Question #98934
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.
1
Expert's answer
2019-11-18T12:11:46-0500

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

We have f(m)f(n)=f(i=1upiki)f(i=1upili)=i=1uf(piki)i=1uf(pili)=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})=

=i=1uf(piki)f(pili)=i=1uf(pimin{ki,li})f(pimax{ki,li})==\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\}})=

=i=1uf(pimin{ki,li})i=1uf(pimax{ki,li})==\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(i=1upimin{ki,li})f(i=1upimax{ki,li})=f((m,n))f([m,n])=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])


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