If $b$ and $c$ are zero, then since $\operatorname{pgcd}(a, b, c)=1$, we must have $a=1$ or $a=-1$. In the first case, there is nothing to do. In the second case, by three successive operations, we can go from $(-1,0,0)$ to $(-1,1,0)$, then to $(1,1,0)$, and finally to $(1,0,0)$.

We can therefore assume that $b$ or $c$ is non-zero.
The goal is to reach a triplet of the form $(1, \mathrm{~s}, \mathrm{t})$ in at most three steps, since with two additional steps, it is then immediate to obtain $(1,0,0)$.

We assume that $\mathrm{c} \neq 0$, the case $\mathrm{b} \neq 0$ being treated similarly.
To achieve the above goal, it suffices to find an integer $n$ such that $c$ and $\mathrm{b}+\mathrm{na}$ are coprime: indeed, suppose for the moment that such an integer $n$ has been determined and see how the conclusion will follow.

We set $b^{\prime}=b+n a$ and, with a first operation, we can go from $(a, b, c)$ to $\left(a, b^{\prime}, c\right)$. Then, by Bézout's theorem, we know that there exist two integers $u$ and $v$ such that $u b^{\prime}+v c=1$. After multiplying by $a-1$, there therefore exist two integers $x$ and $y$ such that $x b^{\prime}+y c=a-1$. We then use a second operation to go from $\left(a, b^{\prime}, c\right)$ to $\left(a-x b^{\prime}, b^{\prime}, c\right)$, and a third to go from $\left(a-x b^{\prime}, b^{\prime}, c\right)$ to $\left(a-x b^{\prime}-y c, b^{\prime}, c\right)=\left(1, b^{\prime}, c\right)$.

To conclude, it only remains to prove that such an integer $n$ exists.
Let $E$ be the set of prime divisors of $c$ that also divide at least one number of the form $b+k a$, where $k \in \mathbb{Z}$.
If $E=\emptyset$ then any integer $n$ is such that $\operatorname{pgcd}(\mathrm{c}, \mathrm{b}+\mathrm{na})=1$.
Otherwise, we note that $E$ is finite because it only contains prime divisors of $c$, with $c \neq 0$. Next, for each $p \in E$, there exists an integer $k_{p}$ such that $b+k_{p} a=0 \bmod [p]$. We note that such a number $p$ does not divide $a$ because otherwise $p$ would also divide $b$, in contradiction with $\operatorname{pgcd}(a, b, c)=1$.

By the Chinese remainder theorem, there exists an integer $n$ such that $n=k_{p}+1 \bmod [p]$ for all $p \in E$. Let's prove that this integer $n$ works:

- If $p$ divides $c$ and $p \notin E$ then $p$ does not divide any number of the form $b+k a$ and therefore, in particular, does not divide $b + na$.
- If $p \in E$ then $b+n a=b+k_{p} a+a=a \neq 0 \bmod [p]$.

Thus, no prime divisor of $c$ divides $\mathrm{b}+\mathrm{na}$, which ensures that $c$ and $\mathrm{b}+\mathrm{na}$ are indeed coprime.
!