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I am trying to prove that the congruence $3x^3+4y^3+5z^3 \equiv 0 \pmod{p}$ is solvable for all primes p. I proved it using primitive root, but my professor in number theory told me that it can be more directly done using the hasse-weil theorem in the theory of elliptic curves, but i cant do it. Would someone kindly show me how to make use of the hasse-weil theorem, please? Thank you in advance.

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possibly related – Old John Nov 10 '13 at 18:24
You did consider $x=y=z=0$ I presume? – WimC Nov 10 '13 at 19:23
@WimC, no, this is nontrivial solutions. Famous example of Selmer on failure of Hasse-Minkowski once we leave quadratic forms. – Will Jagy Nov 10 '13 at 20:26… – Will Jagy Nov 10 '13 at 20:31
apparently in this:… – Will Jagy Nov 10 '13 at 20:48

This can be done in $4$ steps:

  1. Prove it directly for $p=2,3$, i.e., find a non-trivial solution modulo $2$ and one modulo $3$.

  2. Prove that $C: 3x^3+4y^3+5z^3=0$ is a non-singular curve over $\mathbb{F}_p$, where $p>3$ is a prime.

  3. Show the following: if $C:F(x,y,z)=0$ is a non-singular curve given by a homogeneous polynomial of degree $d\geq 1$, then the genus of $C$ is given by $(d-1)(d-2)/2$. In particular, in our case, $g=1$. (This is Exercise 2.7 in Chapter 2 of Silverman's "The Arithmetic of Elliptic Curves", for instance.)

  4. Use the Hasse-Weil bounds.

Note: you cannot use the fact that $C/\mathbb{F}_p$ is an elliptic curve before you find a $\mathbb{F}_p$-rational point on $C$!

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