Gauss' theorem tells us that :
$$ \forall (a, b, c) \in \hspace{0.05em} \mathbb{Z}^3, $$
$$ a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c \qquad (Gauss) $$
Corollary of the Gauss' theorem
Gauss' theorem corollary tells us that :
$$ \forall (a, b, c) \in (\mathbb{Z})^3, $$
$$ a / c \enspace et \enspace b / c, \enspace with \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab/c \qquad \bigl(Gauss \enspace (corollary)\bigr) $$
Let \((a, b, c) \in (\mathbb{Z})^3\) be three integers.
Let us assume that \( a / bc \) and also \( a \) and \( b \) are coprime.
If: \( \Biggl \{ \begin{gather*} a / bc \\ a \wedge b = 1 \end{gather*}\)
By the Bézout's theorem, we know that:
$$ \forall (a, b) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a \wedge b = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = 1 $$
Then,
$$ \exists (k, u, v) \in (\mathbb{Z})^3, \Biggl \{ \begin{gather*} bc = ka \qquad (1) \\ au + bv = 1 \qquad (2) \end{gather*} $$
With \( (2) \), we do have this:
$$ au + bv = 1$$
$$ acu + bcv = c $$
However, thanks to \( (1) \), we know that \( bc = ka \), therefore:
$$ acu + kav = c $$
$$ a \underbrace{(cu + kv)} _\text{\( \in \hspace{0.05em}\mathbb{Z} \)} = c $$
We then have \( a / c \).
And finally,
$$ \forall (a, b, c) \in \hspace{0.05em} \mathbb{Z}^3, $$
$$ a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c \qquad (Gauss) $$
Let \((a, b, c) \in (\mathbb{Z})^3\) be three integers.
Let us assume that \( a / c \), \( b / c \) and also \( a \) and \( b \) are coprime.
If: \(\Biggl \{ \begin{gather*} a / c \\ b / c \end{gather*}\)
Then,
$$ \exists (k, k') \in \hspace{0.05em}\mathbb{Z}^2, \Biggl \{ \begin{gather*} c = ka \qquad (1) \\ c = k'b \qquad (2) \end{gather*} $$
Hence, from \( (1) \) and \( (2) \) we find that:
$$ ka = k'b $$
We then get \( b/ak \), but \( a \wedge b = 1 \).
However, by the Gauss's theorem, we found that:
$$ \forall (a, b, c) \in \hspace{0.05em} \mathbb{Z}^3, \enspace a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c $$
From which we can conclude that \( b/k \), that is to say:
$$ \exists K \in \mathbb{Z}, \enspace k = Kb $$
But \( c = ka \), so,
$$ c = Kba $$
We finally realize that \( ab / c \).
And finally,
$$ \forall (a, b, c) \in (\mathbb{Z})^3, $$
$$ a / c \enspace et \enspace b / c, \enspace with \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab/c \qquad \bigl(Gauss \enspace (corollary)\bigr) $$