# Euler function phi(n) using the Moubilus function

• November 18th 2008, 08:28 AM
bamby
Euler function phi(n) using the Moubilus function
Hi,
How can I do that?
Express Euler function phi(n) using the Moubilus function miu(n)
Thank you!
• November 18th 2008, 11:42 AM
ThePerfectHacker
Quote:

Originally Posted by bamby
Hi,
How can I do that?
Express Euler function phi(n) using the Moubilus function miu(n)
Thank you!

Consider the group $\mathbb{Z}_n$. Every element has order a divisor of $n$. Also if $d|n$ then the number of elements that has order $d$ is $\phi (d)$. Therefore, we have that $\sum_{d|n}\phi (d) = n$.

Since this is true for all $n\geq 1$ by Mobius inversion formula we see that, $\phi (n) = n\left( \sum_{d|n} \frac{\mu (d)}{d} \right)$.