The Gauss' theorem and its corollary

Gauss' theorem tells us that :

$$ \forall (a, b, c) \in \hspace{0.05em} \mathbb{Z}^3, $$

$$ a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c \qquad (Gauss) $$

Gauss' theorem corollary tells us that :

$$ \forall (a, b, c) \in (\mathbb{Z})^3, $$

$$ a / c \enspace et \enspace b / c, \enspace with \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab/c \qquad \bigl(Gauss \enspace (corollary)\bigr) $$

Demonstration

The Gauss' theorem

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,

$$ \forall (a, b, c) \in \hspace{0.05em} \mathbb{Z}^3, $$

$$ a / bc \enspace et \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/c \qquad (Gauss) $$

Corollary of the Gauss' theorem

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,

$$ \forall (a, b, c) \in (\mathbb{Z})^3, $$

$$ a / c \enspace et \enspace b / c, \enspace with \enspace a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab/c \qquad \bigl(Gauss \enspace (corollary)\bigr) $$