Bézout's Identity

Theorem
Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.

Let $\gcd \left\{{a, b}\right\}$ be the greatest common divisor of $a$ and $b$.

Then:
 * $\exists x, y \in \Z: a x + b y = \gcd \left\{{a, b}\right\}$

That is, $\gcd \left\{{a, b}\right\}$ is an integer combination (or linear combination) of $a$ and $b$.

Furthermore, $\gcd \left\{{a, b}\right\}$ is the smallest positive integer combination of $a$ and $b$.

Proof 1
Work the Euclidean Division Algorithm backwards.

Proof 2
Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.

Let $S$ be the set of all positive integer combinations of $a$ and $b$:


 * $S = \left\{{x \in \Z, x > 0: x = m a + n b: m, n \in \Z}\right\}$

First we establish that $S \ne \varnothing$.

We have:

As it is not the case that both $a = 0$ and $b = 0$, it must be that at least one of $\left \vert {a} \right \vert \in S$ or $\left \vert {b} \right \vert \in S$.

Therefore $S \ne \varnothing$.

As $S$ contains only positive integers, $S$ is bounded below by $0$ and therefore $S$ has a smallest element.

Call this smallest element $d$: we have $d = u a + v b$ for some $u, v \in \Z$.

Let $x \in S$. By the Division Theorem, $x = q d + r, 0 \le r < d$.

Suppose $d \nmid x$. Then $x \ne q d$ and so $0 < r$.

But:

which contradicts the choice of $d$ as the smallest element of $S$.

Therefore $\forall x \in S: d \backslash x$.

Thus $d \backslash a \land d \backslash b \implies 1 \le d \le \gcd \left\{{a, b}\right\}$.

However, note that as $\gcd \left\{{a, b}\right\}$ also divides $a$ and $b$ (by definition), we have:

So $\gcd \left\{{a, b}\right\} = d = u a + v b$.

Also known as
This result is also known as Bézout's Identity.

Applications
It is primarily used when finding solutions to linear Diophantine equations, but is also used to find solutions via Euclidean Division Algorithm.

This Identity/Lemma can also be applied to the Extended Euclidean Division Algorithm.

However, there are sources which suggest that this identity was first noticed by Claude Gaspard Bachet de Méziriac.

Bézout's contribution was to prove a more general result, for polynomials.

Source

 * : $\S 3.11$: Theorem $19$
 * : $\S 2.2$: Corollary $2.1$
 * : $\S 23$
 * : $\S 12.3$
 * : $\text{A}.1$: Theorem $\text{A}.2$