# Proof of no rational point on Selmer's Curve 3x^3+4y^3+5z^3=0

The projective curve $3x^3+4y^3+5z^3=0$ is often cited as an example (given by Selmer) of a failure of the Hasse Principle: the equation has solutions in any completion of the rationals $\mathbb Q$, but not in $\mathbb Q$ itself.

I don't think I've ever seen a proof of the latter claim — is someone able to provide an outline? What are the necessary tools?

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My friend has written an introduction to algebraic number theory before, which contains a short proof of this statement, but I didn't check its validity.

[Edit: Update the link of the document] http://www.2shared.com/document/2d6M7kNU/Introduction_to_Algebraic_Numb.html p.41 of the document, or p.45 of the pdf.

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This is certainly a lot more elementary than the other methods mentioned, although it (therefore) does not present a clear example of what the general obstruction is (for the failure of Hasse). Still, thanks a lot - at least this is a proof I can easily follow :-) –  Alon Amit Oct 27 '09 at 7:26
For what it's worth, I once worked out a completely elementary proof that the equation has p-adic solutions for all p and put it on an UG example sheet here: www2.imperial.ac.uk/~buzzard/maths/teaching/04Lent/M4P32/… –  Kevin Buzzard Nov 20 '09 at 23:27
@Ho Chung Siu, do you happen to have a copy of that paper anywhere accessible? –  Alon Amit Apr 18 '11 at 23:26
Hi, the link is updated. –  Ho Chung Siu Apr 19 '11 at 23:44

This problem is in Cassels' book "Local Fields" and I wrote up a solution once along those lines, for an algebraic number theory class. See

but I should advise that it comes out seeming pretty tedious. Solutions that involve elliptic curves are more conceptual. Others have already provided pointers to references for that approach.

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I think I saw a proof of that in Cassel's "Diophantine Equations with special reference to elliptic curves" and in some surveys by Mazur in the Bull. AMS (perhaps this, but I have in the moment no time to look).

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Interesting. The obstruction, as presented by Mazur, is that the curve 60x^3+y^3+z^3=0 does have a rational point, and is a "companion" (Q-twist of) the Selmer curve. I'll need to dig a lot more to understand this. Thanks! –  Alon Amit Oct 27 '09 at 7:22

The "standard" technique for killing the Hasse priniciple for elliptic curves is to show that the Tate-Shafarevich group has a copy of (Z/mZ)^2 for some m - see chapter X in Silverman's the arithmetic of Eliptic curves, both for the theory and examples. All the examples which Silverman presents ar with m = 2. Selmers example requires m = 3, which requires (much) more computations. Poonen has an example on his web page of a family of elliptic curves violating the Hasse principle, and containing Selmers example, but you'd have to dive through a labirinth of references.

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(Hello there :-) ) That's pretty heavy machinery for this humble reader - but an interesting view of the obstruction. Thanks a lot! –  Alon Amit Oct 27 '09 at 7:28
(Hello indeed :) ) The advantage this approach has is certainly not simplicity, it is rather that it can be - and is - mechanised. Google up Hasse Tate Shafarevich and Magma. –  David Lehavi Oct 27 '09 at 8:26

There's a proof in Cassels' little blue book on elliptic curves which the OP might find more to his taste than some others mentioned here.

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