Mathematical Reflections 2016, Issue 6 - Problem O391

Problem:
Find all 4-tuples $(x,y,z,w)$ of positive integers such that $$(xy)^3+(yz)^3+(zw)^3-252yz=2016.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation can be written as $$(xy)^3+(zw)^3+yz(y^2z^2-252)=2016.$$ Observe that it must be $yz(y^2z^2-252) \leq 2016$, so $yz \leq 18$. It follows that $$\begin{array}{rcl} (xy)^3+(zw)^3 & \in & \{2267,2512,2745,2960,3151,3312,3437,3520,3555,\\ & & 3536,3457,3312,3095,2800,2421,1952,1387,720\}. \end{array}$$
A case by case analysis gives $(xy)^3+(zw)^3 \in \{2745,2960\}$. If $(xy)^3+(zw)^3=2745$, we get $xy=1,zw=14,yz=3$ or $xy=14,zw=1,yz=3$, which gives no solutions. If $(xy)^3+(zw)^3=2960$, then $xy=6,zw=14,yz=4$ or $xy=14,zw=6,yz=4$. We get $(x,y,z,w) \in \{(3,2,2,7),(7,2,2,3)\}$.