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Section 6.3 Sums over divisors

If \(m\) or \(n\) is \(1\) then the theorem is trivially true, so suppose that \(m, n\gt 1, \gcd(m, n) = 1,\) and let \(m = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}\) and \(n = q_1^{j_1}q_2^{j_2}\cdots q_s^{j_s}\) be their prime factorizations. Since \(\gcd(m, n) = 1,\) the primes \(p_1,..., p_r, q_1,...,q_s\) are all distinct, and so the prime factorization of \(mn\) is

\begin{equation*} mn = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}q_1^{j_1}q_2^{j_2}\cdots q_s^{j_s}. \end{equation*}

So any divisor \(d\) of \(mn\) is uniquely representable in the form

\begin{equation*} d = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}q_1^{b_1}q_2^{b_2}\cdots q_s^{b_s} \hspace{5mm} 0\leq a_i\leq k_i, 0\leq b_i\leq j_i. \end{equation*}

So we may write \(d = d_1d_2,\) where \(d_1 = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}\) divides \(m\) and \(d_2 = q_1^{b_1}q_2^{b_2}\cdots q_s^{b_s}\) divides \(n.\) Since \(p_i\neq q_j\) for any \(i, j,\) we have \(\gcd(d_1, d_2) =1.\)

Let \(m, n\) be relatively prime positive integers. Then, by the proceeding lemma,

\begin{equation*} F(mn)= \sum_{d\vert mn}f(d)= \sum_{\substack{d_1\vert m \\ d_2\vert n}}f(d_1d_2) \end{equation*}

where \(\gcd(d_1,d_2)=1.\) Since \(f\) is multiplicative \(f(d_1d_2)=f(d_1)f(d_2).\) So

\begin{align*} F(mn) \amp= \sum_{\substack{d_1\vert m \\ d_2\vert n}}f(d_1)f(d_2)\\ \amp= \left(\sum_{d_1\vert m}f(d_1)\right)\left(\sum_{d_2\vert n}f(d_2)\right)\\ \amp= F(m)F(n). \end{align*}