So, I want to prove the theorem in the title.
The implication that if $n$ is product of power of 2 and some Fermat's prime numbers, then $\varphi(n)=2^k$ (where $\varphi$ denotes Euler's totient function and $k$ is some natural number) is, I think, obvious (directly from that formula). The problem is the other implication.
If we have $\varphi(n)=2^l$ then first $2\mid n$ and we could write $n=2^kn_1$ where $k \in \mathbb{N}$ and $2$ does not divide $n_1$. Then, if $p_1,\ldots,p_k$ were all different prime divisors of $n$, then now we have $p_1=2$, while $n_1$ has prime divisors $p_2,\ldots,p_k$, none of which is $2$. Note that if $p_i^{k_i}$ divided $n_1$ for some $i$ , where $k_i>1$, then that $p_i$ divides $\varphi(n)$, and $\varphi(n)$ is not the power of $2$. So, $n_1=p_2p_3\cdots p_n$ and now when we write formula for $\varphi(n)$ we get $$\varphi(n) = 2^k(p_2-1)\cdots(p_k-1)$$ and $\varphi(n)$ is a power of $2$, meaning all its divisors are power of $2$, meaning $p_i = 2^{k_i}+1$ for all $2 \leq i \leq n$ and for some $k_i$, such that $p_i$ is a prime.
The problem is that I can not conclude that $k_i = 2^{m_i}$, which I need for Fermat's numbers (of type $2^{2^n}+1$). Please help.
Best Answer
You can prove that if $2^m+1$ is prime, then $m$ must itself be a power of $2$; see this math.SE thread.