[數論] p58, 第一章第二節, Q17

17. 試證：若 $\left ( a,b \right )=1$, 則 $\left ( d,ab \right )=\left ( d,a \right )\left ( d,b \right )$

證：
$\left ( d,ab \right )=\left ( d,a \right )\left ( \frac{d}{\left ( d,a \right )},\frac{ab}{\left ( d,a \right )} \right )=\left ( d,a \right )\left ( \frac{d}{\left ( d,a \right )},b \right )=\left ( d,a \right )\left ( d,b \right )$

註：

1. 若 $a,b$ 是任意兩個不全為零的整數, $m$ 是任一正整數, 則 $\left ( am,bm \right )=\left ( a,b \right )m$
2. 若 $\left ( a,b \right )=1$, 則 $\left ( a,bc \right )=\left ( a,c \right )$.

$\because \left ( \frac{d}{\left ( d,a \right )},\frac{a}{\left ( d,a \right )} \right )=1$, $\therefore \left ( \frac{d}{\left ( d,a \right )},\frac{ab}{\left ( d,a \right )} \right )=\left ( \frac{d}{\left ( d,a \right )},b \right )$

$\because \left ( \left ( d,a \right ),b \right )=\left ( d,\left ( a,b \right ) \right )=\left ( d,1 \right )=1$, $\therefore \left ( \frac{d}{\left ( d,a \right )},b \right )=\left ( \frac{d}{\left ( d,a \right )}\left ( d,a \right ),b \right )=\left ( d,b \right )$