$$ \forall (a, k) \in (\mathbb{Z}^*)^2, \enspace \forall b \in \mathbb{Z}, $$
$$ ka/kb \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/b $$
$$ \forall (a, b) \in (\mathbb{Z}^*)^2, \enspace \forall c \in \mathbb{Z}, $$
$$ a/b \enspace et \enspace b/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c $$
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(b + c) $$
Linear combination of the dividends
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.05em}\mathbb{Z}^2, \enspace \exists (u , v) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(ub + vc) $$
Recap table of the properties of divisibility
Let \((a, k) \in (\mathbb{Z}^*)^2 \) be two non-zero integers, and \(b \in \mathbb{Z} \) an integer.
If \( ka / kb \), then:
$$ kb = kak' \Longleftrightarrow b = ak'$$
So, \(b / a \).
$$ \forall (a, k) \in (\mathbb{Z}^*)^2, \enspace \forall b \in \mathbb{Z}, $$
$$ ka/kb \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/b $$
Let \((a, b) \in (\mathbb{Z}^*)^2 \) be two non-zero integers, and \(c \in \mathbb{Z} \) an integer.
If \( a/b \) and \( b/c \), then:
$$ \exists (k, k') \in \mathbb{Z}, \enspace \Biggl \{ \begin{gather*} b = ka \\ c = k'b \end{gather*} $$
So,
$$ c = \hspace{0.2em} \underbrace{kk'} _\text{ \( \in \mathbb{Z} \)} a $$
Therefore \( a/c \). We definitely have:
$$ \forall (a, b) \in (\mathbb{Z}^*)^2, \enspace \forall c \in \mathbb{Z}, $$
$$ a/b \enspace and \enspace b/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c $$
Let \(a \in \mathbb{Z^*} \) be a non-zero integer, and \((b , c) \in \hspace{0.05em}\mathbb{Z}^2 \) two integers.
If \( a/b \) and \( a/c \), then:
$$ \exists (k, k') \in \mathbb{Z}, \enspace \Biggl \{ \begin{gather*} b = ka \\ c = k'a \end{gather*}$$
So,
$$ b + c = \hspace{0.2em} \underbrace{(k +k')} _\text{ \( \in \hspace{0.05em} \mathbb{Z} \)} a $$
Therefore \( a/(b + c) \). We definitely have:
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(b + c) $$
We will also have, by extension:
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(b - c) $$
Let \(a \in \mathbb{Z^*} \) be a non-zero integer, \((b , c) \in \hspace{0.05em}\mathbb{Z}^2 \) two integers.
If \( a/b \) and \( a/c \), then:
$$ \exists (k, k') \in \mathbb{Z}, \enspace \Biggl \{ \begin{gather*} b = ka \\ c = k'a \end{gather*}$$
Moreover,
$$ \exists (u, v) \in \mathbb{Z}, \enspace \Biggl \{ \begin{gather*} ub = uka \\ vc = vk'a \end{gather*}$$
So,
$$ ub + vc = \hspace{0.2em} \underbrace{(uk + vk')} _\text{ \( \in \hspace{0.05em} \mathbb{Z} \)} a $$
Therefore \( a/(ub + vc) \). We definitely have:
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.05em}\mathbb{Z}^2, \enspace \exists (u , v) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(ub + vc) $$
We will say that \( a \) divides all linear combinations of \( b \) and \( c \).
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