# summatory function of arithmetic function

 $\sum_{d\,\mid\,n}\varphi(d)\;=\;\sum_{d\,\mid\,n}\varphi\left(\frac{n}{d}% \right)\;=\;n\quad\mbox{for all }\,n\in\mathbb{Z}_{+}.$

Proof.  The first equality follows from the fact that any positive divisor of $n$ is got from $n/d$ where $d$ is a divisor of $n$. Further, let  $1\leq m\leq n$  where  $\gcd(m,\,n)=d$.  Then  $\gcd(m/d,\,n/d)=1$  and  $1\leq m/d\leq n/d$.  This defines a bijection between the prime classes modulo $n/d$ and such values of $m$ in $\{1,\,2,\,\ldots,\,n\!-\!1\}$ for which  $\gcd(m,\,n)=d$.  The number of the latters $\varphi(n/d)$. Furthermore, the only $m$ with  $1\leq m\leq n$  and  $\gcd(m,\,n)=n$ is  $m:=n$,  and  $\varphi(n/n)=\varphi(1)$, by definition.  Summing then over all possible values $d$ yields the second equality.

## References

Title summatory function of arithmetic function SummatoryFunctionOfArithmeticFunction 2013-03-22 19:31:53 2013-03-22 19:31:53 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 11A25 summatory function EulerPhifunction PrimeClass