The Gauss' theorem and its corollary

Gauss' theorem tells us that :

$$ \forall (a, b, c) \in \hspace{0.04em} \mathbb{Z}^3, $$

$$ (a \mid bc) \text{ and } (a \wedge b = 1) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid c \qquad \bigl(\text{Gauss' theorem} \bigr) $$

Corollary of the Gauss' theorem

Gauss' theorem corollary tells us that :

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

$$ (a \mid c) \text{ and } (b \mid c), \enspace \text{with } a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab \mid c \qquad \bigl(\text{Gauss' theorem (corollary)} \bigr) $$

Proofs

The Gauss' theorem

Let $(a, b, c) \in (\mathbb{Z})^3$ be three integers.

Let us assume that $ a \mid bc $ and also $ a $ and $ b $ are coprime.

If: $ \Biggl \{ \begin{gather*} a \mid bc \\ a \wedge b = 1 \end{gather*}$

By the Bézout's theorem , we know that:

$$ \forall (a, b) \in \hspace{0.04em}\mathbb{Z}^2, $$

$$ a \wedge b = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.04em}\mathbb{Z}^2, \enspace au + bv = 1 $$

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:

$$ acu + kav = c $$

$$ a \underbrace{(cu + kv)} _{ \in \hspace{0.1em}\mathbb{Z} } = c $$

We then have $ a \mid c $.

And finally,

$$ \forall (a, b, c) \in \hspace{0.04em} \mathbb{Z}^3, $$

$$ (a \mid bc) \text{ and } (a \wedge b = 1) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid c\qquad \bigl(\text{Gauss' theorem} \bigr)$$

Corollary of the Gauss' theorem

Let $(a, b, c) \in (\mathbb{Z})^3$ be three integers.

Let us assume that $ a \mid c $, $ b \mid c $ and also $ a $ and $ b $ are coprime.

If: $\Biggl \{ \begin{gather*} a \mid c \\ b \mid c \end{gather*}$

Then,

$$ \exists (k, k') \in \hspace{0.04em}\mathbb{Z}^2, \Biggl \{ \begin{gather*} c = ka \qquad (1) \\ c = k'b \qquad (2) \end{gather*} $$

Hence, from $ (1) $ and $ (2) $ we find that:

$$ ka = k'b $$

We then get $ b \mid ak $, but $ a \wedge b = 1 $.

However, by the Gauss's theorem , we found that:

$$ \forall (a, b, c) \in \hspace{0.04em} \mathbb{Z}^3, \enspace (a \mid bc) \text{ and } (a \wedge b = 1) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid c $$

From which we can conclude that $ b \mid k $, that is to say:

$$ \exists K \in \mathbb{Z}, \enspace k = Kb $$

But $ c = ka $, so,

$$ c = Kba $$

We finally realize that $ ab \mid c $.

And finally,

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

$$ (a \mid c) \text{ and } (b \mid c), \enspace \text{with } a \wedge b = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} ab \mid c \qquad \bigl(\text{Gauss' theorem (corollary)} \bigr)$$