I am trying to prove that the equation $3x^3+4y^3+5z^3 \equiv 0 \pmod{p}$ has non-trivial solutions for all primes $p$. I divide it into 3 cases: $p \equiv 0,1,2 \pmod{3}$. The cases $p \equiv 0,2 \pmod{3}$ are easy. The problem is the case $p \equiv 1 \pmod{3}$, can anyone give a help here? thanks!
the method i am using is as follows: $G=\mathbb{F}_p^*=\langle g\rangle$ is cyclic, therefore $G/H= \{H,gH,g^2H\}$, where $H=\{x^3|x \in G\}$. i.e. Every element in $\mathbb{Z}/p\mathbb{Z}$ is either of the form $x^3,gx^3,g^2x^3$.
Look at the first page of these notes by Keith Conrad on Selmer's example.