Bézout's Lemma

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.

$\blacksquare$

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 $b$:

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

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

We have:

 $\displaystyle a > 0$ $\implies$ $\displaystyle \size a = 1 \times a + 0 \times b$ $\quad$ $\quad$ $\displaystyle a < 0$ $\implies$ $\displaystyle \size a = \paren {-1} \times a + 0 \times b$ $\quad$ $\quad$ $\displaystyle b > 0$ $\implies$ $\displaystyle \size b = 0 \times a + 1 \times b$ $\quad$ $\quad$ $\displaystyle b < 0$ $\implies$ $\displaystyle \size b = 0 \times a + \paren {-1} \times b$ $\quad$ $\quad$

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

Therefore $S \ne \O$.

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:

 $\, \displaystyle \exists m, n, \in \Z: \,$ $\displaystyle x$ $=$ $\displaystyle m a + n b$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle x - q d$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {m a + n b} - q \paren {u a + v b}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \paren {m - q u} a + \paren {n - q v} b$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle$  $\displaystyle \paren {r \in S} \land \paren {r < d}$ $\quad$ $\quad$

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

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

In particular:

$d \divides \size a = 1 \times a + 0 \times b$
$d \divides \size b = 0 \times a + 1 \times b$

Thus:

$d \divides a \land d \divides b \implies 1 \le d \le \gcd \set {a, b}$

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

 $\displaystyle \gcd \set {a, b}$ $\divides$ $\displaystyle \paren {u a + v b} = d$ $\quad$ Common Divisor Divides Integer Combination $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \gcd \set {a, b}$ $\divides$ $\displaystyle d$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \gcd \set {a, b}$ $\le$ $\displaystyle d$ $\quad$ $\quad$

Since $d$ is the smallest number in $S$:

$\gcd \set {a, b} = d = u a + v b$

$\blacksquare$

Proof 3

First we establish that:

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

include George E. Andrews: Number Theory coll.2-1

Now to show that $\gcd \left\{ {a, b}\right\}$ is the smallest positive number to satisfy the equation.

We first show that:

$\forall x \in \Z: \exists m, n \in \Z: x = m x + n y \ne 0 \implies d \mathrel \backslash x$

include George E. Andrews: Number Theory coll.2-2

Proof 4

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

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

$J = \left\{{x: x = m a + n b: m, n \in \Z}\right\}$

First we show that $J$ is an ideal of $\Z$

Let $\alpha = m_1 a + n_1 b$ and $\beta = m_2 a + n_2 b$, and let $c \in \Z$

Then $\alpha,\beta \in J$ and :

 $\displaystyle \alpha + \beta$ $=$ $\displaystyle m_1 a + n_1 b + m_2 a + n_2 b$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({m_1 + m_2}\right) a + \left({n_1 + n_2}\right) b$ $\quad$ $\quad$ $\displaystyle$ $\implies$ $\displaystyle \alpha + \beta \in J$ $\quad$ $\quad$

 $\displaystyle c \alpha$ $=$ $\displaystyle c \left({m_1 a + n_1 b}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({cm_1}\right) a + \left({cn_1}\right) b$ $\quad$ $\quad$ $\displaystyle$ $\implies$ $\displaystyle c \alpha \in J$ $\quad$ $\quad$

Thus $J$ is an ideal of $\Z$

$a = 1 a + 0b \land b = 0 a + 1 b \implies a, b \in J$

$a$ and $b$ are not both zero, thus $J \neq \{0\}$

$\exists x_0 > 0 : J = x_0 \Z$

$a \in J \land \{J = x_0 \Z\} \implies x_0 \vert a$

$b \in J \land \{J = x_0 \Z\} \implies x_0 \vert b$

$x_0 \vert a \land x_0 \vert b \implies x_0 \in D(a,b)$

Furthermore

$x_0 \in J \implies \exists r,s \in \Z : x_0 = ra + sb$

Let $x_1 \in D(a,b)$

Then:

 $\displaystyle x_1 \in D(a,b)$ $\implies$ $\displaystyle x_1 \vert a \land x_1 \vert b$ $\quad$ $\quad$ $\displaystyle$ $\implies$ $\displaystyle x_1 \vert (ra + sb)$ $\quad$ $\quad$ $\displaystyle$ $\implies$ $\displaystyle x_1 \vert x_0$ $\quad$ $\quad$ $\displaystyle$ $\implies$ $\displaystyle \vert x_1 \vert \leq \vert x_0 \vert = x_0$ $\quad$ $\quad$

Thus $x_0 = max(D(a,b)) = gcd(a,b) = ra + sb$

Proof 5

Let $\gcd \left\{ {a, b}\right\} = d$.

Let $\dfrac a d = p$ and $\dfrac b d = q$.

$\gcd \left\{{p, q}\right\} = 1$
$\exists x, y \in \Z: p x + q y = 1$

The result follows by multiplying both sides by $d$.

$\blacksquare$

Also known as

This result is also known as Bézout's Identity.

Applications

Bézout's Lemma is primarily used when finding solutions to linear Diophantine equations, but is also used to find solutions via Euclidean Division Algorithm.

This result can also be applied to the Extended Euclidean Division Algorithm.

Source of Name

This entry was named for Étienne Bézout.

Historical Note

There are sources which suggest that Bézout's Lemma was first noticed by Claude Gaspard Bachet de Méziriac.

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