Prove by strong induction that if $\sum_{d\mid n} f(d) =F(n)$, and $F(n)$ is multiplative, then so is $f(n)$

Question

Prove by strong induction that if $\sum_{d\mid n} f(d) =F(n)$, and $F(n)$ is multiplicative, then so is $f(n)$.

Converse of the Mobius inversion formula, I don't know how to start it.

Answer 1

If you meant

With $g(n) = \sum_{d | n} f(d)$, $\ \ $ $g(n)$ is multiplicative iff $f(n)$ is multiplicative

Then look at the minimal $a,b,gcd(a,b)=1$ such that $f(ab) \ne f(a)f(b)$, you get $$g(a)g(b) = \sum_{d | a}\sum_{d'|b} f(d)f(d')=f(a)f(b)-f(ab)+\sum_{d | a,d'|b} f(dd')=f(a)f(b)-f(ab)+g(ab)$$ i.e. (for this minimal $a,b$) $$g(ab) =g(a)g(b)\ \ \Longleftrightarrow \ \ f(ab) =f(a)f(b)$$