Problem statement: Find all integer solutions of equation $5x^3+11y^3+13z^3=0$.
My attempt: I tried to reason that if there's a solution, then, since we have odd powers of variables we have 4 different scenarios to explore:
1) $5x^3 = 11y^3 + 13z^3$
2) $11y^3 = 5x^3 + 13z^3$
3) $13z^3 = 5x^3 + 11y^3$
4) $13z^3 = 5x^3 = 11y^3$. Since there are constant numbers $13$, $5$, and $11$ before variables, the only integer that would make all three equal would be $\boxed{0}$
While my road of thought is leading me to something, I think that I am doing something in a little bit too complicated way.
A few hints:
This question is from a chapter about invariance.
The problem is asked just after $x^3 - 3y^3 - 9z^3 = 0$, which is solved when you show that, if $(x, y, z)$ is a solution, then $(\frac{x}{3^n}, \frac{y}{3^n}, \frac{z}{3^n})$, where $n$ is any natural number, is also a solution. So, it seems that this equation should be solved in a similar manner.
It should not be very complicated - it is just beginning of a high school level book.
Look at cubes modulo $13$ and show that $x$ or $y$ is divisible by $13$, then follow the proof for the equation $x^3 - 3y^3 - 9z^3 = 0$.