Show that every prime number in the form $a+b$ with $a,b$ divisors of $n$ is distinct and not divides $n$ Recently, I have found this problem:

Let $n$ a natural number. Suppose that its positive divisors can be partitioned in tuples of the form $(a,b)$ such that the sum $a+b$ is a prime number. Show that every such prime number is distinct and no of them divides $n$.

I have tried to solve this problem for hours, but I can't completely figure out a solution.
I think that with $n=p^k$ the problem can't be solved because in the set $D=(1,p,p^2,\cdots,p^k)$ can't be partitioned in tuples because every sum ($a+b$) can't be a prime. Any idea of how to proceed?
 A: With the factor of $n$ itself, since any divisor other than $1$ has at least one prime factor in common with $n$, the other divisor must be $1$ itself, i.e., you have $(n,1)$ with $n + 1$ being prime.
Next, consider any prime $p$ where $p \mid n$ and set $a = \frac{n}{p}$. As the question states, there's another divisor $b$ where $a + b$ is prime. If $n$ has more than one factor of $p$, then $a$ has the same set of primes which are factors of $n$ so any $b \gt 1$ (since $1$ is already matched up with $n$) must have at least one prime factor in common with $a$ so $a + b$ can't be prime. This shows $n$ can only have one factor of $p$. Also, since all other prime factors of $n$ divide $a$, this means that $b$ can only be $p$ itself to ensure $a + b$ is prime.
This shows $n$ is square-free, with some $m \ge 1$ distinct primes where
$$n = \prod_{i=1}^{m}p_i \tag{1}\label{eq1A}$$
Note if you have $a$ being $n$ divided by the product of $2$ primes, each of the primes individually has been used before, and no other prime factor may be used since it's a factor of $a$, so the other value, i.e., $b$, must be the product of those $2$ primes. In general, you can prove by induction on the number of primes that due to any smaller # of primes already being used previously, you have each factor being paired with $n$ divided by that factor, e.g., $a = \frac{n}{b}$ for all factors $b$ of $n$, say with $a \gt b$ for uniqueness. I'll leave proving this to you to do.
As for showing the constructed primes are distinct, assume you have $(\frac{n}{b_1},b_1)$ and $(\frac{n}{b_2},b_2)$ with $b_1 \neq b_2$, with
$$\begin{equation}\begin{aligned}
\frac{n}{b_1} + b_1 & = \frac{n}{b_2} + b_2 \\
b_2(n) + b_1^2b_2 & = b_1(n) + b_1b_2^2 \\
b_2(n) - b_1(n) & = b_1b_2^2 - b_1^2b_2 \\
(b_2 - b_1)n & = b_1b_2(b_2 - b_1) \\
n & = b_1b_2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This means $\frac{n}{b_1} = b_2$ and $\frac{n}{b_2} = b_1$ so the two pairs are the same with their values just switched around. This confirms that all of the $a + b$ primes must be unique.
As for showing none of these primes divide $n$, first note that $n + 1 \not\mid n$. As for showing none of the other ones divide $n$, consider that one of them do, so you have for some $b_1$ dividing $n$ and integer $k \ge 1$ that
$$\begin{equation}\begin{aligned}
k\left(\frac{n}{b_1} + b_1\right) & = n \\
kn + kb_1^2 & = nb_1 \\
kn & = b_1(n - kb_1)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since $b_1 \mid n$, this means the RHS has at least $2$ factors of $b_1$. With the RHS, as $n$ has only $1$ factor of $b_1$, this means $k$ must have at least one factor of $b_1$, so $k = rb_1$ for some integer $r \ge 1$. However, this would then give
$$\begin{equation}\begin{aligned}
rb_1\left(\frac{n}{b_1} + b_1\right) & = n \\
rn + rb_1^2 & = n
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
However, with $r \ge 1$, the LHS is $\gt n$, so it's not possible to be equal to $n$. This shows the assumption must be incorrect, which proves $\frac{n}{b_1} + b_1 \not\mid n$, i.e., none of these primes constructed from the sum of factors divide $n$.
As indicated in several question comments by lulu, since $n + 1$ is prime and $n \neq 1$, this means $n$ must be even. Since it's square-free, this means $n = 2q$ for some odd $q$. Several examples which work are $n = 2(5)$ and $n = 2(3)(5)$, although I also don't know if there are infinitely many such $n$.
