<s>[INST] 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:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$
 [/INST] If $x,y,z$ are not coprime, the result is trivial. So We can assume that $x,y,z$ are not all even. Then $t$ must be an odd integer since the bottom 3 numbers are always odd.
$$x^2+y^2+z^2+\dfrac{xy+xz+zx}{2}=\dfrac{t(u+v+w)}{2}$$
$$(x+y+z)^2-\dfrac{3(xy+xz+zx)}{2}=\dfrac{t(u+v+w)}{2}$$
$$(kt)^2-\dfrac{3(xy+xz+zx)}{2}=\dfrac{t(u+v+w)}{2}$$
So $t$ must divide $3(xy+yz+zx)$. Then, $t=\pm 3$ and/or divides $xy+yz+zx$ 
$\implies$
I still have a quadratic equation to deal with to decide that $t$ is an integer. So I will have to keep imposing restrictions to find $t$. I guess .
</s>