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.

Let $(a,b,c)$ be a primitive Pythagorean triple with $a<b<c$. Then $\{a,b\}=\{2mn,m^2-n^2\}$ and $c=m^2+n^2$ for some coprime integers $m>n>0$ of different parity. If $2mn<m^2-n^2$, then $c-b=2n^2$ is twice a square. Moreover, if $m>3n$, then $2mn<(2/3)m^2<(8/9)m^2<m^2-n^2$, so for every $n$ there exists $(a,b,c)$ such that $c-b=2n^2$.
If $m^2-n^2<2mn$, then $c-b=(m-n)^2$, the square of an odd number. Let $k\ge7$ be odd, let $m=2k-2$, $n=k-2$; then $m,n$ are coprime of different parity, $m>n>0$, and $m-n=k$; also, $m^2-n^2=3k^2-4k$, $2mn=4k^2-12k+8$, and $m^2-n^2<2mn$ is equivalent to $k^2-8k+8>0$, which is $(k-4)^2>8$, which is true for $k\ge7$. So, for every odd $k\ge7$ there exists $(a,b,c)$ such that $c-b=k^2$. For $k=1,3,5$, respectively, we can take $(a,b,c)$ to be $(3,4,5)$, $(33,56,65)$, $(65,72,97)$ to get $c-b=k^2$.
Conclusion: for a primitive Pythagorean triple, $c-b$ can take on every odd square, and every twice-a-square, and only those values.
Given any positive integer $t$, the Pythagorean triple $(3t,4t,5t)$ has $c-b=t$, so, if non-primitive triples are allowed, then $c-b$ can take on any positive integer value.
Now, let's bring in $x,y,z$. Using the formulas at The distance from the incenter to an acute vertex of a right triangle, we have
$2x^2=(a+b-c)^2$
$2y^2=a^2+(c-b)^2$
$2z^2=b^2+(c-a)^2$
If $a=2mn<m^2-n^2=b$, $c=m^2+n^2$, then
$$
4(xy)^2=(2mn-2n^2)^2(4m^2n^2+4n^4)=(4n^2)(m-n)^2(4n^2)(m^2+n^2)=16n^4(m-n)^2(m^2+n^2)
$$
so $xy=2n^2(m-n)\sqrt{m^2+n^2}$, and
$$
2z^2=(m^2-n^2)^2+(m-n)^4=2m^4-4m^3n+4m^2n^2-4mn^3+2n^4=2(m-n)^2(m^2+n^2)
$$
so $z=(m-n)\sqrt{m^2+n^2}$. Thus, $xy=2n^2z=(c-b)z$, as requested.
If $a=m^2-n^2<2mn=b$, $c=m^2+n^2$, then
$$
4(xy)^2=(2mn-2n^2)^2((m^2-n^2)^2+(m-n)^4)=8n^2(m-n)^4(m^2+n^2)
$$
so $xy=\sqrt2n(m-n)^2\sqrt{m^2+n^2}$, and
$$
2z^2=4m^2n^2+4n^4=4n^2(m^2+n^2)
$$
so $z=\sqrt2n\sqrt{m^2+n^2}$. Thus, $xy=(m-n)^2z=(c-b)z$, again as requested.
If the triple is not primitive, the common factor cancels out in the end, and the conclusion remains.
