I am trying to prove that the equation $$3x^3 + 4y^3 +5z^3 \equiv 0 \pmod{p}$$ has a non-trivial solution for all primes $p$. I am sure that this is a standard exercise, and I have done the easy parts: treating $p=2, 3, 5$ as special cases (very simple), and then for $p\geq 7$, those for which $p \equiv 2 \pmod{3}$ is also straightforward, as everything is a cubic residue $\pmod{p}$, but I am having a mental block about the remaining cases where $p \equiv 1 \pmod{3}$ and only $(p-1)/3$ of the integers $\pmod{p}$ are cubic residues.
I was hoping to be able to show that the original equation has non-trivial solutions in $\mathbb{Q}_p$, and that this might be an easy first step towards the $p$-adic case.
Any pointers, or references to a proof (I am sure there must be some in the literature) would be most gratefully received.
I think Selmer's example is in some book I own but I cannot find it. It would be a natural footnote in any of my quadratic forms books, but there you go.
Here is some stuff from a book you may not be looking at, page 79, second edition of $p$-adic Numbers by Fernando Q. Gouvea. Related to your example is Problem 121, show your same conditions for $$ (x^2 -2) (x^2 - 17) (x^2 - 34) = 0, $$ which can be checked wit the rational roots theorem. Hmmm. Then $$ x^4 - 2 y^2 = 17. $$ He says non-existence of rational solutions is the hard part in this one. I think this one is accessible from stuff in Mordell's book Diophantine Equations.
The one I wanted to get to is how $x^2 + y^2 + z^3 = n$ has a solution in $\mathbb Z$ for every $n,$ both $n,z$ allowed to be negative when needed, but $$ x^2 + y^2 + z^9 \neq 216 p^3 $$ for positive prime $p \equiv 1 \pmod 4,$ see Integers of the form $a^2+b^2+c^3+d^3$ and MEEEEE