..........

But $2^3$ is not the smallest number with $4$ divisors - $6$ is.

I thought meant that $n$ is the smallest cube that fulfills the requirements..

The answer is yes. Consider a positive integer $n=h^{3}$ and call $k$ the number of its divisors. Since $n$ is a cube, it can be written as $p_{1}^{3a_{1}}\cdot p_{2}^{3a_{2}} \cdot ... \cdot p_{y}^{3a_{y}}$. So $k = (3a_{1}+1)(3a_{2}+1)...(3a_{y}+1)$ and we easily obtain that $k \equiv 1 (mod 3)$. Now suppose that $k$ is divisible by a prime factor $p \equiv 2 (mod3)$. We obtain that $k \equiv 1 \equiv px \equiv 2x (mod3) $, where $x$ is a positive integer, so we have $1 \equiv 2x (mod3) \Leftrightarrow x\equiv 2(mod3)$. For example choose $x=2,p=5, \rightarrow k=10\rightarrow n = p_{1}^{9} = (p_{1}^{3})^{3}$.

Benny140 wrote: The answer is yes. Consider a positive integer $n=h^{3}$ and call $k$ the number of its divisors. Since $n$ is a cube, it can be written as $p_{1}^{3a_{1}}\cdot p_{2}^{3a_{2}} \cdot ... \cdot p_{y}^{3a_{y}}$. So $k = (3a_{1}+1)(3a_{2}+1)...(3a_{y}+1)$ and we easily obtain that $k \equiv 1 (mod 3)$. Now suppose that $k$ is divisible by a prime factor $p \equiv 2 (mod3)$. We obtain that $k \equiv 1 \equiv px \equiv 2x (mod3) $, where $x$ is a positive integer, so we have $1 \equiv 2x (mod3) \Leftrightarrow x\equiv 2(mod3)$. For example choose $x=2,p=5, \rightarrow k=10\rightarrow n = p_{1}^{9} = (p_{1}^{3})^{3}$. For $k=10$ you can take $2^43<2^9$ so your example fails. smy2012 wrote: It seems that for $k=5^2\cdot 7^{10}$ The smallest $n=2^{24}\cdot 3^6\cdot5^6\cdot7^6\cdot\ldots\cdot31^6$. I spent 3hrs trying to prove the contrary, and sought for inspiration from the above example. It occurs to me that the above example is exactly a counter example. It's not clear why $n=2^{24}\cdot 3^6\cdot5^6\cdot7^6\cdot\ldots\cdot31^6$ is smallest such $n$.

Wolowizard: $n$ has to be a cube

@above So $k=10$ won't met the requirement since $2^4\times 3$ is not a cube. And so $10$ can't be the example.

This is one of the most beautiful and surprising number theory problems I've ever seen. It took me quite some time to find the following solution: The answer is negative. Suppose the result is false, then there exists some prime $p$ and some positive integers $q, r \equiv 2 \pmod{3}$ such that $p^{q r - 1} \mid \mid n$. First assume that $r = q$. The key idea is to consider $p_k$ as the largest prime smaller than $p^{q}$ and $p_{k+1}$ as the smallest prime greater than $p^{q}$. Let $p_k^{a-1} \mid \mid n$ and $p_{k+1}^{b-1} \mid \mid n$. Then we can write \[ n = p^{q^2-1} \cdot p_k^{a-1} \cdot p_{k+1}^{b-1} \cdot m \]where $m$ is co-prime with $p, p_k, p_{k+1}$. We observe that the following number also has exactly $k$ divisors: \[ n' = p^{qa-1} \cdot p_k^{q-1} \cdot p_{k+1}^{b-1} \cdot m. \]Thus $n' \geq n$, which is equivalent to $p^{q(a-q)} \geq p_k^{a-q}$. Since $p_k < p^{q}$, we deduce $a \geq q$. Analogously, by considering $n'' = p^{qb-1} \cdot p_k^{a-1} \cdot p_{k+1}^{q-1} \cdot m$, we can deduce $q \geq b$. But by assumption both $a$ and $b$ are $1 \pmod{3}$. So we must in fact have $a > q > b$. Next, consider the following number \[ n''' = p^{ab-1} \cdot p_k^{q-1} \cdot p_{k+1}^{q-1} \cdot m. \]It also has exactly $k$ divisors, so we must have $n''' \geq n$, which is just \[ p_{k+1}^{q-b} \geq p^{q^2-ab} \cdot p_k^{a-q}. \]If $a + b \leq 2q$, we rewrite the above as $(\frac{p_{k+1}}{p_k})^{q-b} \geq p^{q^2-ab} \cdot p_k^{a+b-2q} \geq p^{q^2-ab} \cdot p^{qa+qb-2q^2} = p^{(a-q)(q-b)}$. Thus $\frac{p_{k+1}}{p_k} \geq p^{a-q} \geq p \geq 2$, contradicting Bertrand's postulate. If $a + b > 2q$, we can rewrite and obtain $(\frac{p_{k+1}}{p_k})^{a-q} \geq p^{q^2-ab} \cdot p_{k+1}^{a+b-2q} \geq p^{q^2-ab} \cdot p^{qa+qb-2q^2} = p^{(a-q)(q-b)}$. Hence $\frac{p_{k+1}}{p_k} \geq p^{q-b} \geq p \geq 2$, again contradicting Bertrand's postulate. Either way, we have derived a contradiction assuming that $p^{qr-1} \mid \mid n$ and $r = q$. The case where $r > q$ can be similarly handled.

