Check if an arithmetic function is (completely) multiplicative

Question

In Paul J. McCarthy's book about arithmetic functions, he constructs a completely multiplicative function $g$ by setting $g(p)$ equal to a root of the equation $$X^2+f^{-1}(p)X+f^{-1}(p^2)=0, $$ where $f$ is a multiplicative arithmetic function and $p$ is prime. He does not further explain why this function is multiplicative or even completely multiplicative. I have tried to come up with an explanation for hours but I couldn't. I got to the point where I questioned if I do understand the assertion correctly. I tried showing that for two primes $p,q$ fulfilling $$g(p)^2+f^{-1}(p)g(p)+f^{-1}(p^2)=0,\quad g(q)^2+f^{-1}(q)g(q)+f^{-1}(q^2)=0 $$ it is also true that $$(g(p)g(q))^2+f^{-1}(pq)(g(p)g(q))+f^{-1}((pq)^2)=0. $$ Which would show that $g(pq)=g(p)g(q)$ for primes $p$ and $q$ and thus imply at least multiplicativity. One can use the facts that for multiplicative arithmetic functions $$f^{-1}(p)=-f(p),\quad f^{-1}(p^2)=f(p)^2-f(p^2).$$ I would greatly appreciate any help. I guess that I miss something very obvious.

Answer 1

You can construct any completely multiplicative function $g$ by first defining $g(p)$ for primes $p$ and then the multiplicativity property forces $g(n)$ for general $n$ to be

$$ n=\prod p^e \quad\implies\quad g(n)=\prod f(p)^e. $$

You can also define any (not necessarily completely) multiplicative function $g$ by defining $g(p^e)$ for all prime powers $p^e$ and then $g(n)=\prod f(p^e)$. (Note $g(1)=1$ is assumed.)