Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$

Question

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

Answer 1

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 .

Answer 2

$gcd(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)=1$

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.

Answer 3

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 $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$

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

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

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

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

3. Otherwise, in the range tested,

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