Let $n$ be a positive integer. Prove that
\begin{equation} \sum_{d \vert n} \varphi(d)= n \end{equation}
where $\varphi$ is the totient function of Euler.
Divisibility property of the totient function $\varphi$
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abstract-algebra
group-theory
totient-function
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0see also (http://math.stackexchange.com/q/1373551) – 2017-02-06
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0You can have many www proofs of this result, using the keyword "totient function" as you know it. – 2017-02-06
1 Answers
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For $d\vert n$, let $\mathcal{O}_d$ be the sets of elements of order $d$ in $\mathbb{Z}/n\mathbb{Z}$. Using Lagrange's theorem, $\{\mathcal{O}_d\}_{d\vert n}$ is a partition of $\mathbb{Z}/n\mathbb{Z}$. Howeover, as an element of order $d$ spans a group isomorphic to $\mathbb{Z}/d\mathbb{Z}$, $\mathcal{O}_d$ has cardinality $\varphi(d)$.