Problem 1352

14. Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$.
(i) Proof:
$$\begin{aligned} \sum_{d i p_{s}^{\alpha_{s}}} \varphi(d)= & \varphi(1)+\varphi\left(p_{s}\right)+\varphi\left(p_{s}^{2}\right)+\cdots+\varphi\left(p_{s}^{a_{s}}\right) \\ = & 1+\left(p_{s}-1\right)+\left(p_{s}^{2}-p_{s}\right)+\cdots \\ & +\left(p_{s}^{\alpha_{s}}-p_{s}^{\alpha_{s}-1}\right)=p_{s}^{\alpha_{s}} \end{aligned}$$

Therefore $\square$
$$\begin{aligned} \sum_{d \mid n} \varphi(d) & =\sum_{d_{1} \mid p_{1}^{a_{r}}} \cdots \sum_{d_{k} \mid p_{b}^{a_{k}}} \varphi\left(d_{1} \cdots d_{k}\right) \\ & =\sum_{d_{1} \mid p_{1}^{\alpha_{1}}} \varphi(d!) \sum_{d_{2} \mid p_{2}^{a_{2}}} \varphi\left(d_{2}\right) \cdots \sum_{d_{k} \mid p_{k}^{a_{k}}} \varphi\left(d_{k}\right) \\ & =p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{\alpha_{k}}=n . \end{aligned}$$
(ii) Proof: In the previous problem, take $f(d)=\varphi(d), F(n)=n$, then the conclusion of the previous problem and part (i) of this problem together provide the proof.