jasperE3 wrote: If $f:\mathbb N\to\mathbb R$ is a function such that $$\prod_{d\mid n}f(d)=2^n$$holds for all $n\in\mathbb N$, show that $f$ sends $\mathbb N$ to $\mathbb N$. Using Mobius inversion formulae: $f(n)=\prod_{d\mid n}2^{\frac{n}{d}\mu(d)}=2^{\sum_{d\mid n}\frac{n}{d}\mu(d)}=2^{\phi(n)}$

Induction Number divisors $n$. $$f(n)=2^{\phi(n)}$$