$\{a,b\}$ are generating set of $\mathbb{Z}$ iff $gcd(a,b)=1$

Question

Hello a want to prove that Let $a,b \in \mathbb{Z}$. The set $\{a,b\}$ are generating set of $\mathbb{Z}$ implies that $gcd(a,b)=1$ \

I know that $gcd(a,b)=1$ iff exist unique $s,t \in \mathbb{Z}$ such that $1=as+bt$

In other hand $\{a,b\}$ are generating set of $\mathbb{Z}$ means that if $z \in \mathbb{Z}$ then $z=as+bt$ with $s,t \in \mathbb{Z} $ in particular exist $s_1, t_1$ such that $1=as_1+bt_1$. This comes down to seeing the uniqueness of $s_1$ and $t_1$. Then i supose that exist other pair $s_2$ and $t_2$ such that $1=as_2+bt_2$ then \begin{align} as_1+bt_1=&as_2+bt_2\\ a(s_1-s_2)=&b(t_2-t_1) \end{align} but I don't know how to continue.

Thanks

Answer 1

Since $gcd(a,b)=1$ we know by bezout's lemma that there exists some pair $s_0,t_0\in\mathbb{Z}$ such that $$1=as_0+bt_0$$ Hence we can express any integer $k$ as an integer combination of $a$ and $b$ by multiplying both sides of the equation by $k$. We have $$k=a(s_0k)+b(t_0k)$$ Since $s_0$ and $t_0$ are both integers, so are $s_0k$ and $t_0k$. Hence, we have shown a construction for generating any integer $k$ from the generating set $\{ a,b\}$

Answer 2

I don't understand why you need to prove uniqueness of $s,t$ to prove your statement. Furthermore, if $gcd(a,b) = 1$ then there are an infinite number of solutions for the equation \begin{equation} as + bt = 1 \end{equation} given by \begin{equation} (s,t) \in \left\{ (s_0 + kb , t_0 + ka) : k \in \mathbb{Z}\right\}. \end{equation} where $(s_0,t_0)$ is any particular solution of the equation.