$x^2+y^2=z^2(1+xy)$ prove $z=\min \{x;y;z\}$ (with $x,y,z \in \mathbb{Z^+}$) $x,y,z \in \mathbb{Z^+}$ such that $x^2+y^2=z^2(1+xy)$. Prove $z=\min \{x;y;z\}$

$$x^2+y^2=z^2(1+xy) \iff xy = \frac{x^2+y^2} {z^2} - 1$$. Assum $z>y \implies xy < x^2/z^2$, we have $xy \in Z \implies x> z$ then...stuck
 A: This is a variant of the $ab+1$ problem (an old and famous killer problem from the IMO), and is a corollary of its solution.  For fixed $z$, the solutions $(x,y)$ organize into a chain where the size of the solution can be reduced by moving down the chain until the minimal solution with $xy=0$ is reached. This smallest solution is $(0,z)$ or $(z,0)$. The smallest positive solution is adjacent to this one in the chain, and adjacent solution have one variable the same, so the smallest positive solution is $(z,t)$ or $(t,z)$ for some value of $t$ (larger than $z$, in fact $t = z^3$). This has $\min(x,y)=z$ and the other positive solutions have bigger values of $\min(x,y)$.  For the same reasons, in a positive solution, $\max(x,y) \geq z^3$.  The inequalities are a complicated way of saying that $(x,y)=(z,z^3)$ is the smallest positive integer solution to the equation.
There was a careful analysis of the $ab+1$ problem here:
Seemingly invalid step in the proof of $\frac{a^2+b^2}{ab+1}$ is a perfect square?
Edit: 
Actually the reduction process, applied to any non integer solution with fixed $z$, should eventually reach a pair $(x,y)$ with $0 < x < z \leq y$, so this problem can work only as a Diophantine question and is not derivable from inequalities.  The minimal solution overall will have $xy < 0$ and both variables less than $z$ in absolute size.
