$xy=zw$ substitution In Olympiad math context (mainly AoPS), I often see substitution $(x,\ y,\ z,\ w)=(ab,\ cd,\ ad,\ bc)$ under $xy=zw$.
The following is a construction of $(a,b,c,d)$ which I came up with.

For a prime $p$ and nonzero integer $n$, we let $\nu_p(n)$ denote the largest integer $e$ with $p^e$ dividing $n$.
For an arbitrary prime $p$, let $$(s,\ t,\ u,\ v)=(\nu_p(x),\ \nu_p(y),\ \nu_p(z),\ \nu_p(w))$$ We have $s+t=u+v$.
Without loss of generality we may assume $s=\min\{s,t,u,v\}$.
We let $(a,b,c,d)$ satisfy
\begin{eqnarray}
\nu_p(a)&=&s \\
\nu_p(b)&=&0 \\
\nu_p(c)&=&u \\
\nu_p(d)&=&v-s
\end{eqnarray}
Now we have $\nu_p(a)+\nu_p(b)=\nu_p(x)$, $\nu_p(c)+\nu_p(d)=\nu_p(y)$, etc.
We repeat this operation and the construction is done.

The question is:
Does this substitution have a particular name that one can easily refer to?
 A: I don’t know of a name for the substitution… but I do believe there’s an easier answer/proof than what you’re suggesting. (Coincidentally, I was just writing it up for a paper I’m preparing for publication soon! I’m using the restricted version where $x,y,z,w$ are positive, but the full version is proven in a similar manner.)
Proposition. If $a$, $b$, $c$, and $d$ are positive integers with $ab=cd$, then there exist positive integers $p$, $q$, $r$, and $s$ such that $a=pq$, $b=rs$, $c=pr$, and $d=qs$.
Proof.
We are given $ab=cd$. Set $p=\gcd(a,c)\ge1$, so that $a=pq$ and $c=pr$ for positive integers $q$ and $r$ with $\gcd(q,r)=1$. Hence $pqb = prd$, and dividing both sides by $p \ne 0$ leaves $qb = rd$. As $\gcd(q,r)=1$, the Fundamental Theorem of Arithmetic implies $r \mid b$, say $b=rs$ for an integer $s \ge 1$. Then $qrs=rd$, so $d=qs$.$\square$
A: Your property is related to equidivisibility, which is defined in a more general context.
A monoid $M$ is said to be equidivisible if for each quadruple $(x, y, z, t) \in M^4$ satisfying $xy = zw$, there exists $h \in M$ such that either $x = zh$ and $hy = t$ or $xh = z$ and $y = ht$. 
Thus either 
$$
(x, y, z, t) = (ab, cd, ad, bc),
$$ 
with $a = z$, $b = h$, $c = y$ and $d = 1$, or 
$$
(x, y, z, t) = (ab, dc, ad, bc),
$$ 
with $a = x$, $b = 1$, $c = t$ and $d = h$. Note the difference between the terms $cd$ and $dc$, which of course disappears in a commutative setting.
Levi's lemma shows that any free monoid is equidivisible.
[1] McKnight, J. D., Jr. and Storey, A. J., Equidivisible semigroups, J. Algebra 12 (1969), 24-48.
