Proving uniqueness that if $S = \{as+bt|as+bt>0, a,b \in \Bbb N\}$ and $d, d' \in S$ and $d|a$, $d|b$, $d'|a$, $d'|b$ $\rightarrow$ $d = d'$.

Question

I was looking through proofs of Bezout's identity, and the various posts (1, 2, 3) that $gcd(a,b)$ is a linear combination $as + bt$, but I'm struggling to prove that if $d \in S$ where $S = \{as+bt | as+bt>0, a,b \in \Bbb N\}$ and $d|a$ and $d|b$ that $d$ is necessarily the smallest element in $S$. If not, then $$\exists d,d' \in S,\\ d|a,d|b,\\ d'|a, d'|b, \\d'<d$$ and the smallest element $d \in S$ is not $gcd(a,b)$.

There's some consequence of the proof that I suppose I'm missing, and admittedly, there are infinite choices of $s$ and $t$ which yield the same $d$ smallest element, for example, $d = as + bt \rightarrow d=as+bt+ab-ba=a(s+b)+b(t-a)$ so the choices of $s'=s+b$ and $t'=t-a$ yield $d=as'+bt'=as+bt$, but I'm interested in proving the uniqueness of an element $d\in S$ which divides both $a$ and $b$. Thank you!

Answer 1

Although I forgot I had ever asked this question, and years later am rediscovering it, I was able to figure this out independently.

The idea is to take two strictly positive integers, $d$ and $d'$ in $S$ that divide $a$ and $b$, with $a = q_ad'$, and $b = q_bd'$.

Since $$d = as + bt$$ then $$d = q_ad's+q_bd't$$ $$d = d' * (q_as + q_bt)$$ with the quantity $q_as + q_bt$ a strictly positive integer since $d$ and $d'$ are both strictly positive integers.

Starting at the same point for $d'$, $$d' = as' + bt'$$ results in an equivalent expession: $$d' = d * (q_a's' + q_b't')$$

Where $q_a's' + q_b't'$ is also a strictly positive integer.

With those two relationships established, either $$d = d'$$ $$q_a's' + q_b't' = q_as + q_bt = 1$$

or $$d > d'$$ and $$q_a's' + q_b't' < 1$$

which contradicts the fact that it's a strictly positive integer, and therefore $$d = d'$$