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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!

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    $\begingroup$ It wouldn't be a bad idea to define product-perfect number since that's a relatively uncommon phrase; from context, I'm presuming you mean 'a number that is a product of all its divisors other than 1 and the number', but that's not entirely clear... $\endgroup$ – Steven Stadnicki Feb 13 '15 at 22:07
  • $\begingroup$ Also see OEIS A007422. $\endgroup$ – Jeppe Stig Nielsen Feb 13 '15 at 23:15
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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.

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Let $f(n)$ be the product of the factors of $n$. Can you prove that $f$ is multiplicative, that is that $f(mn)=f(n)f(m)$ for $n,m$ coprime? Also note that $f(n)\ge n$, because $n$ is a factor and there may be more. Now assume that $n$ has at least three primes that divide it and write it as $n=p^aq^br$, with $p,q$ prime and not dividing $r$ (though $r$ may be composite). We have factors of $p^a,q^b,r,p^aq^b,n$ that are enough to show $f(n) \gt n^2$, so $n$ has at most two prime factors. If it has two distinct prime factors, write it as $p^aq^b$ and note that we have factors $p^a,q^b,n$ that multiply to $n^2$, so if either $a,b \gt 1$ we get another factor and are done. Then look at powers of a single prime and find the cube is the only one that works.

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