[proofplan] We translate coprimality into the statement $\gcd(a,b)=1$. In one direction, Bézout's characterisation of the [greatest common divisor](/page/Greatest%20Common%20Divisor) says that the greatest common divisor is the smallest positive integer obtainable as an integer linear combination of $a$ and $b$; when the gcd is $1$, this directly gives a combination equal to $1$. Conversely, if $1$ is an integer linear combination of $a$ and $b$, then the positive gcd divides $1$, forcing it to equal $1$. [/proofplan] [step:Apply Bézout's gcd characterisation when the gcd is $1$] Assume that $a$ and $b$ are coprime. By the meaning of coprimality in this statement, this means \begin{align*} \gcd(a,b)=1. \end{align*} Since $a$ and $b$ are not both zero and $1 \in \mathbb{Z}_{\ge 0}$, the hypotheses of [citetheorem:9892] apply with $c=1$. That theorem states that, for integers $a,b$ not both zero and $c \in \mathbb{Z}_{\ge 0}$, the identity $c=\gcd(a,b)$ is equivalent to saying that $c$ is the smallest positive integer of the form $am+bn$ with $m,n \in \mathbb{Z}$. Applying this equivalence with $c=1$ gives that $1$ is of the form $am+bn$ for some $m,n \in \mathbb{Z}$. Therefore there exist $x,y \in \mathbb{Z}$ such that \begin{align*} ax+by=1. \end{align*} [guided] Assume that $a$ and $b$ are coprime. In the present theorem, coprime means precisely that their positive greatest common divisor is $1$, so \begin{align*} \gcd(a,b)=1. \end{align*} We now use the Bézout form of the gcd characterisation, which speaks directly about integer linear combinations. The cited theorem [citetheorem:9892] says: for integers $a,b$ not both zero and a number $c \in \mathbb{Z}_{\ge 0}$, the equality $c=\gcd(a,b)$ holds if and only if $c$ is the smallest positive integer obtainable in the form $am+bn$ with $m,n \in \mathbb{Z}$. The present theorem assumes that $a$ and $b$ are not both zero, and $1$ belongs to $\mathbb{Z}_{\ge 0}$, so the hypotheses required in that result are satisfied with $c=1$. Since $1=\gcd(a,b)$, the equivalence gives that $1$ is the smallest positive integer obtainable in the form $am+bn$ with $m,n \in \mathbb{Z}$. Being of that form is exactly the assertion that there are integers multiplying $a$ and $b$ whose linear combination equals $1$. Hence there exist $x,y \in \mathbb{Z}$ such that \begin{align*} ax+by=1. \end{align*} [/guided] [/step] [step:Show that a Bézout combination equal to $1$ forces the gcd to be $1$] Conversely, suppose there exist $x,y \in \mathbb{Z}$ such that \begin{align*} ax+by=1. \end{align*} Let \begin{align*} d=\gcd(a,b). \end{align*} Since $a$ and $b$ are not both zero, the greatest common divisor $d$ is a positive integer. By the defining divisibility property of the greatest common divisor, $d \mid a$ and $d \mid b$. Hence there exist $u,v \in \mathbb{Z}$ such that \begin{align*} a=du \end{align*} and \begin{align*} b=dv. \end{align*} Substituting these expressions into the Bézout combination gives \begin{align*} 1=ax+by=d(ux+vy). \end{align*} Since $u,x,v,y \in \mathbb{Z}$, the integer $ux+vy$ belongs to $\mathbb{Z}$. Thus $d \mid 1$. Because $d$ is a positive integer dividing $1$, we must have \begin{align*} d=1. \end{align*} Therefore $\gcd(a,b)=1$, so $a$ and $b$ are coprime. [/step] [step:Conclude the equivalence] The first step proves that coprimality implies the existence of integers $x,y$ with $ax+by=1$. The second step proves the converse implication. Hence $a$ and $b$ are coprime if and only if there exist $x,y \in \mathbb{Z}$ such that \begin{align*} ax+by=1. \end{align*} [/step]