Find all $n$ such that $n/d(n) = p$, a prime, where $d(n)$ is the number of positive divisors of $n$ 
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.

I tried this, but only I could get two solutions.
I proceeded like this - 
Suppose 
$$n = p^r \cdot p_1^{r_1} \cdot \cdots \cdot p_k^{r_k}$$ 
where the $p_i$s are distinct primes.
Given 
$$n=p\cdot d(n)=p(r+1)(r_1+1)\cdots (r_k+1)$$
$$p^r \cdots p_1^{r_1}\cdot \cdots \cdot p_k^{r_k}=(r+1)(r_1+1)\cdots(r_k+1)$$
If $k=0$, then $p^(r-1)=r+1$. Hence $p=2$ and $r=3$, or $p=3$ and $r=2$. I.e., $n=8$, or $n=9$.
But, when I am assuming $k>0$, I am finding no clue.
For this, I need help.
Thanks in advance!
 A: Observation 1. It is easy to prove (e.g. by induction) that:
$$p^n \geq n+1, \forall p\geq2, \forall n\geq 1$$
Thus
$$p^r\cdot \color{blue}{p_1^{r_1}\cdot ... \cdot p_k^{r_k}}=
p(r+1)\cdot \color{blue}{(r_1+1)...(r_k+1)}\geq 
p^r \cdot \color{blue}{(r_1+1)...(r_k+1)}$$
or
$$p(r+1)\geq p^r \iff r+1\geq p^{r-1}$$
These are the only $(r,p)$ combinations possible $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$ and finally $(1,p), \forall p>3, p$ - prime.

Observation 2. Let's check the following case $(1,p), \forall p>3, p$ - prime (i.e. $r=1$).
$$p\cdot \color{blue}{p_1^{r_1}\cdot ... \cdot p_k^{r_k}}=
p\cdot 2\cdot \color{blue}{(r_1+1)...(r_k+1)} \iff \\
p_1^{r_1}\cdot ... \cdot p_k^{r_k}=
2\cdot (r_1+1)...(r_k+1) \iff ...$$
or one of the $p_i=2$, for simplicity let's say $p_1=2$.
$$... \iff 2^{r_1-1}\cdot \color{blue}{p_2^{r_2} ... \cdot p_k^{r_k}}=
(r_1+1)\color{blue}{(r_2+1)...(r_k+1)}\geq 
 2^{r_1-1}\cdot \color{blue}{(r_2+1)...(r_k+1)}$$
or again
$$(r_1+1)\geq 2^{r_1-1}$$
or $r_1 \in \{1,2,3\}$. 
This reduces the problem to the following cases $(1,2), (2,2), (3,2)$ then $(1,3),(2,3)$.
