Bézout's theorem tells us that:
$$ \forall (a, b) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a \wedge b = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = 1 \qquad (Bézout) $$
Corollary of the Bézout's theorem
Bézout's theorem corollary tells us that::
$$ \forall (a, b, c) \in \hspace{0.1em}\mathbb{Z}^3, $$
$$ a \wedge bc = 1 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} \Biggl \{ \begin{align*} a \wedge b = 1 \\ a \wedge c = 1 \end{align*} \qquad \bigl(Bézout \enspace (corollary)\bigr) $$
As well, for a simple product:
$$ \forall (a, b, c, d) \in \hspace{0.05em}\mathbb{Z}^4, $$
$$ ab \wedge cd = 1 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} \left \{ \begin{align*} a \wedge c = 1 \\ a \wedge d = 1 \\ b \wedge c = 1 \\ b \wedge d = 1 \\ \end{align*} \right \} $$
And it is possible to generalize it for any product of integers:
$$ \forall (n,m) \in \hspace{0.05em}\mathbb{N}^2, \ \forall (a_1, a_2,... \ a_n) \in \hspace{0.05em}\mathbb{Z}^n, \ \forall (b_1, b_2,... \ b_m) \in \hspace{0.05em}\mathbb{Z}^m, $$
$$ \left[ \ \prod_{i = 0}^n a_i \ \right] \wedge \Biggl[ \ \prod_{j = 0}^m b_j \ \Biggr] = 1 \Longleftrightarrow \forall (i, j) \in [\![1, n ]\!] \times [\![1, m ]\!], \enspace \Bigl \{a_i \wedge b_j = 1 \Bigr \} $$
Let \((a, b) \in \hspace{0.05em} \mathbb{Z}^2 \) be two integers.
Let us assume that \(a \) and \(b \) are coprime.
In other words,
$$a \wedge b = 1 \Longleftrightarrow PGCD(a , b) = 1 $$
We know from the Bézout's identity that:
$$ \forall (a, b) \in \hspace{0.05em}\mathbb{N}^2, \enspace a > b, $$
$$ \delta = a \wedge b \hspace{0.2em} \Longrightarrow \hspace{0.2em} \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = \delta $$
Since we have assumed that \( a \wedge b = 1\), as a result we get:
$$ \forall (a, b) \in\hspace{0.1em}\mathbb{Z}^2, $$
$$ a \wedge b = 1 \Longrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = 1 $$
Conversely, let now assume that:
$$\exists (u, v) \in\hspace{0.1em}\mathbb{Z}^2, \enspace au + bv = 1 $$
Let consider \( d \), a common divisor of \( a \) and \( b\).
We know from the properties of divisibility that:
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.05em} \mathbb{Z}^2, \enspace \exists (u , v) \in\hspace{0.1em}\mathbb{Z}^2, $$
$$ a/b \enspace et \enspace a/c \hspace{0.2em} \Longrightarrow \hspace{0.2em} a/(ub + vc) $$
\( d \) being a common divisor of \( a \) and \( b\), it divides \( a \), \( b \) as well as all the linear combinations of \( a \) and \( b \).
$$ d/a \enspace and \enspace d/b \hspace{0.2em} \Longrightarrow \hspace{0.2em} d/(au + bv) $$
Well, \(au + bv = 1\), so \( d / 1\).
The only number which divides \( 1\) is itself, then \( d = 1\).
It is the only common divisor of \( a \) and \( b\), so \( a \wedge b = 1 \).
$$ a \wedge b = 1 $$
Thus,
$$ \forall (a, b) \in\hspace{0.1em}\mathbb{Z}^2, $$
$$ \exists (u, v) \in\hspace{0.1em}\mathbb{Z}^2, \enspace au + bv = 1 \Longrightarrow a \wedge b = 1 $$
And as a result of both implications,
$$ \forall (a, b) \in \hspace{0.05em}\mathbb{Z}^2, $$
$$ a \wedge b = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = 1 \qquad (Bézout) $$
Let \((a, b, c) \in \hspace{0.05em} \mathbb{Z}^3 \) be three integers.
