[Lehmer's totient problem][1] asks if there exists a composite number $m$ such that $\phi(m)$ divides $m-1$. Lower bounds on $m$ has been established but we do not know if a solution exists. Clearly, if we find a divisor of $\phi(m)$ which does not divide $m-1$ then $m$ cannot be a solution of the problem. I started counting the number of divisors of $\phi(m)$ which do not divide. Soon I noticed pattern - **Composites of the form :** 

 - $m = 8k+3$ has at least $4$ divisors of $\phi(m)$ which do not divide $m-1$
 - $m = 6k+5$ has at least $5$ divisors of $\phi(m)$ which do not divide $m-1$
 - $m = 12k+7$ has at least $6$ divisors of $\phi(m)$ which do not divide $m-1$
 - $m = 20k+3$ has at least $7$ divisors of $\phi(m)$ which do not divide $m-1$
 - $m = 24k+23$ has at least $8$ divisors of $\phi(m)$ which do not divide $m-1$

$$
\cdots
$$

 - $m = 48k+11$ has at least $14$ divisors of $\phi(m)$ which do not divide $m-1$

In fact each of the above arithmetic progressions is the smallest (lowest coefficient of $k$) for a given $n$ and there could be more than one arithmetic progression with the above property.

> **Claim**: For $n \ge 1$ there is an arithmetic
> progression of odd numbers $m_k = ak + b $ such that if $m_k$ composite then there are 
> at least $n$ divisors of $\phi(m_k)$ which do not divide $m_k-1$.

**Question**: Can we find a counter example i.e. is there an integer $n$ for which at we cannot find such an arithmetic progression?

**Note**: [Question was posted in MSE 3 weeks ago but did not get an answer. Hence posting in MO][2]


  [1]: https://en.wikipedia.org/wiki/Lehmer%27s_totient_problem
  [2]: https://math.stackexchange.com/questions/3317228/how-many-divisors-of-phim-do-not-divide-m-1