\[n = \displaystyle\sum_{d \mid n} \varphi(d) \]
证明：
\[\forall n \in \mathbb{N}_+, f(x) = \displaystyle\sum_{i = 1}^n [\gcd(i, n) = x] \]\[\because \gcd(i, n) = x \]\[\therefore \gcd(\displaystyle\frac{i}{x}, \displaystyle\frac{n}{x}) = 1 \]\[\therefore f(x) = \varphi(\displaystyle\frac{n}{x}) \]\[\because \forall i \in [1, n] \bigcap Z, 1 \leq \gcd(i, n) \leq n \]\[\therefore n = \displaystyle\sum_{i = 1}^{n} f(x) \]\[\because \forall i \nmid n, f(i) = 0 \]\[\therefore n = \displaystyle\sum_{d \mid n} f(d) = \displaystyle\sum_{d \mid n} \varphi(\displaystyle\frac{n}{d}) = \displaystyle\sum_{d \mid n} \varphi(d) \]证毕．
标签：frac,函数,sum,mid,varphi,therefore,displaystyle,欧拉 From： https://www.cnblogs.com/wf715/p/euler.html