We know that if $~n = p_{1}^{a_1} \cdots p_{s} ^ {a_s}~$ then $~\phi(n) = p_1^{a_1 - 1}(p_1 - 1)\cdots p_s^{a_s - 1} (p_s - 1)$.
If $~q~$ is prime dividing $~\phi(n)~$ then there are two situations:
a) $~q = p_k~$ and $~a_k \geq 2~$ for some $~k=1,\dots,s~$
b) $~p_k - 1~$ is divisible by $~q~$ for some $~k = 1, \dots, s~$
The first situation allows to make a guess that $~q~$ has to be below $~\sqrt{n}$. But in general this is not true since for example, $~\phi(7) = 6 = 2\cdot3~$ and $~3 > \sqrt{7}$.
So i have two questions:
Which bounds (except trivial $~q \leq n$) can we put on $~q~$?
Can we characterize somehow the cases when $~q > \sqrt{n}$? The example above shows that it is possible if $~n~$ is prime. Can it happen when $~n~$ is composite? If yes than is there any pattern?
Thanks in advance for any ideas.
Assume that $n$ is odd. There is then the easy bound $q\le \frac{n-1}{2}$. It is not known whether there are infinitely many cases where $q=\frac{n-1}{2}$, since it is not known whether there are infinitely many primes $p$ such that $\frac{p-1}{2}$ is prime (Sophie Germain primes). It has been conjectured that there are infinitely many.
This suggests that there are many non-primes $n$ for which $q$ is substantially greater than $\sqrt{n}$. For we can take a large $p$ such that $\frac{p-1}{2}$ is prime, let $k$ be much smaller than $p$, and let $n=kp$.