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Thread: Euler function phi(n) using the Moubilus function

  1. #1
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    Cool 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!
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    Quote Originally Posted by bamby View Post
    Hi,
    How can I do that?
    Express Euler function phi(n) using the Moubilus function miu(n)
    Thank you!
    Consider the group $\displaystyle \mathbb{Z}_n$. Every element has order a divisor of $\displaystyle n$. Also if $\displaystyle d|n$ then the number of elements that has order $\displaystyle d$ is $\displaystyle \phi (d)$. Therefore, we have that $\displaystyle \sum_{d|n}\phi (d) = n$.

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