For any nonzero integers $(x, y, z)$ such that $x+y+z$ divides $1 - xyz$., it is easy to verify that $\gcd(x+y, z) = 1$.
But what are the nontrivial conditions such that the divisibility holds ?
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4$\begingroup$ I'm not sure what you mean by nontrivial, but if $(x,y,z)$ is a permutation of $(1,n,n^2)$ then the divisibility holds. $\endgroup$– WojowuCommented Dec 23, 2015 at 16:50
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2$\begingroup$ math.stackexchange.com/questions/1586758/… $\endgroup$– Will JagyCommented Dec 23, 2015 at 19:00
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1 Answer
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Partial answer.
Let $p_2(x,y,z)=1-xyz$ and $p_1(x,y,z)=x+y+z$.
You want $p_2/p_1$ to be integer for integers $x,y,z$.
Let $X=y^2 z + yz^2 - y - z + 1$.
$p_2(X,y,z)/p_1(X,y,z)=1-y z$, so $x$ exists for all $y,z$, s.t. $X+y+z \ne 0$.
For the exceptional set, we have $X+y+z=y^2z+yz^2+1=0$. The rational solutions are $y=1/2\,{\frac {-{z}^{2}+\sqrt {{z}^{4}-4\,z}}{z}}$ and $y=-1/2\,{\frac {{z}^{2}+\sqrt {{z}^{4}-4\,z}}{z}}$.
This means $z^4-4z=t^2$ for rational $z,t$.
This is genus $1$ curve and unless I am mistaken, it is birationally equivalent to the Weierstrass model $v^2=u^3+16$, which have finitely many rational solutions, since it it is rank $0$ according to sage.
