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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.

Zev Chonoles
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nikola
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  • "If we have $\varphi(n)=2^l$ then first $2|n$..." that's false for all fermat primes except $2$ –  Apr 02 '17 at 17:36

1 Answers1

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You can prove that if $2^m+1$ is prime, then $m$ must itself be a power of $2$; see this math.SE thread.

Community
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Zev Chonoles
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    Without external links, simply: if $m$ has an odd divisor $d>1$, then $2^d+1$ is a divisor of $2^m+1$. It follows that $2^m+1$ being a prime implies that $m$ has no non-trivial odd divisors, i.e. $m=2^k$. – Jack D'Aurizio Apr 02 '17 at 17:36