This question might be a duplicate, if so, I apologise in advance. It is simple, but answering it is probably harder :)
Is it true, that the $\phi(n)$ function(Euler's totient function) takes on all of it's values, when $n$ is an odd integer?
I obviously tried it for the first some $n$:
$\phi(1) = 1, \phi(3) = 2, \phi(5) = 4, $ and so on.
It is obvious, that if $n$ is a prime, than $\phi(n) = n-1$, so we cover all the $p-1$ numbers, where $p$ is a prime. I just can't really prove if any number is missing on this list.
The question can be asked in this way too: Is it true, that if we use the totient function with only odd integers, we get all the values from it. I hope you can understand it, if not, just comment below, and I try to answer. :) Thanks for any comments!