For what values $m \in \mathbb{N}$, $\phi(m) | m$, where $\phi(m)$ is the Euler function. I am working with elementary number theory and, although in theory the $\phi$ Euler function seems easy to understood, I am having some problemas making the exercises. 
For example, in this question:
For what values $m \in \mathbb{N}$, $\phi(m) | m$, where $\phi(m)$ is the Euler function,
I know two expressions to the function $\phi$ but, I tried to use them to solve this problem and I failed. Could someone help-me?
Thanks a lot.
Here is what I tried:
What I did was:
$\phi(m) = (p_1^{\alpha_1} - p_1^{\alpha_1 - 1})\ldots (p_k^{\alpha_k} - p_k^{\alpha_k - 1}) \Rightarrow$
$\frac{m}{\phi(m)} = \prod_k \frac{p_k^{\alpha_k}}{p_k^{\alpha_k}(1-\frac{1}{p_k})}= \prod_k \frac{p_k}{- 1 +p_k}.$
Then, I can't follow from here.

$m=1$ works. Let $m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ with $2\le p_1<p_2<\cdots <p_k$. Then 
$$\phi(m)=p_1^{\alpha_1-1}p_2^{\alpha_2-1}\cdots p_k^{\alpha_k-1}\left(p_1-1\right)\left(p_2-1\right)\cdots\left(p_k-1\right)$$
$$\phi(m)\mid m\iff \frac{p_1p_2\cdots p_k}{\left(p_1-1\right)\left(p_2-1\right)\cdots\left(p_k-1\right)}\in\Bbb Z$$
If $k=1$, then $$p_1-1\mid p_1\implies p_1-1\mid p_1-\left(p_1-1\right)\implies p_1-1\mid 1\implies p_1=2$$
Then $m=2^{\alpha_1}$, and indeed it's a solution.
If $k\ge 2$, then we'll prove $p_1=2$.
Assume for contradiction $p_1\ge 3$. Then exists a prime $q$ such that 
$$q\mid p_1-1\mid p_1p_2\cdots p_k\implies q\mid p_1p_2\cdots p_k\implies$$
$$ \left(\left(q\mid p_1\right) \text{ or } \left(q\mid p_2\right) \text{ or } \ldots \text{ or } \left(q\mid p_k\right)\right)\implies q\in\{p_1,p_2,\ldots, p_k\}$$
But then $q$ is too large to divide $p_1-1$. Contradiction.
Therefore $p_1=2$. Then $$\frac{2p_2\cdots p_k}{\left(p_2-1\right)\cdots\left(p_k-1\right)}\in\Bbb Z$$
Let $p$ be a prime divisor of $p_2-1\ge 2$. Then
$$p\mid p_2-1\mid 2p_2\cdots p_k\implies p\mid 2p_2\cdots p_k\implies p\in\{2,p_2,\ldots, p_k\},$$
so $p=2$, because if $p\ge p_2$, then $p\nmid p_2-1$.
Therefore $p_2-1=2^h$ for some $h\in\Bbb Z^+$, so
$$2^h=p_2-1\mid 2p_2\cdots p_k\implies 2^{h-1}\mid p_2\cdots p_k$$
But $p_2\cdots p_k$ is odd, so $h=1$, so $p_2=3$.
Assume for contradiction $k\ge 3$. Then $4$ divides the denominator of $\frac{2p_2\cdots p_k}{\left(p_2-1\right)\cdots\left(p_k-1\right)}$ but not the numerator, so $\frac{2p_2\cdots p_k}{\left(p_2-1\right)\cdots\left(p_k-1\right)}$ is not an integer. Therefore $k=2$.
Then $m=2^{\alpha_1}3^{\alpha_2}$, which is indeed a solution.
Answer: $m\in\{1,2^t,2^a3^b\},\, t,a,b\ge 1$.
