(This question is related to the latest comment underneath this MSE question.)
Let $$\sigma(x)=\sum_{d \mid x}{d}$$ be the sum of the divisors of $x$.
Denote the abundancy index of $x$ by $$I(x)=\frac{\sigma(x)}{x}.$$
Here is my question:
If $I(x)=\dfrac{p+2}{p}$, where $p$ is an odd prime, does it follow that $x$ is an odd square?
From the equation $$p\sigma(x)=(p+2)x$$ I can only prove that $x$ is an odd square when $x$ is odd. But what about if $x$ is even?
That is, if $I(x) = \dfrac{p+2}{p}$, where $p$ is an odd prime, must it necessarily be the case that $x$ is odd?
First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$ I(ab) = I(a) I(b). $$
Suppose there is an even number $n = 2^k \cdot l$, where $k \geqslant 1$ and $l$ odd, such that$$ I(n) = \frac{p + 2}{p}. $$
Case 1: $k \geqslant 2$. Then$$ \frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geqslant I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geqslant \frac{7}{4}, $$ which implies $4(p + 2) \geqslant 7p$, contradictory to $p \geqslant 3$.
Case 2: $k = 1$ and $p \geqslant 5$. Then, analogously,$$ \frac{p + 2}{p} \geqslant 2 - \frac{1}{2^k} = \frac{3}{2}, $$ which implies $2(p + 2) \geqslant 3p$, contradictory to $p \geqslant 5$.
Case 3: $k = 1$ and $p = 3$. Then$$ \frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}. $$
If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathrel{\|} l$, then$$ \frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geqslant I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\ = 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geqslant 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9}, $$ a contradiction. Now suppose the prime factorization of $l$ is$$ l = \prod_{i = 1}^s p_i^{a_i}, $$ then $p_i \geqslant 11$. Because$$ \frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}}, $$ then$$ 9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right., $$ contradictory to $p_i \geqslant 11$.
Therefore, there does not exist an even positive integer $n$ such that$$ I(n) = \frac{p + 2}{p}. $$