angiland wrote: This is one of the most beautiful and surprising number theory problems I've ever seen. It took me quite some time to find the following solution: The answer is negative. Suppose the result is false, then there exists some prime $p$ and some positive integers $q, r \equiv 2 \pmod{3}$ such that $p^{q r - 1} \mid \mid n$. First assume that $r = q$. The key idea is to consider $p_k$ as the largest prime smaller than $p^{q}$ and $p_{k+1}$ as the smallest prime greater than $p^{q}$. Let $p_k^{a-1} \mid \mid n$ and $p_{k+1}^{b-1} \mid \mid n$. Then we can write \[ n = p^{q^2-1} \cdot p_k^{a-1} \cdot p_{k+1}^{b-1} \cdot m \]where $m$ is co-prime with $p, p_k, p_{k+1}$. We observe that the following number also has exactly $k$ divisors: \[ n' = p^{qa-1} \cdot p_k^{q-1} \cdot p_{k+1}^{b-1} \cdot m. \]Thus $n' \geq n$, which is equivalent to $p^{q(a-q)} \geq p_k^{a-q}$. Since $p_k < p^{q}$, we deduce $a \geq q$. Analogously, by considering $n'' = p^{qb-1} \cdot p_k^{a-1} \cdot p_{k+1}^{q-1} \cdot m$, we can deduce $q \geq b$. But by assumption both $a$ and $b$ are $1 \pmod{3}$. So we must in fact have $a > q > b$. Next, consider the following number \[ n''' = p^{ab-1} \cdot p_k^{q-1} \cdot p_{k+1}^{q-1} \cdot m. \]It also has exactly $k$ divisors, so we must have $n''' \geq n$, which is just \[ p_{k+1}^{q-b} \geq p^{q^2-ab} \cdot p_k^{a-q}. \]If $a + b \leq 2q$, we rewrite the above as $(\frac{p_{k+1}}{p_k})^{q-b} \geq p^{q^2-ab} \cdot p_k^{a+b-2q} \geq p^{q^2-ab} \cdot p^{qa+qb-2q^2} = p^{(a-q)(q-b)}$. Thus $\frac{p_{k+1}}{p_k} \geq p^{a-q} \geq p \geq 2$, contradicting Bertrand's postulate. If $a + b > 2q$, we can rewrite and obtain $(\frac{p_{k+1}}{p_k})^{a-q} \geq p^{q^2-ab} \cdot p_{k+1}^{a+b-2q} \geq p^{q^2-ab} \cdot p^{qa+qb-2q^2} = p^{(a-q)(q-b)}$. Hence $\frac{p_{k+1}}{p_k} \geq p^{q-b} \geq p \geq 2$, again contradicting Bertrand's postulate. Either way, we have derived a contradiction assuming that $p^{qr-1} \mid \mid n$ and $r = q$. The case where $r > q$ can be similarly handled. Can you post a solution for $q < r$? It seems not so trivial by using the argument of $q = r$

