Integer solutions to $xyz = w^2(x+y+z)$ I'm looking for a way to enumerate all positive integer solutions of the equation
$xyz = w^2(x+y+z)$
where $w \le W$ and $1 \le x \le y \le z$. 
Could anyone provide a hint at how to approach this?
 A: Generalizing individ's answer,
in
$xyz=w^2(x+y+z)
$,
if
$x+y = z$,
then
$xy(x+y)=2w^2(x+y)
$
or
$xy = 2w^2$.
If
$r(x+y) = z$
where $r$ is rational,
then
$xyr(x+y)=(1+r)w^2(x+y)
$
or
$rxy = (1+r)w^2$
or
$xy = (1+\frac1{r})w^2
$.
Therefore,
for each $w^2$,
for any $r$
such that
$w^2/r$
is an integer,
look at all the factorizations of
$w^2(1+1/r)=xy$.
In particular,
if $r = \frac1{n}$
for integer $n$,
let $xy$ go over the
factors of
$(n+1)w^2$,
choose those where
$n$ divides $x+y$,
and let
$z = (x+y)/n$.
To verify,
$xyz
=(n+1)w^2(1/n)(x+y)
=(n+1)w^2(x+y)/n
$
and
$w^2(x+y+z)
=w^2(x+y+(x+y)/n)
=w^2(x+y)(n+1)/n
$.
If $r=1$,
this gives the first solution.
If $r=2$,
we look at the
factors of $xy=3w^2$
with the same parity
(this implies that
$w$ is even),
and let
$z = (x+y)/2$.
For general $n$,
we want
$n | x+(n+1)w^2/x$.
It's getting late
and I'm tired,
so I'll stop here.
A: I ended up using the following method for computing the answer: first, multiply both sides by $x$ and factor the equation as
$$(xz - w^2)(xy - w^2) = w^2(x^2 + w^2)$$
Then iterating through all $x$ and $w$ I factorize the right-hand side, find all its divisors $f$ and leave only those that yield integer values
$$z = (w^2 + f)/x, \qquad y = w^2(x^2+w^2 + f)/(xf)$$
A: $$xyz=w^2(x+y+z)$$
Decompose into factors of the number.
$$tp=2w^2$$
Then the decision on the record.
$$x=t$$
$$y=p$$
$$z=t+p$$