If \(a \wedge bc = 1 \), then with the Bézout's theorem, we do have this:
$$ a \wedge bc = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bcv = 1 $$
As a result we do notice that:
$$ au + b (cv) = 1 \Longleftrightarrow a \wedge b = 1 $$
$$ au + c (bv) = 1 \Longleftrightarrow a \wedge c = 1 $$
And finally,
$$ \forall (a, b, c) \in \hspace{0.1em}\mathbb{Z}^3, $$
$$ a \wedge bc = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \Biggl \{ \begin{align*} a \wedge b = 1 \\ a \wedge c = 1 \end{align*} $$
If we do have this: \(\Biggl \{ \begin{align*} a \wedge b = 1 \\ a \wedge c = 1 \end{align*} \)
Then, still with the Bézout's theorem,
$$ \exists (u, v, u', v') \in \hspace{0.05em}\mathbb{Z}^4, \enspace \Biggl \{ \begin{align*} au + bv = 1 \hspace{2.8em} (1) \\ au' + cv' = 1 \qquad (2) \end{align*} $$
Performing the product \((1) \times (2) \), we do have this:
$$ (au + bv)(au' + cv') = 1 $$
$$ auau' + aucv' + bvau' + bv cv' = 1 $$
$$ a \hspace{0.1em} \underbrace { (uau' + ucv' + bvu')} _\text{ \(= \hspace{0.1em} U \)} \hspace{0.1em} + \hspace{0.1em} bc \hspace{0.1em} \underbrace { (vv') } _\text{ \(= \hspace{0.1em} V \)} \hspace{0.1em} = 1 $$
$$ a U + bcV = 1 \Longleftrightarrow a \wedge bc = 1 , \enspace with \enspace \Biggl \{ \begin{align*} U = uau' + ucv' + bvu'\\ V = vv' \end{align*} $$
Thus,
$$ \forall (a, b, c) \in \hspace{0.1em}\mathbb{Z}^3, $$
$$ \Biggl \{ \begin{align*} a \wedge b = 1 \\ a \wedge c = 1 \end{align*} \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \wedge bc = 1 $$
And as a result of both implications, we do obtain the following equivalence:
$$ \forall (a, b, c) \in \hspace{0.05em}\mathbb{Z}^3, $$
$$ a \wedge bc = 1 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} \Biggl \{ \begin{align*} a \wedge b = 1 \\ a \wedge c = 1 \end{align*} \qquad \bigl(Bézout \enspace (corollary)\bigr) $$
If we now take the hypothesis where \(ab \wedge cd = 1 \), then again with the Bézout's theorem, we do have:
$$ ab \wedge cd = 1 \Longleftrightarrow \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace abu + cdv = 1 $$
Now, we can apply the theorem four times from this equivalence and:
$$ \left \{ \begin{align*} a(bu) + c(dv) = 1 \Longleftrightarrow a \wedge c = 1 \\ a(bu) + d(cv) = 1 \Longleftrightarrow a \wedge d = 1 \\ b(au) + c(dv) = 1 \Longleftrightarrow b \wedge c = 1 \\ b(au) + d(cv) = 1 \Longleftrightarrow b \wedge d = 1 \end{align*} \right \}$$
As a result we obtain,
$$ \forall (a, b, c, d) \in \hspace{0.05em}\mathbb{Z}^4, $$
$$ ab \wedge cd = 1 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \left \{ \begin{align*} a \wedge c = 1 \\ a \wedge d = 1 \\ b \wedge c = 1 \\ b \wedge d = 1 \\ \end{align*} \right \} $$
If we retrieve the previous ipplication and we try to verify its reciprocal, we do startfrom this hypothesis:
$$ \left \{ \begin{align*} a \wedge c = 1 \\ a \wedge d = 1 \\ b \wedge c = 1 \\ b \wedge d = 1 \\ \end{align*} \right \} $$
By applying the Bézout's theorem, we do have this:
$$ \left \{ \begin{align*} a \wedge c = 1 \\ a \wedge d = 1 \\ b \wedge c = 1 \\ b \wedge d = 1 \\ \end{align*} \right \} \Longleftrightarrow $$
