Local solutions of a Diophantine equation 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.
 A: I would like to tell you about my approach to this problem; clearly enough, it can easily be adapted to yield an even more general result.
Let $p \equiv 1 \, \, (\mathrm{mod} \, \, 3)$ be a prime number and let us denote with $\mathbf{J}$ the number of solutions of the congruence $3x^{3}+4y^{3}+5y^{3} \equiv 0 \, \, (\mathrm{mod} \, \, p)$. Then , by basic properties of the complex exponential function it follows that
\begin{eqnarray*}
\mathbf{J} &=& \frac{1}{p} \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \sum_{z=0}^{p-1} \sum_{\lambda=0}^{p-1} e^{2\pi i \frac{3x^{3}+4y^{3}+5z^{3}}{p} \lambda} \\
&=& \frac{1}{p} \sum_{\lambda=0}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right)
\left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right)\\
&=& p^{2} + \frac{1}{p} \sum_{\lambda=1}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right)
\left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right).
\end{eqnarray*}
Appealing to the well-known estimate $ \left| \sum_{x=0}^{p-1} e^{2\pi i \frac{\omega}{p}x^{3}}\right| \leq 2 \sqrt{p}$ (which is valid for every $\omega$ coprime with $p$), we obtain that
$$ \mathbf{J} > p^{2}-8p^{1.5}.$$
Since $p^{2}-8p^{1.5}>1$ for every $p \geq 65$, the problem has been reduced to verifying that the given congruence has a non-trivial solution for every $p \in E:=\{7, 13, 19, 31, 37, 43, 61\}$. I list below a non-trivial solution for the corresponding congruence for every $p \in E$ (even though at first sight they may strike you as having been chosen without rhyme or reason, they all were obtained by resorting to an idea of the Princeps Mathematicorum):
\begin{eqnarray*}
p&=&07: \quad  (x=3, y=6, z=0)\\
p&=&13: \quad (x=1, y=4, z=5)\\
p&=&19: \quad (x=13,y=0, z=3)\\
p&=&31: \quad (x=3, y=2, z=3)\\
p&=&37: \quad (x=1, y=4, z=0)\\
p&=&43: \quad (x=25,y=0, z=42)\\
p&=&61: \quad (x=46,y=32, z=41)\\
\end{eqnarray*}
QED.
A: 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
