Splitting sum into two sums

Question

Assuming that $f$ is a multiplicative arithmetic function. Let $n_1,n_2\in \mathbb{N}$ with $gcd(n_1,n_2)=1$. Consider the sum $$\large S=\sum_{a\mid n_1n_2}f(a).$$

Can I split the sum $S$ into two parts, one over divisors of $n_1$ and the second over divisors of $n_2$?.

Thanks.

Answer 1

Denote the set of divisors of $n$ by $D(n)$. Then since $n_1$ and $n_2$ are coprime there is a bijection $D(n_1) \times D(n_2) \to D(n_1n_2)$ given by $(a,b) \mapsto ab$. It follows that $$ S = \sum_{x \in D(n_1n_2)} f(x) = \sum_{a \in D(n_1)} \sum_{b \in D(n_2)} f(ab) = \sum_{a \mid n_1} \sum_{b \mid n_2} f(a)f(b) = \sum_{a \mid n_1} f(a) \sum_{b \mid n_2} f(b). $$ We have $f(ab)=f(a)f(b)$ since $a$ and $b$ are coprime for $a \in D(n_1)$, $b \in D(n_2)$.

Answer 2

Since $\gcd(n_1,n_2) = 1$, the divisors of of $n_1 n_2$ are of the form $d_1 d_2$, where $d_1 | n_1$ and $d_2 | n_2$. Moreover, $\gcd(d_1,d_2) = 1$ for every such $d_1$ and $d_2$. Hence, $f(d_1d_2) = f(d_1) f(d_2)$ for those $d_1 | n_1$ and $d_2 | n_2$. Thus $$S = \sum_{d_1 | n_1} \sum_{d_2 | n_2} f(d_1 d_2) = \sum_{d|n_1} \sum_{d|n_2} f(d_1) f(d_2) = \sum_{d|n_1} f(d) \sum_{d|n_2} f(d)$$