Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$? This question was raised in comments of
Is the $(3,4,5)$ triangle the only rectangular triangle with this property?
and I was suggested to ask it as a separate one.
First some notation, let's write $(a,b,c)$ where $a<=b<c$ for a pythagorean triple, and let's write $(x,y,z)$ where $x<=y<z$ for the distances from the vertices to the center of the inner circle of the corresponding right triangle.

($XYZ$ should be lowercase but geogebra did not allow these labels)
The answer to above question proved (partially left to reader) the property:
$x * y = z$ if and only if $c - b = 1$
In comments was asked if a similar property would exist for $c - b = 2$
and it was confirmed (and proof was left to reader) that:
$x * y = 2 * z$ if and only if $c - b = 2$
Somewhat natural question then was raised (by me) if one can generalize for other values of $n >= 1$, that is:
for which $n$ (perhaps all) holds:
$x * y = n * z$ if and only if $c - b = n$
?
Thanks to @heropup for the suggestion (and the answers for $n=1,2$)
update
A simple computer programmed enumeration seems to confirm equivalence. At least for all $(a,b,c)$ with maximum $c <= 10000$. Note that it is not known to me (but perhaps it is known to others) if all $c - b$ cover all $n >= 1$.
So asking for all $n$ is a bit ambiguous since some $n$ might never occur.
An alternative, perhaps better, question rephrase is:
Prove equality
$$x * y = (c - b) * z$$
for all pythagorean triples.

Hint: For all primitive Pythagorean triples (right triangles with integer sides),
$\space C-B=(2n-1)^2,n\in\mathbb{N}$
This can be seen at a glance using a formula I developed in $2009.$
\begin{align*}
&A=(2n-1)^2+&&2(2n-1)k\\
&B=               &&2(2n-1)k+2k^2\\
&C=(2n-1)^2+&&2(2n-1)k+2k^2
 \end{align*}
This formula generates all triples where $\space GCD(A,B,C)\space$ is an odd square. This includes all primitives where $GCD(A,B,C)=1.$
For Pythagorean triples, it appears that
$\quad z^2=(2n-1)^2C$