To fill the slight details left by the great @angiland's solution (since it required changes of variables and bounds, please allow me to re-write the whole argument): Suppose to the contrary. There exist a prime $p$ and positive integers $q,r\equiv 2\pmod{3}$ that $p^{qr-1}\mid \mid n$ and $r\geqslant q$. Let $p_k$ be the greatest prime that is smaller than $p^q$ and let $p_{k+1}$ be the smallest prime that is larger than $p^r$. Let $n=p^{qr-1}\cdot p_k^{a-1}\cdot p_{k+1}^{b-1}\cdot m$ where $\gcd (m,p\cdot p_k\cdot p_{k+1})=1$. Recall $p_k<p^q<p^r<p_{k+1}$ and $3$ divides all of $q-2,r-2, a-1,b-1$. Consider the following positive integers: \begin{align*} & n_1=p^{qa-1}\cdot p_k^{r-1}\cdot p_{k+1}^{b-1}\cdot m\\ & n_2=p^{rb-1}\cdot p_k^{a-1}\cdot p_{k+1}^{q-1}\cdot m\\ & n_3=p^{ab-1}\cdot p_k^{r-1}\cdot p_{k+1}^{q-1}\cdot m\\ \end{align*}Note that they all have the same number of divisors with $n$. From $n_1\geqslant n$ we get $p^{q(a-r)}\geqslant p_k^{a-r}\implies a\geqslant r$. From $n_2\geqslant n$ we get $p^{r(b-q)}\geqslant p_{k+1}^{b-q}\implies b\leqslant q$. We conclude that $a>r\geqslant q>b$. From $n_3\geqslant n$ we get $p_{k+1}^{q-b}\geqslant p^{qr-ab}\cdot p_k^{a-r}$. By Bertrand's postulate, note that $p_{k+1}<2\cdot p^r$ and $p_k>p^q/2\implies p_{k+1}/p_k<4\cdot p^{r-q}$. We then divide into two cases: If $a+b\leqslant q+r\implies q> b$. We get $$\left( \frac{p_{k+1}}{p_k}\right)^{q-b}\geqslant p^{qr-ab}\cdot p_k^{a+b-q-r}\geqslant p^{qr-ab}\cdot p^{q(a+b-q-r)}=p^{(a-q)(q-b)}.$$This implies $p^{a-q} \leqslant p_{k+1}/p_k<4\cdot p^{r-q}\implies 4>p^{a-r}$ but $a-r\equiv -1\pmod{3}\implies p^{a-r}\geqslant 2^{a-r}\geqslant 4. \quad \perp$ If $a+b>q+r$. We get $$\left( \frac{p_{k+1}}{p_k}\right)^{a-r}\geqslant p^{qr-ab}\cdot p_{k+1}^{a+b-q-r}\geqslant p^{qr-ab}\cdot p^{r(a+b-q-r)}=p^{(a-r)(r-b)}.$$This implies $p^{r-b}\leqslant p_{k+1}/p_k<4\cdot p^{r-q}\implies 4>p^{q-b}\underset{q>b}{\implies} q-b=1$ and $p\in \{ 2,3\}$. Since $3\mid q-2$ we get that either $q=2$ or $q\geqslant 5\implies p^q\geqslant 2^5=32\implies \lfloor p^q\cdot 5/6\rfloor \geqslant 25$. In the former case we have $(p_k,p^q)\in \{ (3,4),(7,9)\} \implies p_k/p^{q-1}\in \{ 3/2,7/3\} \implies p_k/p^{q-1}\geqslant 3/2$. In the latter case, by Nagura's theorem, we get that $p_k\geqslant p^q\cdot 5/6\implies p_k/p^{q-1}\geqslant p\cdot 5/6\geqslant 3/2$. Back to the inequality $p_{k+1}^{q-b}\geqslant p^{qr-ab}\cdot p_k^{a-r}$ we get the bound $$p_{k+1}\geqslant p^r\cdot \left( \frac{p_k}{p^{q-1}}\right)^{a-r}\geqslant p^r\cdot \left( \frac{3}{2}\right)^{a-r}\geqslant p^r\cdot \left( 1+\frac{a-r}{2}\right)\geqslant p^r\cdot 2.$$This contradicts Bertrand's postulate and we are done.

Suppose k exist, let p_i be all prime divide n and k=(a1+1)..(am+1) and 3|ai forall i, by minimality of n we have a1>=..>=am Lemma: Assume a_r+1=ab with a,b<>1. If p_s<p_r^a<p_{s+1} then a_s>=b-1>=a_{s+1} Proof: Consider n1=p_r^{(a_s+1)a-1}.p_s^{b-1}.. has k divisors so n1>=n but (p_r^a/p_s)>=1 mean a_s>=b-1 similar define n' we have a_{s+1}<=b-1. Consider r max that a_r+1=ab with a=b=2[3], let s,t satisfy p_s<p_r^a<p_{s+1} and p_t<p_r^b<p_{t+1}. By Bertrand, p_r^{a-1}<p_s<p_r^a<p_{s+1}<p_r^{a+1} similar for b. It follows that s,t>r and similarly |s-t|<>1. Now a_s>b-1>a_{s+1} and a_t>a-1>a_{t+1}. Choose n2=p_r^[(a_s+1 (a_{t+1}+1)-1]. p_s^(b-1).p_{t+1}^(a-1) has k divisors, n2>=n hence by p_{t+1}<p^(b+1) and p_s>p_r^{a-1}, we have 1<=p_r^[1-(a_s-b)(a-2-a-a_{t+1}) hence (a_s-b)(a-2-a_{t+1})<=0 so a_{t+1}=a-2 but a_r+1 is odd so 2|a similar 2|b so a_r=4c-1. From 2|a_m we have a_m>3>1>a_{m+1}=0 so apply lemma for (a,b)=(2,2c) and (4,c) gives us p_r^2c and p_r^c both in (p_m,p_{m+1}) which is contradict with bertrand postulate hence q.e.d