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

**Proof.**Since $p$ is odd, $p > 2$. 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 lemma is actually a theorem from this preprint (to be updated soon). (We omit the proof.)

**Lemma 4.**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":

**"Theorem".**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 4**, we have the bounds

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

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

$$D(n) = p\left(\sigma(n) - n\right) - \sigma(n).$$

We compute:

$$n + D(n) = (p - 1)\left(\sigma(n) - n\right) \Longleftrightarrow 1 + \frac{D(n)}{n} = (p - 1)\left(I(n) - 1\right)$$

and

$$2n + D(n) = (p - 1)\left(\sigma(n) - n\right) + n \Longleftrightarrow 2 + \frac{D(n)}{n} = (p - 1)\left(I(n) - 1\right) + 1.$$

Thus, we have the lower bound

$$\frac{2}{(p - 1)\left(I(n) - 1\right)} < I(n)$$

which implies that (upon setting $I(n) = \frac{p+2}{p}$)

$$\frac{2}{(p - 1)\left(\frac{p+2}{p} - 1\right)} < \frac{p + 2}{p},$$

from which we obtain

$$2p^2 < 2(p - 1)(p + 2) = 2(p^2 + p - 2) = 2p^2 + 2p - 4.$$

Lastly, this implies

$$2 < p.$$

No contradictions so far. We now consider the upper bound.

We have the upper bound

$$I(n) < \frac{(p - 1)\left(I(n) - 1\right) + 1}{(p - 1)\left(I(n) - 1\right)},$$

which implies that

$$I(n) - 1 < \frac{1}{(p - 1)\left(I(n) - 1\right)}.$$

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

$$\frac{2}{p} < \frac{1}{(p - 1)\frac{2}{p}}.$$

This implies that

$$4(p - 1) < p^2.$$

Thus, $p^2 - 4p + 4 = (p - 2)^2 > 0$. Still no contradictions.

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 $D(n)>1$.