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.