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Let $a,b,c$ be integers, with $a$ and $b$ relatively prime. Then $$ a\mid bc\implies a\mid c. $$

The proof starts as follows:

Let $x,y\in\mathbb Z$ with $xa+yb=1$. From $a\mid bc$ we get $a\mid xac+ybc=\dots$

I can follow the rest of the proof. However, I don't understand why $$ a\mid bc\implies a\mid xac+ybc. $$ I can see that $a\mid bc\implies a\mid ybc.$ But How do they get $xac$?

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    $a$ trivially divides $xac$. Since it also divides $ybc$, it has to divide their sum. – 2017-02-14
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    $$\require{cancel}{x\cancel{a}c \over \cancel{a}}$$ – 2017-02-14
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    $\color{#c00}{a\mid bc}\ \Rightarrow\, \begin{align} &\ \ xac+ybc\\ =\ &a(xc + y(\color{#c00}{bc/a}))\end{align}\ \ $ – 2017-02-14

2 Answers 2

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$a\mid xac$ trivially since $a=xc\cdot a$. Hence $a\mid xac$ and $a \mid ybc$. This implies $a \mid xac+ybc$. As simple as that.

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First notice $a|xac$ because $xac=a(xc)$. Then notice $a|ybc$ because $a|bc$ so if $bc=ak$ then $ybc=y(ak)=a(yk)$.

Now that we know that $a|xac$ and $a|ybc$ we have $a|xac+ybc$.