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### Which $n$ will satisfy $\phi(n)=2n$?

Mostly, yes. More precisely, your conclusion is that the formulation of $\phi$ given shows if $n$ has any prime divisors, then $\phi(n) < n$. But that doesn't mean that, for your final candidate of ...
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Step 1: show the number $z := \zeta_n^k$ is a root of unity with order $m := n/(k,n)$. Step 2: use transitivity of the norm mapping on field extensions to show $${\rm N}_{\mathbf Q(\zeta_n)/\mathbf Q}... • 48.2k 1 vote ### Sum of modular multiplicative units The idea is like If (a,n)=1, so is (-a,n). Therefore if a\in\mathbb{Z}_n^*, so is n-a, and they will always have a sum of n. \mathbb{Z}_n^* has exactly \phi(n) elements by definition. • 1,539 1 vote ### Arithmetical Functions Sum, \sum\limits_{d|n}\sigma(d)\phi(\frac{n}{d}) and \sum\limits_{d|n}\tau(d)\phi(\frac{n}{d}) Here is a proof that relies only on the techniques and theorems from that and previous chapters of the same book.$$\sum_{d|n}\sigma(d)\phi\left(\frac{n}{d}\right)=\sum_{d|n}\sigma(\frac{n}{d})\phi(d)=...
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