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 $4,5,9$ 
 A: Outline:


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*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.
A: COMMENT.-(This is not an answer) It seems that there are not many Pythagorean triples with $x, y, z$ coprimes answering your problem. The only one that I have been able to locate is $(9,40,41)$. 
You must bear in mind that when speaking of integers that are the sum of two squares, $0$ is considered to be a square so we have $9=3^2 + 0^2,\space40=6^2 + 2^2,\space41=5^2 + 4^2$ and it is clear that $180|xyz$. 
On the other hand, taking into account that the product of the sums of two squares is stable for multiplication, we can obtain an infinity of desired  Pythagorean triples but these will not be primitive (i. e. not coprime). In fact, $180$ is a sum of two squares and then $180 ^ n$ is also . Indeed $180 = 12 ^ 2 + 6 ^ 2$. This infinite set of examples is
given by
$$ (x_n,y_n,z_n)=(180^n\cdot9,\space180^n\cdot40,\space180^n\cdot41)$$
A: $$x^2= y^2+z^2\;\;\;(*)$$


*

*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 $3$.

