If $\,gcd(b,c)=1\,$ then $\,a\mid kb,kc\Rightarrow\, a\mid k,\,$ for all $\,a,b,c,k\in\Bbb Z$ Let $a,b,c,k \in \mathbb{Z}$ and $a \mid k·c$, $a \mid k·b$ and $gcd(c,b)=1$. Prove that $a \mid k$.
 A: Bezout's identity implies that there exists $x,y\in\mathbb{Z}$ s.t. $cx+by=1$. Multiply both sides of the equation by $k$ and conclude.
A: As in the proofs given below: $ $ if $\, \color{#c00}{(b,c)=1},\,$ then $\, a\mid kb,kc\ \Rightarrow\ a\mid (kb,kc) = k\color{#c00}{(b,c)} = k$
Euclid's Lemma in Bezout form, gcd form and ideal forms
$\smash[t]{\!\begin{align}\\ \\ 
Ax\!+\!ay=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\  Ab,ab\, \Rightarrow\,  A\,\mid Abx\!\!+\!aby\! =\, (\!\overbrace{Ax\!+\!ay}^{\large\color{#c00} 1}\!) b = b\\
(A,\ \ \ a)=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\  Ab,ab\, \Rightarrow\,  A\,\mid (Ab,\ \ ab) = (A,\ \ \ a)\ \ b =\, b\\
A\!+\!(a)=&\,\color{#c00}1,\,\ A\supseteq\! (ab)\, \Rightarrow\, A \supseteq\! (b).\,  {\bf Proof}\!:\,  A \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(A\!+\!(a))b =\! (b)\\
A +{\cal A}\ =&\,\color{#c00}1,\,\ A\supseteq {\cal A B}\, \Rightarrow\, A \supseteq\, {\cal B}.\,\   {\bf Proof}\!:\,  A\, \supseteq\! A{\cal B},\!{\cal AB}\!\Rightarrow\!\! A\supseteq A{\cal B}\!+\!\!{\cal AB} =(A+{\cal A}){\cal B} = {\cal B} 
\end{align}}$
