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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.

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This is Selmer's example of the failure of Hasse-Minkowski for cubic forms. Let me try to find something online. en.wikipedia.org/wiki/Hasse_principle –  Will Jagy Aug 4 '12 at 19:51
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Thanks, @WillJagy - that was just the clue I needed to find it (i.e. the name Selmer!) –  Old John Aug 4 '12 at 19:54

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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

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Many thanks for these pointers, I had done the exercise about $(x^2 -2) (x^2 - 17) (x^2 - 34) = 0$ from another book (maybe Borevich, Shafarevich), using quadratic residues. The help you gave earlier by giving me the search term "Selmer" has enabled me to find some other references, and I think I have the cubic residue part sorted out in my mind now - just need to get the details written down. A fascinating area! –  Old John Aug 5 '12 at 8:20

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