I want to find a perfect number in the form $p^2q$ where $p$ and $q$ are both prime numbers.
I get the diophantine equation :$p^2q=p^2+pq+p+q+1$ from the definition of the perfect number.
So I assume that there is one positive integer solution which is $(p,q)$ and $c=p+q$ which is a constant.
Rearranging the equation, I get $p^3-cp^2+cp+c+1=0$
So $p^3+1=(p+1)(p^2-p+1)=c(p^2-p-1)$
Thus $c=\frac{(p+1)(p^2-p+1)}{p^2-p-1}=p+1+\frac{2p+2}{p^2-p-1}$
Obviously $c-p-1$ is a positive integer since $p,q$ are prime numbers which are greater than $1$.
So assume positive integer $k=\frac{2p+2}{p^2-p-1}$
Thus $kp^2-(k+2)p-(k+2)=0$
Apply the root formula, we get $p=\frac{k+2+\sqrt{(k+2)^2+4k(k+2)}}{2k}$
so $(k+2)^2+4k(k+2)=n^2$ where $n$ is a positive integer.
Thus $n^2-(k+2)^2=(n-k-2)(n+k+2)=4k(k+2)$
Then what I can do next? I try to list all of the situations but $k$ is not a prime number.
Solving $p^2q=p^2+pq+p+q+1$ for $q$ we get $$ q = \frac{p^2 + p + 1}{p^2 - p - 1} = 1 + \frac{2p+2}{p^2 - p - 1} $$ The last fraction needs to be an integer and so we need $2p+2\ge p^2 - p - 1$. This gives $p \in \{0,1,2,3\}$. The only solution is $p=2$ and $q=7$, which gives $p^2q=28$.