\sigma(p)=\sigma(n)=p+2.$$

This is a contradiction.                        QED

Remarks.
In fact, one can show that, if $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then
$$\sigma(n)=\left(\frac{n}{p}\right)\cdot(p+2)$$
and
$$n=\left(\frac{\sigma(n)}{p+2}\right)\cdot{p}.$$
(Note that
$$\frac{n}{p}$$
and
$$\frac{\sigma(n)}{p+2}$$
are integers because of Lemma 2.)  Consequently,
$$\gcd\left(n,\sigma(n)\right) = \frac{n}{p} = \frac{\sigma(n)}{p+2}.$$
More is actually true.  One can also show that
$$p\left(2n-\sigma(n)\right)=(p-2)n$$
so that
$$D(n)=\left(p-2\right)\cdot\left(\frac{n}{p}\right)=\left(p-2\right)\cdot\left(\frac{\sigma(n)}{p+2}\right) = \left(p-2\right)\cdot\gcd\left(n,\sigma(n)\right).$$
We therefore conclude that
$$\frac{D(n)}{n}=\frac{p-2}{p}=\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$
To facilitate easier reference later, we compute:
$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$
and
$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

The following lemma is actually a theorem from this preprint.  (We omit the proof.)

Lemma 6.  If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy index of $n$ in terms of the deficiency of $n$:
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}.$$

We now attempt to prove the following "result":

"Conjecture".   If $p$ is an odd prime, then $\frac{p+2}{p}$ is an abundancy outlaw.
"Proof Attempt".
Let $p$ be an odd prime, and assume to the contrary that $I(n)=\frac{p+2}{p}$.

By Lemma 1, $n$ is deficient (so that $D(n) \neq 0$).

By Lemma 3, $D(n) \neq 1$.

Thus, by Lemma 6, we have the bounds

$$\frac{2}{1 + \frac{D(n)}{n}}= \frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}}.$$

By the Remarks, $I(n) = \frac{p+2}{p}$ implies that

$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$

and

$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

We compute:
$$I(n) > \frac{2n}{n+D(n)} = \frac{2}{1 + \frac{D(n)}{n}} = \frac{p}{p-1} > I(p),$$
(no contradictions so far), and
$$I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}} = \frac{\left(\frac{3p-2}{p}\right)}{\left(\frac{2(p-1)}{p}\right)} = \frac{3p-2}{2p-2} = \frac{4p-4}{2p-2} - \frac{p-2}{2(p-1)} = 2 - {\frac{1}{2}}\cdot\left(\frac{p-2}{p-1}\right),$$
(still no contradictions, per this WolframAlpha computational verification).

It thus seems fruitful to try to improve on the (trivial?) bounds

$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}$$

for $n$ satisfying $D(n)>1$.