The question is :

What positive integers has exactly (and prove your result) (a) one positive divisor; (b) two positive divisors; (c) three positive divisors; (d) four positive divisors; (e) five positive divisors.

I think that,

for (a), the answer should be only integer 1.

for (b), that answer should be all the primes

for (c), that answer should be all the integer a satisfy a = prime^2

(d) and (e) should be similar but I am not sure.

My confusion is about how to prove these five cases. Any hints or sample prove for part (c) will be pretty helpful.

  • $\begingroup$ Are (c) the square of a prime number? $\endgroup$ – Sylvain Biehler Mar 4 '14 at 14:48
  • $\begingroup$ Yes you are right. I have corrected that. Thank you $\endgroup$ – SonicFancy Mar 5 '14 at 19:45

If $n=p_1^{k_1} \cdots p_r^{k_r}$ is the prime factorization of $n$ then $n$ has $d$ divisors where $$d=(k_1+1)(k_2+1)\cdots (k_r+1).$$ Thus you need to take the goal number of divisors and factor it in all ways as a product of one or more factors each greater than 1.

This means for example in case (d) where you want 4 divisors, the possibilities are (4) and (2)(2), which means going back that $n$ is either a single prime cubed i.e. $p^3$ or else $n$ is the product of two different primes each to the first, i.e. $pq$ where $p,q$ are distinct primes.

Note that if you apply this to case (c) you can see that it must be $n=p^2$ for $p$ a prime, and not "prime x prime" (unless you meant the same prime twice).

See this page for an explanation of the above formula for the number of divisors $d$, but note on that page it is notated as $\sigma_0(n)$ where the subscript $0$ means one is summing the $0$th powers of the divisors, i.e. counting them.


Well, any positive integer $\ge 1$ can be written as the product of primes, like $$ n = p_1^{e_1}\cdot p_2^{e_2}\cdot p_3^{e_3}\cdot ... p_n^{e_n} = \prod_{i=0}^n p_i^{e_i} $$ so, all the divisors of $n$ are all the possible values of $p_1^{e_1}\cdot p_2^{e_2}\cdot p_3^{e_3}\cdot ... p_n^{e_n}$, which are evidently $$(e_1+1)\cdot (e_2+1)\cdot (e_3+1) ... (e_n+1) $$ because any of the divisors of $p_i$ (from 1 to $p_i^{e_i}$) divides also $n$. So if the number of the divisors of $n$ is $d$, $d$ must be the product of some integer number greater than 2.

  • a) $d=1$, there are no such $n \ge 1$ that have only 1 divisor (at least 2: 1 and $n$); by the way 1 is the only positive integer that has only 1 divisor
  • b) $d=2$, $n$ must be a prime, because it can be divisible only by 1 and itself
  • c) $d=3$= product of integers $\ge 2$ ($\to$ only solution: 3), means that $n$ must be written as $p^2$, having 3 different divisors (1,$p$ and $p^2$)
  • d) $d=4=2\cdot 2=1\cdot 4$, it occurs that $n=p\cdot m$ with $p$ and $m$ prime numbers or $n=p^3$ are the 2 only solutions
  • e) $d=5$,the only solution is $n = p^4$, as the divisors are 1, $p$, $p^2$, $p^3$ and $p^4$
  • $\begingroup$ You left out the second case for part d. $\endgroup$ – Charles Mar 4 '14 at 15:35
  • $\begingroup$ sorry...I'm correcting.. $\endgroup$ – sirfoga Mar 4 '14 at 15:51

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