Does $\gcd(a,bc)$ divides $\gcd(a, b)\gcd(a, c)$?

Solution 1:

Let $d=\gcd(a,bc)$ and $d_1=\gcd(a,b) $ and $d_2=\gcd(a,c)$. Then there exits $ s_1, s_2,t_1, \text{ and } t_2 \in \mathbb{Z}$ such that $$ d_1=s_1 a+t_1b$$ and $$d_2= s_2a+t_2 c $$then $$ d_1d_2 = s_1s_2 a^2 +s_1 t_2ac +t_1 s_2 ba + t_1 t_2 bc$$ but $d\mid a $ and $d\mid bc$ so $d\mid a^2$, $d\mid ac$, $d\mid ba$ and $d\mid bc$. Hence $d \mid d_1d_2$.

Solution 2:

$$(\gcd(a,b)\cdot\gcd(a,c))\mathbb Z= \gcd(a,b)\mathbb Z\cdot\gcd(a,c)\mathbb Z=(a\mathbb Z+b\mathbb Z)\cdot(a\mathbb Z+c\mathbb Z)\subseteq a^2\mathbb Z+ac\mathbb Z+ab\mathbb Z+bc\mathbb Z\subseteq a\mathbb Z+bc\mathbb Z=\gcd(a,bc)\mathbb Z\Rightarrow\gcd(a,bc)\mid\gcd(a,b)\cdot\gcd(a,c).$$

Edit. The property holds for $a,b,c\in R$ with $R$ a GCD domain.

Let $d=(a,b)$ and $e=(a,c)$. Then $a=da_1$, $b=db_1$ with $(a_1,b_1)=1$, and $a=ea_2$, $c=ec_2$ with $(a_2,c_2)=1$. We have to show that $(a,bc)=(da_1,db_1c)=d(a_1,b_1c)\mid de$, that is, $(a_1,b_1c)\mid e$. Set $f=(a_1,b_1c)$. Since $f\mid a_1$ and $(a_1,b_1)=1$ we have $(f,b_1)=1$. Then $f\mid c$ and $f\mid a$, so $f\mid e$.

Solution 3:

$d:=(a,bc)\mid \color{#c00}a,bc\, \Rightarrow\, d\mid \color{#c00}aa,\color{#c00}ab,\color{#c00}ac,bc\, \Rightarrow\, d\mid (aa,ab,ac,bc)=(a,b)(a,c)$