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On multiplicative functions

Let \mu denote the Möbius function and \varphi the Euler’s totient function. Prove that

    \[\frac{n}{\varphi(n)}=\sum_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}\]

Solution

Since both functions are multiplicative it suffices to prove the identity for prime numbers. Hence for n=p^k we have

    \begin{align*} \sum_{d \mid p^k} \frac{\mu^2(d)}{\varphi(d)} &=1+\frac{1}{\varphi(p)} \\ &=\frac{p}{p-1} \\ &=\frac{p^k}{p^{k-1}(p-1)} \\ &=\frac{p^k}{\varphi(p^k)} \end{align*}

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