If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$ This is a problem based on Pythagorean Triples.

If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$

Any ideas of  how to start solving it?

I tried substituting $x = a^2 + b^2 \  , y = c^2 + d^2 \ , z = e^2 + f^2$ and so on and I am not getting any idea how to show it is divisible by $180$ , or divisible by $4,5,9$
Outline:


By considering $x^2+y^2=z^2$ modulo $8$, deduce that one of $x,y,z$ must be a multiple of $4$.
By considering $x^2+y^2=z^2$ modulo $5$, deduce that one of $x,y,z$ must be a multiple of $5$.
By considering $x^2+y^2=z^2$ modulo $3$, deduce that one of $x,y,z$ must be a multiple of $3$.
Confirm that any multiple of $3$ that is the sum of two squares must be a multiple of $9$. (This is the only step where being a sum of two squares is used.)


These four results combine to give you the divisibility by $180$ you desire.
$$x^2= y^2+z^2\;\;\;(*)$$


Divisibilty with $4$: 


If $y$ and  $z$ are even then we are done. 
If $y$ and  $z$ are odd then $x $ is even so $x=2a$ and $y=2b+1$ and $z=2c+1$ so $$4a^2 = 4b^2+4b+4c^2+4c+2$$ which is impossible
If $y=2b+1$ and  $z=2c$ then $x=2a+1$ so $$4a^2+4a+1= 4b^2+4b+1+4c^2$$ so $$\underbrace{a(a+1)}_{even}-\underbrace{b(b+1)}_{even}=c^2\implies 2\mid c$$ and we are done since $4\mid z$.






Divisibilty with $5$: For each $n$ we have $$n^2\equiv 0,1,-1 \pmod 5$$. So if $x,y, z\ne 0$ then we can not have $(*)$





Divisibilty with $9$: Hint, observe the equation modulo $3$.