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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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    $\begingroup$ 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 $\endgroup$ – Will Jagy Aug 4 '12 at 19:51
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    $\begingroup$ Thanks, @WillJagy - that was just the clue I needed to find it (i.e. the name Selmer!) $\endgroup$ – 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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  • $\begingroup$ 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! $\endgroup$ – Old John Aug 5 '12 at 8:20
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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.

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  • $\begingroup$ This looks very interesting but I don't understand even the first step. Why is J equals to this nested sum? $\endgroup$ – i9Fn Mar 4 '17 at 15:23
  • $\begingroup$ @i9Fn: The innermost sum is equal to $p$ if $p \mid 3x^{3}+4y^{3}+5z^{3}$ and $0$ otherwise. $\endgroup$ – José Hdz. Stgo. Mar 5 '17 at 22:44
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    $\begingroup$ Jose, I think you would like Arithmetic of Finite Fields by Charles Small, published 1991. I made copies of the chapter on quadratic forms. i would guess he also does other homogeneous forms; the next chapter is on Gauss sums and the like. $\endgroup$ – Will Jagy Mar 6 '17 at 3:44
  • $\begingroup$ 1. Where I can find proof for your estimate? 2. Why is it $1.5$ and not $0.5$? $\endgroup$ – i9Fn Mar 6 '17 at 10:20
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    $\begingroup$ 1) It follows from a well-known bound that you can find on page 23 of Korobov's "Exponential sums and their applications": if it is not available in your local library, I would suggest that you Google "bounds on cubic Gauss sums". 2) Each of the $p-1$ terms of that sum is bounded by $(2\sqrt{p})^{3}$... $\endgroup$ – José Hdz. Stgo. Mar 6 '17 at 17:16

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