Answer to Question #98934 in Combinatorics | Number Theory for Dorothy

Question

If f is a multiplicative arithmetic function then prove that
f([m, n])f((m, n)) = f(m)f(n)
for all positive integers m and n.
Here [m, n] is the least common multiple of m and n and (m, n) is the
greatest common divisor of m and n.

Answer 1

Let $m=\prod\limits_{i=1}^u p_i^{k_i}$ and $n=\prod\limits_{i=1}^u p_i^{l_i}$.

We have $f(m)f(n)=f(\prod\limits_{i=1}^u p_i^{k_i})f(\prod\limits_{i=1}^u p_i^{l_i})=\prod\limits_{i=1}^u f(p_i^{k_i})\prod\limits_{i=1}^u f(p_i^{l_i})=$

$=\prod\limits_{i=1}^u f(p_i^{k_i})f(p_i^{l_i})=\prod\limits_{i=1}^u f(p_i^{\min\{k_i,l_i\}})f(p_i^{\max\{k_i,l_i\}})=$

$=\prod\limits_{i=1}^u f(p_i^{\min\{k_i,l_i\}})\prod\limits_{i=1}^u f(p_i^{\max\{k_i,l_i\}})=$

$=f(\prod\limits_{i=1}^u p_i^{\min\{k_i,l_i\}})f(\prod\limits_{i=1}^u p_i^{\max\{k_i,l_i\}})=f((m,n))f([m,n])$