[Algebra]
Problem: Compute the ordered triple $(x, y, z)$ representing the farthest lattice point from the origin that satisfies $x y-z^{2}=y^{2} z-x=14$.

Solution: First, eliminate $x: y\left(y^{2} z-x\right)+\left(x y-z^{2}\right)=14(y+1) \Rightarrow z^{2}-y^{3} z+14(y+1)=0$. Viewed as a quadratic in $z$, this equation implies $z=\frac{y^{3} \pm \sqrt{y^{6}-56(y+1)}}{2}$. In order for $z$ to be an integer, the discriminant must be a perfect square. Because $y^{6}=\left(y^{3}\right)^{2}$ and $\left(y^{3}-1\right)^{2}=y^{6}-2 y^{3}+1$, it follows that $|56(y+1)| \geq 2\left|y^{3}\right|-1$. This inequality only holds for $|y| \leq 5$. Within that range, the only values of $y$ for which $y^{6}-56 y-56$ is a perfect square are -1 and -3 . If $y=-1$, then $z=-1$ or $z=0$. If $y=-3$, then $z=1$ or $z=-28$. After solving for the respective values of $x$ in the various cases, the four lattice points satisfying the system are $(-15,-1,-1),(-14,-1,0),(-5,-3,1)$, and $(-266,-3,-28)$. The farthest solution point from the origin is therefore $(-\mathbf{2 6 6}, \mathbf{- 3 , - 2 8})$.

Final answer: $(-266,-3,-28)$