Consequence of Linear Diophantine Equation

Question

Let $a,b \in \mathbb{Z}$. By linear Diophantine equations we know that if $\gcd(a,b) = g$ then there exist $x,y \in \mathbb{Z}$ such that $ax+by = g$.

That said, is the following true? Let $a,b \in \mathbb{Z}$. Suppose that there exist integer solutions to the equation $ax+by = 1$. Does that mean that $\gcd(a,b) = 1$? That is, is the Linear Diophantine property bidirectional?

Answer 1

This is similar to Bezout's identity, where if $d = \gcd(a,b)$, then there exist integers $r,s$ such that $ar+bs = d$.

Assume first that for $r,s \in \mathbb{Z}$ we have $ar+bs = 1$, then assume that $d$ is a common divisor of both $a$ and $b$. Then $d |(ar+bs)$ and so $d|1$, this means that $d=1$ and hence $\gcd(a,b) = 1$.

Answer 2

I'm posting as answer because I'm not allowed to comment.

gcd of $a,b$ is the least positive integer that can be written as linear combination of $a,b$.
so yes :
$$\gcd(a,b)=1 \iff \exists x,y\in \mathbb{Z}: ax+by=1$$

Answer 3

Yes, it is.

Suppose $m|a$ and $m|b$. Then $m|ax+by=1,$ so $m=\pm 1.$ Therefore $\gcd(a,b)=1.$