Are there infinitely many integers that do not divide any totient number?
My try:
If $a|b$ then $\phi(a)|\phi(b)$, so the main question would be equivalent to asking wether there are infinitely many integers that do not divide $\phi(p)$. We also know that $\phi(p)=p-1$. Then,
Are there infinitely many integers that do not divide any number of the form $p-1$? Would Dirichlet's Theorem on Arithmetic Progressions be enough to disprove it?
Thank you.
Marking this community wiki as I'm just fleshing out the comments:
No, there aren't infinitely many integers that do not divide any totient number, in fact, there are none. Unless maybe you want to count 0 as such a number.
But yes, Dirichlet's theorem on arithmetic progressions is enough to disprove the existence of totient nondivisors, since, as you note, $\phi(p) = p - 1$ for $p$ prime.
Consider the number 8 for example. Obviously 9 is not prime. But Dirichlet tells us there are infinitely many primes of the form $8k + 1$. So if $8k + 1$ is prime, then $\phi(8k + 1) = 8k$, confirming that 8 divides infinitely many totients of primes.
Of course that's not the only way a number can be a totient. 8 is itself the totient of 15, 16, 20, 24, 30, none of which are primes.