Canonical decompositions and product of primes

Question

Let $S$ be the set of natural numbers $n$ that have exactly $9$ positive divisors. Describe all possible canonical decompositions (as products of primes) of elements of $S$.

Answer 1

If $$ n=p_1^{k_1}\cdots p_m^{k_m} $$ then the number of divisors is $$ N=(1+k_1)\cdots(1+k_m). $$ So if $N=9$, then either $$ N=p^8, $$ for some prime $p$ or $$ N=p^2q^2, $$ where $p$, $q$ are distinct primes.

Answer 2

Do you count 1 and itself? If yes then it leaves 7 divisors.

Now consider the product of two prime numbers, p and q:

divisors: 1, pq,p,q -->4

Now with three prime number p,q,r:

divisors: 1,pqr, p, q, r, pq, pr, qr --> 8, not possible

So it has at most two different prime numbers, since a power increases the number of divisor

So n=p^a*q^b

we want suppose both a and b are >1. For the same reason, say that a=2, b=2

divisors: 1, p^2*q^2, p^2, q^2, p, q, p^2*q, pq^2, pq --> not possible

So at least a or b equals 1. Let's say a=1

n=p*q^b

divisors: 1, n, p, pq, pq^2,..,p*q^(b-1)

--> b=5

So these are the integers that are of the for p*q^5 , with p and q prime numbers

Answer 3

Hint: If $m = \prod_i p_i^{n_i}$ then the number of divisors of $m$ is $$ \prod_i (n_i + 1). $$ So the question really asks for which sequences $1 \leq n_1 \leq \cdots \leq n_t$ (for some $t$) is it the case that $(n_1+1)\cdots(n_t+1) = 9$. The relevant ways of factoring $9$ are $9 = 9 = 3 \cdot 3$, and this should give you the answer.