is there an elementary proof of $\sum\limits_{d|n}\phi(d)=n$?

Question

this question has been asked for several times but I need an elementary solution without advanced techniques

there are some links that can help you

1-$\sum_{d|n}\phi(d)=n$

2-$\sum_{d|n}\phi(d)=n$

3-$\sum_{d|n}\phi(d)=n$

4-$\sum_{d|n}\phi(d)=n$

Answer 1

I was unable to check the links, so I apologise if this proof has already been suggested, but hopefully it meets your needs. By the way, "elementary" means "without using complex numbers", not "without using advanced techniques".

Call this sum $S(n)$ so if $n=n_1 n_2$ with $(n_1,\,n_2)=1$ then $S(n)=\sum_{d_i|n_i}\phi(d_1)\phi(d_2)=S(n_1)S(n_2)$. We may thus assume $n=p^k$ with $p\in\mathbb{P},\,k>0$ (since $n=1$ is a trivial case). Then $S=1+\sum_{i=1}^k p^{i-1}(p-1)=p^k=n$ by telescoping.