Show that if $(b,c)=1$, then $(a,bc)=(a,b)(a,c)$.

Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]

This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$

But how should I prove it using that hint or at least a similar proof.

Solution 1:

Your proof can be done more efficiently as below. More generally see this answer.

$(a,b)(a,c) = (aa,ab,ac,bc) = (a\color{#c00}{(a,b,c)},bc) = (a,bc),\ $ by $\ (b,c)=1\,\Rightarrow\,\color{#c00}{(a,b,c)= 1}$

where we used GCD polynomial arithmetic, i.e. associative, commutative, distributive laws.

To use the hint reduce to case $\,(\bar a,\bar b)\!=\!1\,$ via cancel $\,\color{#0a0}{d \!=\! (a,b)}\,$ from both sides to get equivalent

$$ (\bar a,\bar bc) = (a,c),\ \ \ \bar a = a/d,\, \bar b = b/d$$

$d\mid \bar a,\bar bc\,\Rightarrow\, d\mid \bar a\mid a\ $ & $\ d\mid \bar ac,\bar bc\,\Rightarrow\,d\mid (\bar ac,\bar bc)=(\bar a,\bar b)c = c\,$ so $\,d\mid a,c$. $d\mid a,c\,\Rightarrow\, d\mid c\mid \bar bc\ $ & $\ d\mid\!\overbrace{\!\!\bar ad,\!\!\!}^{\large a}\,\bar ac\,\Rightarrow\, d\mid (\bar ad,\bar ac)=\bar a(\color{#0a0}d,c)=\bar a\overbrace{(\color{#0a0}{a,b},c)}^{\large \color{#c00}{=\,\bf 1}}=\bar a$ so $\,d\mid \bar a,\bar bc$

So $\,\bar a ,\bar bc\,$ and $\,a,c\,$ have the same common divisors $\,d,\,$ so the same greatest common divisor.