Show $\sum\limits_{d|n}\phi(d) = n$. Show $\sum\limits_{d|n}\phi(d) = n$. 
Example : $\sum\limits_{d|4}\phi(d) = \phi(1) + \phi(2) + \phi(4) = 1 + 1 + 2 = 4$
I was told this has a simple proof. Problem is, I can not think of a way to show this in a very simple and straightforward way. 
 A: The simplest proof I know is this:
Consider all proper fractions of the form $a/n$. There are $n$ of those. When you consider their reduced forms you get fractions of the form $b/d$ with $d|n$ and $(b,d)=1$. By definition, there are $\phi(d)$ of those. The result follows.
A: Here's one approach:
Consider the polynomial $x^n-1$ as a polynomial over the complex numbers $\mathbb{C}$. You can easily see, almost tautologically, that 
$$x^n-1=\prod_{d\mid n}\Phi_d(x)\qquad(\ast)$$
where $\Phi_d(x)$ is the polynomial whose roots are the primitive $d^{\text{th}}$ roots of unity (i.e. solutions to $x^d-1$ that do not satisfy $x^c-1$ for $0<c<d$). 
But, rephrasing the definition, $\Phi_d(x)$ is the polynomial with roots $e^{\frac{2\pi i r}{d}}$ where $e^{\frac{2\pi r s}{d}}\ne 1$ for $0<s<d$. But, this is saying that $sr\not\equiv 0\mod d$ for $0<s<d$, or that $d$ and $r$ have no common factors. Since we only need to care about $r$ which are less than or equal to $d$ we see that, in fact, the are $\phi(d)$ such roots of $\Phi_d(x)$. 
This allows us to conclude that $\deg \Phi_d(x)=\phi(d)$. Indeed, since $\Phi_d(x)\mid x^n-1$, and $x^n-1$ has no repeated roots in $\mathbb{C}$ (it's coprime to its derivative) we see that $\Phi_d(x)$ has no repeated roots. Thus, $\deg\Phi_d(x)$ is the number of roots of $\Phi_d(x)$ which is $\phi(d)$.
So, by comparing degrees in $(\ast)$ we get:
$$n=\sum_{d\mid n}\deg\Phi_d(x)=\sum_{d\mid n}\phi(d)$$
A: Consider the set $[n]=\{1,2,\ldots,n\}$ of size $n$. Define an equivalence relation as follows $$m\sim m'\iff (m,n)=(m',n)$$ That this is an equivalence relation should be clear. The equivalence classes are of the form $$\widehat m=\{0<m'\leqslant n:(m',n)=d\}$$ where $d=(m,n)\mid n$.  Now, what is the size of this class? It contains all positive elements $m'\leqslant n$ such that $(m',n)=d$, or what is the same, all positive elements $$\frac{m'}d\leqslant \frac nd$$ such that $$\left(\frac{m'}{d},\frac nd \right)=1$$
This amounts to $\varphi\left(\dfrac nd\right)$ elements. But there is one class for each positive divisor of $n$, thus we have paititioned our set getting $$n=\sum_{d\mid n}\varphi\left(\dfrac nd\right)$$
But as $d$ runs through all divisors of $n$; so does $\dfrac nd$, so that $$n=\sum_{d\mid n}\varphi(d)$$
A: Let $G=\langle x\rangle$ be the cyclic group of order $n$ generated by $x$. Then,
\begin{align}
n &= |G| \\
&= \sum_{1\leq d\leq n} |\{\text{elements of order } d\}| \\
&= \sum_{d|n}|\{\text{elements of order } d\}|
\end{align}
by Lagrange's Theorem. But $G$ has a unique cyclic subgroup of order $d$ for each $d|n$, namely $\langle x^{n/d}\rangle$. Moreover, each such subgroup has $\varphi(d)$ generators, so
$$n = \sum_{d|n}\varphi(d)$$
A: Hint: You can show that $\sum\limits_{d|n}\phi(d) = \sum\limits_{d|n}\phi(\frac{n}{d}) $ but $\phi(\frac{n}{d})$ is all numbers with GCD of $d$ with $n$. Therefore the last sum counts all numbers till $n$ and is equal to $n$.
