## Multiplicatively Perfect

#### Problem

The proper divisors of a positive integer, $n$, are all the divisors excluding $n$ itself. For example, the proper divisors of $6$ are $1, 2,$ and $3$.

A number, $n$, is said to be multiplicatively perfect if the product of its proper divisors equals $n$. The smallest such example is six: $6 = 1 \times 2 \times 3$ the next such example is eight: $8 = 1 \times 2 \times 4$.

Determine the nature of all multiplicatively perfect numbers.

#### Solution

We shall begin by exploring numbers of the form, $n = p^k$, where $p$ is prime.

Thus we are looking for $n = p^k = 1 \times p \times p \times p^2 ... \times p^{k-1} = p^{1 + 2 + ... + (k-1)}$.

$$\begin{align}\therefore 1 + 2 + ... + (k-1) = \frac{1}{2}k(k -1) &= k\\k^2 - k &= 2k\\\therefore k(k - 3) &= 0 \Rightarrow k =3\end{align}$$Hence the cube of all primes will be multiplicatively perfect.

It should be clear that any number of the form, $n = pq$, where $p$ and $q$ are distinct prime, will also be multiplicatively perfect, as the proper divisors will be $1, p$, and $q$.

Moreover we can prove that where there are at least two prime factors it is only numbers that have exactly two distinct prime factors that are multiplicatively perfect.

Suppose that $n = mpq$, where $m$ is some integer greater than 1 which could be prime or composite.

The proper divisors of $n$ will be at least $1, m, p, q$, but also $mp, mq, pq$, and possibly more if $m$ is composite. Therefore the product of proper divisors will exceed $n$ and it cannot be multiplicatively perfect.

Hence $n$ will be multiplicatively perfect iff it is of the form $p^3$ or $pq$.