About phi function 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.
 A: HINT: Use the following fact-
For any positive integer $n$ with prime factorisation $$n=\prod_i^k p_i^{a_i}$$ where $p_i$'s are the prime factors of $n$, $$\phi(n)=n\prod_i^k\left(1-\frac{1}{p_i}\right)$$ $$=\prod_i^k p_i^{a_i-1}\left(p_i-1\right)$$
So, since you have $\phi(n)|n$, the required condition is $$\color{red}{\prod_i^k \left(p_i-1\right)|\prod_i^k p_i}$$
A: 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, $n=2^m3^j$
A: $n=1$ is a solution. If $n\ge 2$, then let $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ be the unique prime factorization of $n$ with $p_1<p_2<\cdots < p_k$ and $\alpha_1,\alpha_2,\ldots,\alpha_k\ge 1$.
$$\phi(n)=p_1^{\alpha_1-1}p_2^{\alpha_2-1}\cdots p_k^{\alpha_k-1}(p_1-1)(p_2-1)\cdots (p_k-1)$$
$$\phi(n)\mid n\iff \frac{p_1p_2\cdots p_k}{(p_1-1)(p_2-1)\cdots (p_k-1)}\in\mathbb Z$$
If $k=1$, then $p_1-1\mid p_1$, so $p_1-1\mid p_1-(p_1-1)=1$, so $p_1=2$ and $n=2^{\alpha_1}$, which is a solution for all $\alpha_1\in\mathbb Z^+$.
If $k\ge 3$, then the denominator is divisible by $4$ but the numerator is not.
If $k=2$, then $(p_1-1)(p_2-1)\mid p_1p_2$. Also $\gcd(p_2-1,p_2)=1$ and $p_2-1>1$, so $p_2-1=p_1$, so $p_2=3$, $p_1=2$, so $n=2^{\alpha_1} 3^{\alpha_2}$, which is a solution for all $\alpha_1,\alpha_2\in\mathbb Z^+$.
