mruczek wrote: Determine all trios of integers $(x, y, z)$ which are solution of system of equations $\begin{cases} x - yz = 1 \\ xz + y = 2 \end{cases}$ First equation gives $x=yz+1$ and second equation becomes $y=\frac{2-z}{z^2+1}$ This implies : Either $z>2$ and $z^2+1\le z-2$, impossible Either $z=2$ and so $y=0$ and $x=1$ Either $z<2$ and so $z^2+1\le 2-z$ and so $z\in\{-1,0\}$, from which only $z=0$ fits, giving $y=2$ and $x=1$ Hence the result $\boxed{(x,y,z)\in\{(1,0,2),(1,2,0)\}}$

First and second equation to the second power, sum them, and get $(x^2+y^2)(z^2+1)=5$. Because we search for trios of integers $(x, y, z)$ we are left with two systems of equations. $z^2+1=5\implies ...\implies (x,y,z)=(1,0,2)$ $z^2+1=1\implies z=0\implies x=1\wedge y=2$.

This problem was proposed by Burii.

oralayhan wrote: $\boxed{(x,y,z)\in\{(1,0,2),(1,2,0)\}}$ Not integer solution When I was young student, many many years ago, $0,1,2\in\mathbb Z$ Are you claiming that some of these numbers no longer are integers ($\notin\mathbb Z$) ?