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$n$ is integer with exactly 4 different positive divisors. I need to find all possibilities for the number of prime factors of n.

I know that each positive divisor of n is from the form:

$d = p_1 ^{a_1}p_2 ^{a_2}...p_k ^{a_k} $ but I guess maybe I don't totally understand the question becuase I don't know how to go from here.

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I think you're trying to jump to full generality too quickly.

Can it have $1$ prime factor? Sure, if $n=p^3$ then its divisors are $1$, $p$, $p^2$ and $p^3$.

Can it have $2$ prime factors?

Can it have $3$ prime factors?

Can it have more than $3$ prime factors?

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  • $\begingroup$ If I understood than it can have 2 prime factors - p1,p1 and than we have the divisors 1,p1,p2,p1p2 BUT it can't have 3 prime factors Is that correct? $\endgroup$ – baaa12 Apr 4 '13 at 14:22
  • $\begingroup$ Yes that's right. Because if $n=p^aq^br^c$, where $p$, $q$ and $r$ are distinct primes, then there are way more than four positive divisors; at the very least you have eight: $1, p, q, r, pq, pr, qr, pqr$. $\endgroup$ – Clive Newstead Apr 4 '13 at 14:40
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hint: if $d = p_1 ^{a_1}p_2 ^{a_2}...p_k ^{a_k} $, then the number of divisors of $d$ is $(a_1+1)(a_2+1) \cdots (a_k+1)$.

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Hint: when you write $n=p_1^{a_1} \cdots p_k^{a_k}$, how many prime factors of $n$? The answer should be in term of $a_1 ,\ldots, a_k$.

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