Mathematical Reflections 2017, Issue 2 - Problem O406

Problem:
Solve in prime numbers the equation $$x^3-y^3-z^3+w^3+\dfrac{yz}{2}(2xw+1)^2=2017.$$

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

Solution:
Since $x^3-y^3-z^3+w^3$ and $2017$ are integers, then also $\dfrac{yz}{2}(2xw+1)^2$ must be an integer. Since $(2xw+1)^2$ is odd, then $2 \ | \ yz$. Since $y$ and $z$ are primes, then $y=2$ or $z=2$. By symmetry, we can assume $y=2$. We have
$$x^3-z^3+w^3+z(2xw+1)^2=2025.$$ Since $2025$ is odd, then the LHS must be odd and it's easy to see that $x,z,w$ cannot be all odd. Moreover, the given equation can be written as $$x^3+w^3+z(2xw-z+1)(2xw+z+1)=2025.$$ Since $z(2xw-z+1)(2xw+z+1)$ is always even, then $x^3+w^3$ must be odd, which implies that at least one between $x$ and $w$ is even, i.e. $x=2$ or $w=2$. By symmetry, we can assume $x=2$. We have $$w^3+z(4w-z+1)(4w+z+1)=2017.$$ If $4w-z+1 \geq 0$, then $w^3 \leq 2017$, so $w \leq 11$. An easy check gives $w=7$ and $z=2$.
So, $(x,y,z,w) \in \{(2,2,2,7),(7,2,2,2)\}$.
The case $4w-z+1 < 0$ remains open.