Divisors of the Euler totient function For which values of $m$ do there exist infinitely many $n$ such that $m$ divides $\phi(n)$?
Now, I know $\phi(n)$ is even for $n >2$, so clearly $m$ can’t be an odd number (except $1$). 
So $m=2$ clearly satisfies this, so I was wondering do all even integers satisfy this? Or just even integers which are of the form $p-1$ for a prime $p$?
 A: Every $m$ works.  
Pf:  if $m=\prod p_i^{a_i}$  then $m$ divides $\varphi(n)$ for any $n$ of the form $n=\prod p_i^{b_i}$ with $b_i>a_i$ for all $i$.
A: " I know ϕ(n) is even for n>2, so clearly m can’t be an odd number (except 1)"
Odd numbers can't be divided BY even numbers (not even 1), but odd numbers can divide even numbers. If $m$ is odd then $m\mid 2m$, of course.
I'm afraid once you realize this question is what numbers can divide Totient values, and not the other way-- what Totient numbers can divide other numbers-- this question becomes trivial:
All numbers.
If $m = \prod p_i^{k_i}$ is the prime factorization of $m$.
And $n = \prod q_i^{m_i}$ is the prime factorization of $n$ then $\phi(n) =\prod (q_i - 1) \prod q_i^{m_i - 1}$.  
Obviously we can set the $q_i = p_i$ and then $m_i > k_i$ and $m|\phi(n)$.  And that's not the only way.  Clearly it is always possible that some $p_j^{k_j}|\prod (q_i -1)$ as well.
As an example:
Let $m = \prod p_i^{k_i}$.  Let $n = \prod p_i^{k_i+a_i}$.
Then $\phi(n) = \prod (p_i-1)\prod p_i^{k_i + a_i-1} = m*\prod (p_i - 1)\prod p_i^{a_i - 1}$.
......
