Show that if $F$ is multiplicative, then $f$ is multiplicative Suppose that $F(n) = \sum_{d\mid n} f(d)$ for all $n$. Show that if $F$ is multiplicative, then $f$ is also multiplicative.
I'm sorry for my weird notation. I'm not used to this type of syntax but this is what I have so far: Let $F(n) = \sum_{d\mid n} f(d)$. Suppose $F$ is multiplicative. Let $n= n_1n_2$ with $n_1, n_2 \in \mathbb Z^+$ such that $\gcd(n_1, n_2) = 1$. So $F(n) = F(n_1)F(n_2)$. By the Möbius inversion formula, we have 
\begin{align*} 
f(n) &= \sum_{d\mid n} \mu(d)F(n/d)\\ &= \sum_{d\mid n_1n_2} \mu(d)F(n_1n_2/d)\\
&= \sum_{d\mid n_1n_2} \mu(d)F(n_1/d)F(n_2/d)\\ 
&= \sum_{d|n_1} \mu(d)F(n_1/d)\cdot \sum_{d\mid n_2} \mu(d)F(n_2/d)\\
&= f(n_1)f(n_2).
\end{align*}
 So $f$ is multiplicative. 
 A: A few critical points:


*

*You must use the fact that $\mu$ is multiplicative also, otherwise this line of proof cannot work.

*You are in too much of a hurry to use the fact that $F$ is multiplicative.  Look closely; is $n_1/d$ even an integer?

*In the second to last step, how did you get your $d$ to mutate into two different $d$'s?  And why are there two sums now, instead of one?


Still, you understand the basic structure that the proof should take.  But you should address the hardest part first: how are you going to turn a single sum into a double sum?  A wise mathematician would begin this way:
$$\sum_{d \mid n_1 n_2} \mu(d) F(\frac{n}{d}) = \sum_{d_1 \mid n_1} \sum_{d_2 \mid n_2} \mu(d_1 d_2) F(\frac{n_1n_2}{d_1d_2})$$
(Note: When justifying this step, it is absolutely necessary to mention the fact that $\gcd(n_1,n_2)=1$.)
From here, I think you know enough to finish the proof.  
A: If $n_1, n_2$ are relatively prime, the divisors of $n_1n_2$ can be written in a unique way as $d_1d_2$, where $d_1$ is a divisor of $n_1$ and $d_2$ is a divisor of $n_2$. Moreover, $d_1$ and $d_2$ are relatively prime. Therefore
$$\sum_{d\mid n_1n_2} f(d) = \sum_{d_1 \mid n_1}\sum_{d_2 \mid n_2} f(d_1d_2) = \sum_{d_1 \mid n_1}\sum_{d_2 \mid n_2} f(d_1)f(d_2) = \left(\sum_{d_1 \mid n_1}f(d_1)\right)\left(\sum_{d_2 \mid n_2}f(d_2)\right).$$
