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The regular $n$-gon is constructible by ruler and compass precisely when the odd prime factors of $n$ are distinct Fermat primes. Prove that in this case, $φ(n)$ is a power of $2$. Where $φ(n)$ is Euler's totient function.

I am having difficulty with this, I used the hints from here but still can't seem to get it. It would be useful if someone could provide me their version of the proof or tell me what I have done wrong.

My attempt: I think $n$ has divisors $1,2^i,2^{2^i}+1,n$ So since $φ(n)$ is multiplicative we can write $φ(2^{2^i}+1)φ(2^i)$

Then using the fact $φ(p^k)=p^{k-1}(p-1)$ I get:

$2^{2^i}(2^{2^i}+1)2^{i-1}(2^i-1)$

Then I try to exapand but can't arrive at showing it is a power of two so feel I must be wrong somewhere.

2 Answers 2

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You are simply misapplying the formula $\phi(p^k) = p^{k-1} (p-1)$. Just think carefully about what $p$ and $k$ should be in the two cases $p^k = 2^{2^i}+1$ and $p^k = 2^i$. You'll find that $\phi(2^{2^i}+1)$ is not $2^{2^i}(2^{2^i}+1)$ and $\phi(2^i)$ is somewhat simpler than $2^{i-1}(2^i - 1)$.

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    So should I get: $\phi(2^{2^i}+1)=2^{2^i}$ and $\phi(2^i)=2^{i-1}$? Then $\phi(2^{2^i}+1)\phi(2^i)= 2^{{2^i}+i-1}$ which still isn't quite a power of two is it? – 2017-02-25
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    @harry55 $2^x$ is a power of two whenever $x$ is an integer. Is $2^i+i-1$ an integer? – 2017-02-25
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For a fermat prime $p:=2^{2^n}+1$ , we have $\phi(p)=2^{2^n}$ which is a power of $2$. Moreover, the $\phi$-function is multiplicative when the arguments are coprime. Take it from here.

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    I end up with $\phi(n)=2^{{2^i}+i-1}$ but this isn't a power of two is it? – 2017-02-25
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    Of course, it is a power of $2$. The base is $2$ and the exponent is a positive integer. – 2017-02-25