Does the totient function reach all its values when restricted to odd numbers? 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!

The answer is no. Let $m = 2^kn$, with $k=3$ or $4<k<15$ and $n$ odd. Then $\phi(m) = \phi(2^k)\phi(n)=2^{k-1}\phi(n)$. I claim there is no odd number $x$ with $\phi(x) = \phi(m) = 2^{k-1}\phi(n)$.
Let $x$ be an odd number. It can be written as $\Pi_{i=1}^n p_i^{e_i}$, where because $x$ is odd, none of the $p_i$ is $2$. The formula for the totient given a prime factorization yields $\phi(x) = \Pi_{i=1}^n p_i^{e_i-1}(p_i-1)$. 
$p_i^{e_i-1}$ is clearly not a power of two for $p_i \neq 2$, and $(p_i-1)$ is a power of two only for $p_i$ a Fermat prime. The only known such primes are $3,5,17,257, 65537$. This means the only known powers of two that can appear in $\phi(x)$ are $2,4,16,256,65536$, or $2^1,2^2,2^4,2^{16}$. So if $k =4$ or $5<k<17$, $\phi(x)$ cannot equal $\phi(2^kn)$.
