Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done: 
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$

I first noticed that $(x,y,z)$ are not interchangeable, and found only $gcd=1$ and $gdg=3$, with $(x,y,z)$ in the range $1$ to $999$ and co-prime in pairs.
I’ll use $(f_1,f_2,f_3,f_4)$ for $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$


*

*When $x=3a+1,y=3b+1,z=3c+1$


$$f_1=3*(a+b+c+1)$$
$$f_2=3*(3a^2+3ab+3a+b+3c^2+2c+1)$$
$$f_3=3*(3b^2+3bc+3b+3c^2+3c+1)$$
$$f_4=3*(3a^2+3ac+3a+3c^2+3c+1)$$


*When $x=3a+2,y=3b+2,z=3c+2$


$$f_1=3*(a+b+c+2)$$
$$f_2=3*(3a^2+3ab+6a+2b+3c^2+4c+4)$$
$$f_3=3*(3b^2+3bc+6b+3c^2+6c+4)$$
$$f_4=3*(3a^2+3ac+6a+3c^2+6c+4)$$


*Otherwise, in the range tested, 


$$gcd(f_1,f_2,f_3,f_4)=1$$
