$\DeclareMathOperator\lcm{lcm}$This is a rather severe revision of a question I asked recently. We know over the integers that $\gcd(a^2,b^2)=\gcd(a,b)^2$. We might prove this via unique factorization. In building the theory of prime factorization we use the fact that $\gcd(a,b)$ exist. This fact is sometimes proved with the slick (to a beginner) move that there must be a minimal element in $\lbrace d\mid d>0 \text{ and }d=as+bt\rbrace$ and it must be a common divisor and indeed the greatest such. This also shows that there is a linear combination $\gcd(a,b)=as+bt$. I'm sure standard terminology exists but allow me to call any such pair $(s,t)$ Bézout cofactors for the pair $(a,b)$. Of course with the extra condition $\lvert s\rvert<b$ we also have $\lvert \rvert|<a$ and two cofactor pairs, one with $s<0<t$ and one with $t<0<s$. The same things are true in more general settings such as polynomials. The slick argument above gives no indication how to find $s$, $t$ but we do know the extended Euclidean algorithm. To hone in on my question I'll stick to the case that $\gcd(a,b)=1$: Given that $a,b$ are integers with cofactors $s$, $t$ such that $as+bt=1$, there are also cofactors $s'$, $t'$ for $a^2$, $b^2$. One proof would be that evidently $\gcd(a,b)=1$ so also $\gcd(a^2,b^2)=1$ (via a small bit of theory) and hence the slick argument gives that $s'$, $t'$ exist (and the Euclidean algorithm will only take about twice as long to find $s'$, $t'$ as it will for $s$, $t$). However it is easy to check that $a^2\cdot s^2(as+3bt)+b^2\cdot t^2(3as+bt)=1$. And this establishes the existence of $s'$, $t'$ constructively given only the fact that $a$, $b$, $s$, $t$ belong to some ring, pairwise commute, and satisfy $as+bt-1=0$. (An aside: I believe I can prove that no $s'$ and $t'$ cubic in $a,b,s,t$ make $a^2s'+b^2t'=1$.) My question has to do with similar translations to polynomial identities. I'll start with a specific instance based on $ab=\gcd(a,b)\lcm(a,b)$ and then attempt to state my general question.
The following is true over the integers: if $\gcd(u,v)=1$ and $au=bv$ then there is a $w$ with $a=vw$. Is the following true as well: if $A,B,U,V,S,T$ are commuting variables and we are given the expressions $US+VT-1$ and $AU-BV$, is there an (explicit) expression $W=W(A,B,S,T,U,V)$ such that $A-VW$ is in the ideal of $\mathbb{Z}(A,B,U,V,S,T)$ generated by $AS+BT-1$ and $AU-BV$?
Note that $au=bv$ would be $\lcm(a,b)$, also $b=uw$ and $w=\gcd(a,b)$.
Consider theorems of integer divisibility whose premises and conclusions can be written as multinomial equations ($d \mid a$ becomes $a-da'=0$ , $\gcd(a,b)=1$ becomes $as+bt-1=0$ etc.). Is there always (or when is there) a derivation of the conclusion purely from manipulation of $\mathbb{Z}$ multinomials?
This is partly idle curiosity, but I also find that sometimes an explicit constructive solution is very useful to improve results.
