The Gauss' theorem and its corollary

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: \( \Biggl \{ \begin{gather*} a / bc \\ a \wedge b = 1 \end{gather*}\)

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:

However, thanks to \( (1) \), we know that \( bc = ka \), therefore:

We then have \( a / c \).

And finally,

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,

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

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

We then get \( b/ak \), but \( a \wedge b = 1 \).

From which we can conclude that \( b/k \), that is to say:

But \( c = ka \), so,

We finally realize that \( ab / c \).

And finally,