2
$\begingroup$

I have to demonstrate that $$\phi^{-1}(n)= \prod_{p|n}(1-p)$$ where $\phi(n)$ is the Euler's totient function.

I know that I can write $\phi$ in terms of the Mobius function $\mu$ as$$\phi(n)= \sum_{n|d} \mu(d) \cdot \left(\frac{n}{d}\right)$$ and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- \frac{1}{f(1)} \sum_{d|n \\d<n} f \left(\frac{n}{d}\right)f^{-1}(d) $$ but I can't find a solution.

Can anyone give me a hint?

$\endgroup$
2
$\begingroup$

Since $\phi(n)$ is multiplicative, i.e., satisfies $\phi(nm) = \phi(n)\phi(m)$ for all $m,n$ with $gcd(n,m) = 1$, it follows from $\phi(p^k)= p^{k-1}(p-1)$ that $$\sum_{d | n} \phi(d) = n, \qquad \phi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$$ by the Moebius inversion formula. Thus $$\phi^{-1}(n) = \sum_{d | n} d \mu(d)$$ by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that $$ \sum_{d\mid n}\mu(d)f(d)=\prod_{p\mid n}(1-f(p)). $$ Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain $$\phi^{-1}(n)= \prod_{p|n}(1-p).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.