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:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$

$gcd(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)=1$
If $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$ shared a common factor, the existence of an algebraic factor would soon be apparent after trying a few numerical examples.
So I used $(x,y,z)=(17,23,31),(2,3,5),(3,7,11)$ and found $gcd=1$ in each case.
I realise this method might seem unscientific, but I’ve yet to find an example where it fails.
