Mathematical Reflections 2014, Issue 5 - Problem J316

Problem:
Solve in prime numbers the equation $$x^3+y^3+z^3+u^3+v^3+w^3=53353.$$

Proposed by Titu Andreescu.

Solution:
Assume $x \leq y \leq z \leq u \leq v \leq w$. Since $53353$ is odd, then $x=2$. We have
$$y^3+z^3+u^3+v^3+w^3=53345.$$ Moreover, $5w^3 \geq 53345 \geq w^3$, so $w \in \{23,29,31,37\}$. A case by case analysis gives $(x,y,z,u,v,w)=(2,3,5,7,13,37)$.