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Let $d(n)$ be the number of positive divisors of $n$. Find all $n$ such that $\frac{n}{d(n)}=p$, a prime.

If $n=\prod_{1\leq i \leq k} p_i^{r_i}$, then \begin{eqnarray*} n&=&p_1\cdot d(n)=p_1(r_1+1)(r_2+1)...(r_k+1)\\ &\ & \implies p_1^{r_1-1}p_2^{r_2}...p_k^{r_k}=(r_1+1)(r_2+1)...(r_k+1). \end{eqnarray*} But I do not get any clue here. (I previously posed a question which asked only for primes but that got only one answer, which was just an observation. But now i have deleted that account of mine and have no means for it to gain attention). Can you please help me? Thanks!

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  • $\begingroup$ Good start! Can you show from your identity (or otherwise) that $p_1^{r_1-1} \le r_1+1$? Can you find all solutions to that inequality? $\endgroup$ – Greg Martin Apr 25 at 16:55
  • $\begingroup$ Only possibility i guess occurs in $r_1=2, p_1=3$, et cetera $\endgroup$ – user665856 Apr 25 at 16:56
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Let $n=\prod_{i=1}^kp_i^{r_i}$ be given, where the $p_i$ are distinct primes and the $r_i$ are positive integers. You're off to a good start with the identity $$d(n)=\prod_{i=1}^k(r_i+1).$$ From this it follows that $$\frac{n}{d(n)}=\frac{\prod_{i=1}^kp_i^{r_i}}{\prod_{i=1}^k(r_i+1)}=\prod_{i=1}^k\frac{p_i^{r_i}}{r_i+1}.$$ It is clear that if $\tfrac{n}{d(n)}=p$ is prime then $p\mid n$, so without loss of generality $p=p_1$. Then $$1=\frac{n}{p_1d(n)}=\frac{p_1^{r_1-1}}{r_1+1}\prod_{i=2}^k\frac{p_i^{r_i}}{r_i+1}.$$ Now prove by induction that

For all primes $p$ and all positive integers $r$ you have $\tfrac{p^r}{r+1}\geq 1$, with equality iff $(p,r)=(2,1)$.

Can you then finish the proof using this result?

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