$\DeclareMathOperator\lcm{lcm}$This is a rather severe revision of [a question I asked recently](https://mathoverflow.net/questions/48888/). 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.



Explicit Bézout cofactors
The answer to the question posed by Aaron Meyerowitz to darij grinberg in the [comments](https://mathoverflow.net/questions/49035/explicit-b%c3%a9zout-cofactors#comment119719_49035) is unfortunately negative, even in the integers, by taking $a=c=u=v=w=0$ but $b=d=1$. However, it has a positive answer when $u\neq 0$.


Assume $ab=cd$, $ax+cy=u$, $uv=a$, and $uw=c$. Using the latter two equalities, we can solve for $a$ and $c$. The first two equalities now become
$$
uvb=uwd\qquad \text{ and }\qquad uvx+uwy=u.
$$
As $u\neq 0$, and we are working in an integral domain, we obtain
$$
vb=wd\qquad\text{ and }\qquad vx+wy=1.
$$
We find
$$
b=b(vx+wy)=(vb)x+bwy=(wd)x+bwy=w(dx+by).
$$
Similarly, $d=v(dx+by)$. Thus, taking $t:=dx+by$ works.