$$ \exists (u_1, v_1, u_2, v_2, u_3, v_3, u_4, v_4) \in \hspace{0.05em}\mathbb{Z}^8, \enspace \left \{ \begin{align*} au_1 + cv_1 = 1 \\ au_2 + dv_2 = 1 \\ bu_3 + cv_3 = 1 \\ bu_4 + dv_4 = 1 \\ \end{align*} \right \}$$
Indeed, by performing the product of these four expressions, we do obtain:
$$(au_1 + cv_1)(au_2 + dv_2)(bu_3 + cv_3)( bu_4 + dv_4) = 1 $$
Now, when distributing all this, we do obtain a kind of tree, when each term will contain either \(ab\), nor \(cd\):
$$ \begin{align*} au_1 \hspace{4em} cv_1 \\ \hspace{1em} \downarrow \hspace{0.05em} \searrow \hspace{0.05em} \swarrow \hspace{0.05em} \downarrow \\ \hspace{1em} \downarrow \hspace{0.05em} \swarrow \hspace{0.05em} \searrow \hspace{0.05em} \downarrow \\ au_2 \hspace{4em} dv_1 \\ \hspace{1em} \downarrow \hspace{0.05em} \searrow \hspace{0.05em} \swarrow \hspace{0.05em} \downarrow \\ \hspace{1em} \downarrow \hspace{0.05em} \swarrow \hspace{0.05em} \searrow \hspace{0.05em} \downarrow \\ bu_3 \hspace{4em} cv_3 \\ \hspace{1em} \downarrow \hspace{0.05em} \searrow \hspace{0.05em} \swarrow \hspace{0.05em} \downarrow \\ \hspace{1em} \downarrow \hspace{0.05em} \swarrow \hspace{0.05em} \searrow \hspace{0.05em} \downarrow \\ bu_4 \hspace{4em} dv_4 \\ \end{align*} $$
$$ \begin{align*} \ \ \ \ \textcolor{#A65757}{a} u_1 au_2 \textcolor{#A65757}{b} u_3 b u_4 \qquad (1^{st} \ column) \\ + \ \textcolor{#A65757}{a} u_1 au_2 \textcolor{#A65757}{b} u_3 d v_4 \qquad (3 \ first \ elements \ of \ the \ 1^{st} \ column - \ 4^{th} \ element \ of \ the \ 2^{nd}) \\ + \ \textcolor{#A65757}{a} u_1 au_2 c v_3 \textcolor{#A65757}{b} u_4 \qquad (2 \ first \ elements \ of \ the \ 1^{st} \ column - \ 3^{rd} \ element \ of \ the \ 2^{nd} - \ 4^{th} \ element \ of \ the \ 1^{st} ) \\ + \ ... \\ + \ \textcolor{#446e4f}{c} v_1 \textcolor{#446e4f}{d} v_1 b u_3 dv_4 \qquad (2 \ first \ elements \ of \ the \ 2^{nd} \ column - \ 3^{rd} \ element \ of \ the \ 1^{st} - \ 4^{th} \ element \ of \ the \ 2^{nd} ) \\ + \ \textcolor{#446e4f}{c} v_1 \textcolor{#446e4f}{d} v_1 cv_3 b u_4 \qquad (3 \ first \ elements \ of \ the \ 2^{nd} \ column - \ 4^{th} \ element \ of \ the \ 1^{st}) \\ + \ \textcolor{#446e4f}{c} v_1 \textcolor{#446e4f}{d} v_1 cv_3 dv_4 \qquad (2^{nd} \ column) \\ \end{align*} $$
such as, after having gathered all these terms according to \(ab\) or \(cd\): \(\exists (U, V) \in \hspace{0.05em} \mathbb{Z}^2, \ abU + cdV = 1\)
And as a result of both implications, we do obtain the following equivalence:
$$ \forall (a, b, c, d) \in \hspace{0.05em}\mathbb{Z}^4, $$
$$ ab \wedge cd = 1 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} \left \{ \begin{align*} a \wedge c = 1 \\ a \wedge d = 1 \\ b \wedge c = 1 \\ b \wedge d = 1 \\ \end{align*} \right \} $$
By applying the previous reasoning, but for any product of integers, we can generalize this corollary by claiming that:
$$ \forall (n,m) \in \hspace{0.05em}\mathbb{N}^2, \ \forall (a_1, a_2,... \ a_n) \in \hspace{0.05em}\mathbb{Z}^n, \ \forall (b_1, b_2,... \ b_m) \in \hspace{0.05em}\mathbb{Z}^m, $$
$$ \left[ \ \prod_{i = 0}^n a_i \ \right] \wedge \Biggl[ \ \prod_{j = 0}^m b_j \ \Biggr] = 1 \Longleftrightarrow \forall (i, j) \in [\![1, n ]\!] \times [\![1, m ]\!], \enspace \Bigl \{a_i \wedge b_j = 1 \Bigr \} $$
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