SAUDI ARABIAN IMO Booklet 2023

Adding the first two equations yields $$0=xy+z+yz+x=(x+z)(y+1).$$ So $y=-1$ or $z=-x$.

If $y=-1$ then the system reads $z=x-30$ and $xz=-17$. It follows that $x^2-30x+17=0$ and hence $x=15\pm 4\sqrt{13}$.

If $z=-x$, then the system becomes $xy-x=-30,y=x^2-18$ so $$x^3-19x+30=0,$$ whose solutions are $x=2,x=3$ and $x=-5$. Therefore, we have five solutions $(x,y,z)$ in total: $(15+4\sqrt{13},-1,15-4\sqrt{13})$, $(15-4\sqrt{13},-1,15+4\sqrt{13})$, $(2,-14,-2)$, $(3,-9,-3)$, $(5,7,-5)$.

$\square$