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

math enthusiastic
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  • $\displaystyle n = \sum_{k \le n} 1 = \sum_{d| n} \sum_{k \le n, gcd(n,k)=d} 1 =\sum_{d| n} \sum_{m\le n/d, gcd(n,dm)=d} 1 =\sum_{d| n} \sum_{m\le n/d, gcd(n,m)=1} 1= \sum_{d| n} \phi(n/d)$ – reuns Aug 11 '17 at 11:51
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    My answer in link 3 is completely elementary. Isn't that what you're looking for? – lhf Aug 11 '17 at 12:01
  • @lhf: can you explain a little more about the last part we get ϕ(d)fractions having denominator d but what happens next? I couldn't understand the last part how did you come to the conclusion – math enthusiastic Aug 11 '17 at 12:47
  • Try also this other answer of mine: https://math.stackexchange.com/a/504088/589 – lhf Aug 11 '17 at 13:42

1 Answers1

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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.

J.G.
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