# 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?

## 5 Answers

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: updated 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.

• 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
• The link in the answer is now broken. – KConrad Apr 16 '11 at 16:32
• @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.

• Ten years too late, but I thought I'd point it out: the section headings in this paper are flipped! You wrote "No local solutions" and then "Global solutions". Regardless, thank you for this, and your numerous other awesome papers. – Alon Amit Apr 13 '20 at 20:26
• @AlonAmit fixed. You can email me directly the next time you find a typo. – KConrad Apr 14 '20 at 2:39

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).

• 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
• "This" is now dead, but it appears to have been a link to Mazur - On the passage from local to global in number theory, with the relevant result beginning on p. 22. – LSpice May 14 '20 at 22:26

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.

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.

• (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