Let $f_1, f_2$ multiplicative functions such that $f_1 \ne 0$ and $f_2 \ne 0$.
I want to show that $f_1 + f_2$ is multiplicative if and only if $f_1 = -f_2$
I tried starting the "only if" part as follows:
$$f_1(n) + f_2(n) = (f_1 + f_2)(n) = (f_1 + f_2)(p_1^{k_1}) \cdot ... \cdot (f_1 + f_2)(p_r^{k_r})$$
And then opening the binomial-like expression but I couldn't get it to be zero so that $f_1(n) = -f_2(n)$
Any help would be appreciated.
Let $x,y$ be coprime. $$(f_1+f_2)(xy)=f_1(xy)+f_2(xy)\\=f_1(x)f_1(y)+f_2(x)f_2(y)=_{demand}(f_1+f_2)(x)\cdot(f_1+f_2)(y)\\=f_1(x)f_1(y)+f_2(x)f_2(y)+f_1(x)f_2(y)+f_1(y)f_2(x)$$ The demanded equality above will hold if and only if $f_1(x)f_2(y)=-f_1(y)f_2(x)$ for all such $x,y$. Can you finish the proof?