# Bézout's Identity

(Redirected from Bezout's Lemma)

Bézout's Identity states that if $x$ and $y$ are nonzero integers and $g = \gcd(x,y)$, then there exist integers $\alpha$ and $\beta$ such that $x\alpha+y\beta=g$. In other words, there exists a linear combination of $x$ and $y$ equal to $g$.

Furthermore, $g$ is the smallest positive integer that can be expressed in this form, i.e. $g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}$. In particular, if $x$ and $y$ are relatively prime then there are integers $\alpha$ and $\beta$ for which $x\alpha+y\beta=1$.

## Proof

Let $x = gx_1$, $y = gy_1$, and notice that $\gcd(x_1,y_1) = 1$.

Since $\gcd(x_1,y_1)=1$, $\text{lcm}(x_1,y_1)=x_1y_1$. So $\alpha=y_1$ is smallest positive $\alpha$ for which $x_1\alpha\equiv 0\pmod{y}$. Now if for all integers $0\le a,b, we have that $x_1a\not\equiv x_1b\pmod{y_1}$, then one of those $y_1$ integers must be 1 from the Pigeonhole Principle. Assume for contradiction that $x_1a\equiv x_1b\pmod{y_1}$, and WLOG let $b>a$. Then, $x_1(b-a)\equiv 0\pmod {y_1}$, and so as we saw above this means $b-a\ge y_1$ but this is impossible since $0\le a,b. Thus there exists an $\alpha$ such that $x_1\alpha\equiv 1\pmod{y_1}$.

Therefore $y_1|(x_1\alpha-1)$, and so there exists an integer $\beta$ such that $x_1\alpha - 1 = y_1\beta$, and so $x_1\alpha + y_1\beta = 1$. Now multiplying through by $g$ gives, $gx_1\alpha + gy_1\beta = g$, or $x\alpha+y\beta = g$.

Thus there does exist integers $\alpha$ and $\beta$ such that $x\alpha+y\beta=g$.

Now to prove $g$ is minimum, consider any positive integer $g' = x\alpha'+y\beta'$. As $g|x,y$ we get $g|x\alpha'+y\beta' = g'$, and as $g$ and $g'$ are both positive integers this gives $g\le g'$. So $g$ is indeed the minimum.

## Generalization/Extension of Bézout's Identity

Let $a_1, a_2,..., a_m$ be positive integers. Then there exists integers $x_1, x_2, ..., x_m$ such that $$\sum_{i=1}^{m} a_ix_i = \gcd(a_1, a_2, ..., a_m)$$ Also, $\gcd(a_1, a_2, ..., a_m)$ is the least positive integer satisfying this property.

### Proof

Consider the set $P = \{n \in \mathbb{Z}^{+}|n= \sum_{i=1}^{m} a_iu_i: u_1, \dots, u_m \in \mathbb{Z}\}$. Obviously, $P \neq \emptyset$. Thus, because all the elements of $P$ are positive, by the Well Ordering Principle, there exists a minimal element $d \in P$. So $$d=a_1x_1 +a_2x_2 + \dots +a_mx_m$$

if $n >d$ and $n \in S$ then $$n=a_1u_1 +a_2u_2 + \dots +a_mu_m$$ But by the Division Algorithm: $$n=qd +r \Longrightarrow r=n-qd$$ $$= \sum_{i=1}^m a_i(u_i-qx_i)$$ $$\Longrightarrow r\in P$$

But $0 \le r so this would imply that $r \in P$ which contradicts the assumption that $d$ is the minimal element in $P$. Thus, $r=0$ hence, $d|n$. But this would imply that $d|a_i$ for $i \in \{1, 2,\dots,m\}$ because $a_i = a_i \cdot1 + \sum_{k=1; k \neq i}^{m}(a_k\cdot0) \Longrightarrow \{a_1, a_2, \dots, a_m \} \subset P$. Now, because $d|a_i$ for $i \in \{1, 2,\dots,m\}$ we have that $d|\gcd(a_1, a_2,\dots, a_m)|a_i$. But then we also have that $\gcd(a_1, a_2,\dots, a_m)|\sum_{i=1}^m a_iu_i =d$. Thus, we have that $\boxed{d=\gcd(a_1, a_2,\dots, a_m)}$ $\Box$