Sum of Euler Phi Function over Divisors

Theorem
Let $n \in \Z_{>0}$ be a strictly positive integer.

Then $\displaystyle \sum_{d \mathop \backslash n} \phi \left({d}\right) = n$

where:
 * $\displaystyle \sum_{d \mathop \backslash n}$ denotes the sum over all of the divisors of $n$
 * $\phi \left({d}\right)$ is the Euler $\phi$ function, the number of integers less than $d$ that are prime to $d$.

That is, the total of all the totients of all divisors of a number equals that number.

Proof
Let us define $S_d = \left\{{m \in \Z: 1 \le m \le n, \gcd \left\{{m, n}\right\} = d}\right\}$.

That is, $S_d$ is all the numbers less than or equal to $n$ whose GCD with $n$ is $d$.

Now from Divide by GCD for Coprime Integers we have:
 * $\gcd \left\{{m, n}\right\} = d \iff \dfrac m d, \dfrac n d \in \Z: \dfrac m d \perp \dfrac n d$

So the number of integers in $S_d$ equals the number of positive integers no bigger than $\dfrac n d$ which are prime to $\dfrac n d$.

That is, by definition of the Euler phi function:


 * $\left|{S_d}\right| = \phi \left({\dfrac n d}\right)$

From the definition of the $S_d$, it follows that for all $1 \le m \le n$:


 * $\exists d \mathrel \backslash n: m \in S_d$

Therefore:


 * $\displaystyle \left\{{1, \ldots, n}\right\} = \bigcup_{d \mathop \backslash n} S_d$

Moreover, it follows from the definition of the $S_d$ that they are pairwise disjoint.

Now from Corollary to Cardinality of Set Union, it follows that:

But from Sum Over Divisors Equals Sum Over Quotients:
 * $\displaystyle \sum_{d \mathop \backslash n} \phi \left({\dfrac n d}\right) = \sum_{d \mathop \backslash n} \phi \left({d}\right)$

and hence the result.