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

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    i think the GCD is equal to $1$ – 2017-01-08
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    Please let me know why. – 2017-01-08
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    @Dr.SonnhardGraubner It's not 1, take $x=y=z=1$, you'll get 3 – 2017-01-08
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    There is no restriction for $x,y,z$ other than of being integers? – 2017-01-08
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    They are pairwise coprime ( to keep things interesting). – 2017-01-08

3 Answers 3

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

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

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

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

  1. Otherwise, in the range tested,

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

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    Thanks Peter. You did not consider the case where one of $x,y,z$ is divisible by 3. Anyways, so far that seems to confirm my suspicion. I just wish I could make a stronger case to prove it. – 2017-01-15
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    I looked at ALL cases in the range fitting the bill. Gcd=3 only when $x,y,z$ all 1 mod 3 or all 2 mod 3. But I'm only human, so please give me an example & I'll check. – 2017-01-15
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    I really appreciate your help. I actually thought it is the easiest case $x=3a+1, y=3b+2, z=3c$ because the focus can be solely on 2 variables instead of 3. Since we know that 3 is already one of the prime factors, that'd automatically force the other 2 factors to be divisible by 3 as well. A contradiction since $x,y$ are not divisible by 3. – 2017-01-15
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    Well, there are 27 cases (mod 3), and I've done the maths on two of them. I think you can easily reduce this to just 6 cases, using common factors and $f_1$, to prove we have covered all for gcd=3. However, I'm not sure how to tackle the rest of the primes. – 2017-01-16
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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 .

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

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.

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    sorry to say it but clearly 3 can be their $\gcd$. Just try: $$x=2$$ $$y=5$$ $$z=11$$ – 2017-01-13
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    @NumThcurious Thank you for your interest. Yes, indeed, $(x,y,z)=(2,5,11)$ produces a $gcd=3$, and I expect there are other examples where $gcd>1$. My method needs, at least, one more test than the highest power. My argument is if $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$ had a common algebraic factor then finding a $gcd=1$ would be the exception rather than the rule. My method dates back to before the time of freely available computerised algebraic engines, and I used it mainly as an add to factorising algebraic expressions. Perhaps I’ve just misunderstood the question. – 2017-01-14
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    i was trying to figure out whether there can be other prime factors other than 3. – 2017-01-14
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    Your pairwise co-prime restriction makes coding a search messy, but if I can find the time I'll try and find out. – 2017-01-14
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    Thank you.That'd be greatly appreciated. – 2017-01-14
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    @NumThcurious I finally found time to look at this; there is too much to post as a comment. I’ve not yet had time to check the formulae with values, sorry. – 2017-01-15