Perfect Number of the form $p^2q$

Question

Are there any perfect numbers of the form $p^2q$, with $p$ and $q$ positive integer primes? Obviously the proper divisors of the number would be $1, p, q, pq, \text{ and } p^2$ and therefore we want to consider the diophantine equation $$p^2q=p^2+pq+p+q+1$$

Of course, there's an easy solution if you apply the Euclid-Euler theorem for even perfect numbers, but that's not totally general. How can you find all positive integer prime solutions to this diophantine equation?

Answer 1

The equation $p^2q=p^2+pq+p+q+1$ can be rewritten as

$$(p^2-p-1)q=p^2+p+1=(p^2-p-1)+2p+2$$

which implies $(p^2-p-1)$ divides $2p+2$, which in turn implies $p^2-p-1\le2p+2$, or $p^2-3p-3\le0$, which implies $p\le3$. Of the two possibilities, $p=2$ gives a solution: $(4-2-1)q=4+2+1$ implies $q=7$, which is prime. But $p=3$ does not: $(9-3-1)q=9+3+1$ implies $q=13/5$, which is not an integer, much less prime.

Answer 2

You may be able to use your method, but here's a more analytic approach I would use.

Graph $x^2y = x^2+xy+x+y+1$ and see

You're only interested in positive integer solutions. It's trivial to show with the rational root theorem that for $x=1$ there is no valid positive integer $y$. $x=2$ yields the solution $(2, 7)$, and $x=3$ has a rational root (at $\frac{13}{5}$ which you can easily see is valid, unique and nonintegral with, say, the rational root theorem again) is the only positive rational solution and therefore yields no solutions.

This function is linear in $y$, and therefore you can solve for $y$ and show that this is a function $\mathbb{R}\to\mathbb{R}$ for say $x \geq 2$ (or anywhere beyond the vertical asymptote). From here, I leave it as an exercise in inequalities (there's the analysis) that $1 < y <2$ for $x \geq 4$.

Once you've filled in the details, this is a very intuitive way to see that $(2, 7)$ and therefore $\boxed{28} $ is the only solution.