I've just figured out that a product perfect number is a number N such that either N=pq or $N=q^3$. But how do I prove this?
Attempt: I know that any number of this form is a product perfect number since p and q cannot be factored any further so N being product perfect is just restating the prime factorization of N.
I've tried to prove the other direction but I came across too many cases to consider and some of which I can't solve.
Any help would be great. Thanks!
Best Answer
It is almost Valentine's Day. So if $a$ and $b$ are positive integers such that $a\ne b$ and $ab=N$, call $a$ and $b$ lovers. There are two cases to consider, (i) $N$ is not a perfect square and (ii) $N$ is a perfect square.
Case (i): If $N$ is not a perfect square, then the positive divisors of $N$ are divided into pairs of lovers. The number of pairs is $\frac{d(N)}{2}$, where $d(N)$ is the number of positive divisors of $N$.
The product of the divisors of $N$ is therefore $N^{d(N)/2}$. This is $N^2$ (meaning that $N$ is multiplicatively perfect) precisely if $d(N)=4$. This is the case if $N$ is the product of $2$ distinct primes or if $N$ is the cube of a prime. Here we have used the fact that if $N=p_1^{a_1}\cdots p_k^{a_k}$, where the $p_i$ are distinct primes, then $d(N)=(a_1+1)\cdots (a_k+1)$, though one can do it with less machinery.
Case (ii): Let $N$ be a perfect square, say $m^2$. Then the divisors of $N$ consist of pairs of lovers, together with the solitary divisor $m$. So the product of the divisors is $N^{(d(N)-1)/2}\sqrt{N}$, which is $N^{d(N)/2}$. For this to be $N^2$ we want either $N=1$ or $\frac{d(N)}{2}=2$, which is impossible if $N$ is a perfect square. So in addition to the examples you mentioned, $1$ is also multiplicatively perfect.
\