Pf. (推论之证明)
对于 $q=p_1^{\ell_1}p_2^{\ell_2}\cdots p_s^{\ell_s}$, 令 $q_1=p_1^{\ell_1}$, $q_2=q/p_1^{\ell_1}=p_2^{\ell_2}p_3^{\ell_3}\cdots p_s^{\ell_s}$. 显然, $(q_1,q_2)=1$, 根据引理,
\[
\begin{split}
S(q,f(x))&=S(q_1 q_2,f(x))=S(q_1,\frac{f(q_2 x)}{q_2})\cdot S(q_2,\frac{f(q_1 x)}{q_1})\\
&=S(p_1^{\ell_1},\frac{f(qx/p_1^{\ell_1})}{q/p_1^{\ell_1}})\cdot S(p_2^{\ell_2}\cdots p_s^{\ell_s},\frac{f(p_1^{\ell_1} x)}{p_1^{\ell_1}})
\end{split}
\]
令 $g(x)=\dfrac{f(p_1^{\ell_1} x)}{p_1^{\ell_1}}$, 应用引理, 得
\[
\begin{split}
S(p_2^{\ell_2}\cdots p_s^{\ell_s},\frac{f(p_1^{\ell_1} x)}{p_1^{\ell_1}})&=S(p_2^{\ell_2}\cdots p_s^{\ell_s},g(x))\\
&=S(p_2^{\ell_2},\frac{g(p_3^{\ell_3}\cdots p_s^{\ell_s}\cdot x)}{p_3^{\ell_3}\cdots p_s^{\ell_s}})\cdot S(p_3^{\ell_3}\cdots p_s^{\ell_s},\frac{g(p_2^{\ell_2}x)}{p_2^{\ell_2}})\\
&=S(p_2^{\ell_2},\frac{f(p_1^{\ell_1}\cdot p_3^{\ell_3}\cdots p_s^{\ell_s}\cdot x)}{p_1^{\ell_1}\cdot p_3^{\ell_3}\cdots p_s^{\ell_s}})\cdot S(p_3^{\ell_3}\cdots p_s^{\ell_s},\frac{f(p_2^{\ell_2}p_1^{\ell_1}x)}{p_2^{\ell_2}\cdot p_1^{\ell_1}})
\end{split}
\]
因此, 不断分裂下去(即使用归纳法), 可得
\[
S(p_1^{\ell_1}p_2^{\ell_2}\cdots p_s^{\ell_s},f(x))=\prod_{i=1}^{s}S(p_i^{\ell_i},\frac{qx/p_i^{\ell_i}}{q/p_i^{\ell_i}}).
\]
\[
\]
