We want to develop a version of the Euclid-Bézout algorithm for triples of natural numbers.
Let $n_1$, $n_2$, $n_3 \in \mathbb{N}$ but not all three zero. We define $\gcd(n_1, n_2, n_3) \in \mathbb{N}$ as the maximum natural number $d$ such that $d\mid n_1$ and $d\mid n_2$ and $d\mid n_3$. For example, $\gcd(70, 105, 770) = 35$, because $35$ is a common divisor and there is no a common divisor which is larger. Let three natural numbers be given (where not all three are $0$ as before, and the order does not matter) We will modify this triple and obtain another triple as follows: Let's call $a_1$ the smallest non-zero number among the three, and let's call $a_2$, $a_3$ the other two numbers. Let $r_2$ be the remainder of $a_2$ after division-with-remainder by $a_1$, and $r_3$ the remainder of $a_3$ after division- with-remainder by $a_1$. Let $b_1:= a_1$, $b_2:= r_2$, $b_3:= r_3$. So from the triple $a_1$, $a_2$, $a_3$ we get another triple $b_1$, $b_2$, $b_3$. For example the triple $70, 105, 770$ becomes the triple $70, 35, 0$. I have shown that $\gcd(a_1, a_2, a_3) = \gcd(b_1, b_2, b_3)$ and if $b_2 = b_3 = 0$ then $\gcd(b_1, b_2, b_3) = b_1$.
(i) Let $d = \gcd(a_1, a_2, a_3)$. If $s_1, s_2, s_3$ are three integers such that $s_1b_1 + s_2b_2 + s_3b_3 = d$, find three integers $t_1, t_2, t_3$ such that $t_1a_1 + t_2a_2 + t_3a_3 = d$. For instance, we choose $s_1 = 1$, $s_2 = 0$, $s_3 = 0$ if $b_2 = b_3 = 0$.
Can someone please help me on how to find those three integers?
