TITLE: About phi function
QUESTION [6 upvotes]: Find all positive integers n such that $ \phi(n) |n $
I find  $n=2^k ,2^k ×3^j$ is answer ,I can't find another answers.

REPLY [4 votes]: Using the formula for $\phi (n)$ as pointed out be @ScrondingersCat, 
the $\phi (n) |n$ is equivalent to 
$s=(p_1-1)(p_2-1)....(p_k-1)|p_1p_2p_3.....p_k=t$
Assume WLOG $p_1=2$ (else $t$ is odd which is not possible except when $n=1$
Since,  we must have $v_2(s) \leq v_2(t)=1$.
So, $v_2(s)=1$ which gives $k=2$ or $v_2(s)=0$ which gives $n=2^m$
Now,for $k=2$ we have $p_2-1|2p_2$. So, $p_2=3$
So, $n=2^m3^j$