solve the diophantine equation: $x^3-3xy^2=z^3$

Question

Let $ x,y,z$ be 3 integers greater than 1,if $x$ and $y$ are relatively prime, solve the diophantine equation: $x^3-3xy^2=z^3$.

Answer 1

$$x^3-3xy^2=z^3 \Rightarrow x^3-z^3=3xy^2 \Rightarrow 3 \mid x^3-z^3 \Rightarrow x^3 \equiv z^3 \pmod 3$$

From Fermat's theorem:

$$x^3 \equiv x \pmod 3 \\ z^3 \equiv z \pmod 3$$

So,we have:

$$x \equiv \ z \pmod 3 \Rightarrow 3 \mid x-z \Rightarrow x=z+3k,k \in \mathbb{Z}$$

Replace at $x^3-3xy^2=z^3$ and take into consideration the fact that $(x,y)=1$.