<s>[INST] 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 $4,5,9$ 
 [/INST] $$x^2= y^2+z^2\;\;\;(*)$$


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*Divisibilty with $4$: 


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*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 $4\mid z$.






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*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 $(*)$



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

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