To prove $\sum_{dn}\phi(d)=n$. What is the easiest proof for this to tell my first year undergraduate junior. I do not want any Mobius inversion etc only elementry proof. Tthanks!

See also http://math.stackexchange.com/q/504063/. – lhf Dec 02 '14 at 14:15
4 Answers
I think the easiest proof is to consider all fractions $\frac k n$ with $1\le k\le n$.
On the one hand, there are $n$ of those.
On the other hand, after reducing each fraction to lowest terms, you get $\phi(d)$ fractions having denominator $d$. The possible denominators are exactly the divisors of $n$.
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Not sure why this is not mentioned: a function $f(n)$ from the positive integers to, for example, the positive integers, is called "multiplicative" in the number theory sense if $$ \gcd(a,b) = 1 \Longrightarrow f(ab) = f(a) f(b). $$ This definition also works if the values of $f$ are allowed to be fractions, real numbers, complex numbers, whatever.
Proposition: if $f(n)$ is multiplicative, then $$ g(n) = \sum_{dn} f(d) $$ is also multiplicative
Proposition: a multiplicative function is determined completely by its values on primes and prime powers
Corollary: two multiplicative functions that agree at primes and prime powers agree for all numbers.
Proposition: Euler's totient function $\phi(n)$ is multiplicative.
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Write $n = \prod p^{a_p}$
The divisors of $p$ are $$ \prod p^{b_p}, b_p \le a_p $$ so the sum on the left hand side is $$ \sum_{b_2=0}^{a_2}\dots \sum_{b_P=0}^{a_P} \phi(\prod_{p \text{ prime}, pn, =2}^P p^{b_p}) = \sum_{b_2=0}^{a_2}\dots \sum_{b_P=0}^{a_P} \prod_{p \text{ prime}, pn, b_p>0} \left(1\frac 1p\right) p^{b_p} \\ = \prod_{p \text{ prime}, pn, =2}^P \left[1 + \sum_{b_p = 1}^{a_p} \left(1\frac 1p\right) p^{b_p} \right] = \prod_{p \text{ prime}, pn, =2}^P p^{a_p} = n $$
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2Yes, this requires arguing first that $\phi$ is multiplicative. But if we do this, then it may be cleaner to argue in general that if $f$ is multiplicative, so is $F$ given by $F(n)=\sum_{d\mid n}f(d)$, so that verifying the result reduces to proving it for powers of primes, and this can be done a bit more transparently by writing $\sum_{d\mid p^a}\phi(d)=\sum_{n=0}^a\phi(p^n)=1+\sum_{n=1}^a(p^np^{n1})=p^a$. – Andrés E. Caicedo Dec 02 '14 at 15:49

No. It only requires the explicit form of $\phi(n) = n\prod_{pn} 11/p$ (plus the fundamental theorem of arithmetics of course) – mookid Dec 02 '14 at 16:07

Yes, but any proof of this essentially argues the function is multiplicative. – Andrés E. Caicedo Dec 02 '14 at 16:52
$d\mid n \land S_d=\{m:\gcd(m,n)=d\} \implies \left\lvert S_d \right\rvert=\phi\left(\dfrac{n}{d}\right)$
$S_{d_i}\cap S_{d_j}=\emptyset\ \forall i\neq j$
$\displaystyle\bigcup_{d \mid n}S_d=\{1,2,\ldots,n\}$
$\therefore \left\lvert\displaystyle\bigcup_{d \mid n}S_d\right\rvert=\displaystyle\sum_{d\mid n}\left\lvert S_d\right\rvert\implies n=\displaystyle\sum_{d\mid n}\phi\left(\dfrac{n}{d}\right)=\displaystyle\sum_{d\mid n}\phi\left(d\right)$

There is a mistake here. For example, $n=20,d=5$ gives us $S_d=\{5,15\}$ so $S_d$ has $2\ne\phi(5)$ elements. In fact, using the [other argument](http://math.stackexchange.com/a/1048215/462) we have that $S_d=\phi(n/d)$. – Andrés E. Caicedo Dec 02 '14 at 15:45
