There are many facts about integer gcds which can be proved by appealing to unique prime factorization (up to sign). for example $\gcd(a^2,b^2)=\gcd(a,b)^2$. One way to get the machinery (if that is not too strong a word) about primes and unique factorization is to start with the fact that for all integers $a,b$ there are is a linear combination $d=as+bt$ which divides both $a$ and $b$. That is, to show that the integers are a *Bezout Domain* and then follow the familiar series of results we all learned and perhaps forget (but then relearned when we taught it). An important tool is that the cofactors $s,t$ can be computed via the Euclidean Algorithm. There are other domains with an integer norm where the same path can be followed: $\mathbb{Z}[\sqrt{k}]$ for certain $k$ such as $2$ and $-1$ for example. However, many of the facts about divisibility can be derived without all that machinery. In a recent answer I pointed out that $as+bt=1$ implies after cubing and simplifying that $a^2(as+3bt)s^2+b^2(3as+bt)t^2=1$ so that $\gcd(a,b)=1$ implies $\gcd(a^2,b^2)=1$. That can be fixed up to a proof that $\gcd(a^2,b^2)=\gcd(a,b)^2$. My question is how far one can get assuming only that a commutative ring (with 1) is a Bezout Domain? Lest I be accused of not having a question I'll ask the following (but feel free to tell more).

Is there any nice treatment of the facts about divisibility which follow only from the assumption that we are in a Bezout Domain? And are there any nice Bezout Domains which show some facts which don't follow?

I suppose all the premises would be equations as would the conclusions. Can we say anything about the increase in complexity? In the equation $a^2(as+3bt)s^2+b^2(3as+bt)t^2=1$ we can use $as+bt=1$ to replace $as+3bt$ with $2bt+1$ or $as+bt+2$ although this is still the same value. Must the cofactors for $s',t'$ for $a^2s'+b^2t'=1$ be at least quartic in $a,b,s,t$ or is there a nicer equation? Over the integers the $s'$ and $t'$ I mention are usually much bigger than they need to be. For example $a,b=11,18$ gives $as+bt=11\cdot 5+18\cdot (-3)=1$ which yields $11^2\cdot (-2675)+18^2\cdot 999=1$ But we prefer $11^2\cdot (-83)+18^2\cdot 31=1$ or at least $11^2\cdot 241+18^2\cdot (-90)=1$.