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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.

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  • $\begingroup$ I give a (somewhat long, probably far from optimal) proof using your hint in 18.781 (Spring 2016): Floor and arithmetic functions ( cip.ifi.lmu.de/~grinberg/floor.pdf ) (proof of Proposition 1.2.10). That said, your proof is a lot nicer! $\endgroup$ – darij grinberg May 15 at 6:10
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The first 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 $\ (b,c)=1\,\Rightarrow\,\color{#c00}{(a,b,c)= 1}$


To prove it by the hint first reduce to case $\,{(a,b)=1}\,$ by cancelling $\,(a,b).\ $ So it becomes

$$ \color{#0a0}{(a,b)=1}\ \Rightarrow\ (a,bc) = (a,c) $$

$\ \begin{align}(a,c)\mid a\ \,\\ \ (a,c)\,\mid c\mid bc\end{align}\,\Rightarrow\ \ (a,c)\,\mid\, (a,bc).\ $ For the reverse divisibility we have

$\begin{align}(a,bc)\mid a\mid ac\\ (a,bc)\mid bc\end{align}\,\Rightarrow\, \color{#c00}{(a,bc)}\mid (ac,bc) = \color{#0a0}{(a,b)}c = \color{#c00}c,\ $ so $\,\begin{align}(a,bc)\mid a\\\color{#c00}{(a,bc)\mid c}\end{align}\Rightarrow\,(a,bc)\mid (a,c)$

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