Solve in integers the system of equations
$\begin{array}{lll} xy-\dfrac{z}{3} = xyz+1 \\ yz-\dfrac{x}{3} = xyz - 1 \\ zx-\dfrac{y}{3} = xyz-9. \end{array}$
Proposed by Titu Andreescu.
Solution:
It's easy to see that $x,y,z$ must be divisible by $3$. Summing up all the equations and factorizing, we obtain $$(x-1)(y-1)(z-1)(xy+yz+zx+1)=8.$$ Then $x-1=\pm1,\pm 2, \pm 4, \pm 8$, i.e. $x=0,2,-1,3,-3,5,-7,9$ and for what we said, we reduce to $x \in \{-3,0,3,9\}$. Now, consider the second equation of the system $$yz-\dfrac{x}{3} = xyz - 1.$$ We have four cases.
(i) If $x=-3$, then $2yz=-1$, so no integer solution.
(ii) If $x=0$, then $yz=-1$, but the first equation gives $z=-3$, so no integer solution.
(ii) If $x=0$, then $yz=-1$, but the first equation gives $z=-3$, so no integer solution.
(iii) If $x=3$, then $yz=0$. If $y=0$ we get $z=-3$ from the first equation of the system and these values satisfy also the third equation, i.e. $(3,0,-3)$ is a solution. If $z=0$, the first equation gives $3y=1$, i.e. no integer solution.
(iv) If $x=9$, then $4yz=-1$, so no integer solution.
Therefore, $(3,0,-3)$ is the only integer solution to the given system.
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