How can " $\small {n \over \varphi(n) } \text{ is integer only if } n=2^r \cdot 3^s $ " simply be shown? I've tried to answer the  question concerning  $\small { 3^n-2^n\over n } $ and wanted to show this simply by referring to the property that 
$$  3^{\varphi (n)}-2^{\varphi (n)}\equiv 0 \pmod n  $$
 and 
$$  3^{\varphi (n) \cdot m}-2^{\varphi (n) \cdot m}\equiv 0 \pmod n  $$
and thus that it is required that $$ n = \varphi(n) \cdot m $$ and then $$  {n \over \varphi(n)}=m$$ is integer. After a short consideration I got aware that I could not really prove easily, that this is in general impossible. The wikipedia-page does not have a short statement about this and a short heuristic indicates, that this is possible only for $ n=2^r 3^s \gt 1 $ where then $ m \in \{ 2 , 3\}$.        
Q: What is a short and simple way to prove this?
(Remark: in the referred question there is an answer like "apply Fermat's little theorem repeatedly" - here I'd like an answer which is less abstract)
 A: $\frac{n}{\phi(n)}$ is multiplicative, so if $n=\prod p_i^{a_i}$ is the prime factorization of $n$, then
$$
\frac{n}{\phi(n)} = \prod \frac{p_i^{a_i}}{p_i^{a_i-1}(p_i-1)} = \prod \frac{p_i}{p_i-1}
$$
If $n$ is divisible by two odd primes then the denominator is divisible by $4$ but the numerator cannot be so the result is not an integer. If $n$ is divisible by one odd prime $p$ then the denominator is divisible by $2$ and we must have $n=2^rp^s$ and $n/\phi(n) = 2p/(p-1)$. If $p=3$ this is an integer, otherwise $1<p/(p-1)<3/2$ and it cannot be. If $n$ is not divisible by an odd prime then $n=2^r$.
A: It comes down to counting $2$'s. It turns out that usually $\phi(n)$ has more $2$'s in its prime factorization than $n$ does.   
Let $2^e$ be the highest power of $2$ that divides $n$. We have $\frac{2^e}{\varphi(2^e)}=2$. Every distinct odd prime factor $p$ of $n$ contributes at least one $2$ to the prime power factorization of $\varphi(n)$. So if $\varphi(n)$ divides $n$,  there cannot be two distinct odd primes that divide $n$. 
Thus our only candidates have shape  $n=2^e p^k$, where $p$ is an odd prime and $k \ge 0$. 
If $p-1$ is not a power of $2$, then $p-1$ cannot divide $2^ep^k$. So $p-1$ is a power of $2$. If $p-1$ is a power of $2$ greater than $2$, then the power of $2$ in $\varphi(n)$ is greater than the power of $2$ in $n$. So $p-1=2$.  
