MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Fermat's proof of FLT(4) is an example of infinite descent as is Euler's (or whoever you attribute it to's) proof of FLT(3). There are similar proofs to Fermat's for Diophantine equations like $x^4 + y^4 = 2z^2$.

I have unsuccessfully tried to view these proofs in terms of group homomorphisms on conics and elliptic curves but it is not at all clear whether this is possible.

Can we reinterpret these infinite descent proofs geometrically, in terms of curves?

share|cite|improve this question
In the case of FLT(4), I was thinking of considering the conic $X^2 + Y^2 - 1$ and there is a subset of square points on it, "descent" would prove that the only square points are the trivial solutions. – Quanta May 14 '11 at 1:26

(That picture in your avatar is Weil, right? You should start by reading Weil's Number Theory: an approach through history).

FLT(3) is the assertion that the curve $x^3+y^3=1$ has three rational points (including the point at infinity). The standard process of putting an elliptic curve in Weierstrass form shows that this curve is $y^2=x^3-432$ if I remember correctly. Now use descent on this elliptic curve (maybe an isogeny of degree 3) to show that the Mordell-Weil group is finite of order three (see, e.g. Silverman for the general theory). This may be Euler's proof, maybe it's discussed in Weil's book.

FLT(4) is weaker than the statement that the equation $x^4+y^4=z^2$ has only the obvious solutions. Again, this becomes the problem of finding the rational points on $y^2=x^4+1$ which is again an elliptic curve and Fermat's proof is a 2-descent showing that the Mordell-Weil group is finite of order four. I'm pretty sure this is in Weil's book.

I have no idea what conics have to do with any of this.

share|cite|improve this answer
I think the photo is of Schrödinger. – S. Carnahan May 13 '11 at 22:46
According to Elkies , Euler's proof for FLT(3) indeed amounts to a $3$-descent between elliptic curves of conductor 27. – François Brunault May 13 '11 at 23:10
Scott seems to be right about the photo. Incidentally, google pictures seems to be confused between Andre Weil, Andrew Weil (the alternative doctor) and Andrew Wiles. – Felipe Voloch May 13 '11 at 23:12
(For FLT(4) see and for FLT(7) see ) – François Brunault May 13 '11 at 23:14
Conics come in when you construct the descent map for elliptic curves of the type $y^2 = x^4 + ax^2 + b$: one approach is looking at a parametrization of the conic $y^2 = z^2 + az + b$. Most elementary proofs of FLT4, for example, use the parametrization of the unit circle (Pythagorean triples). – Franz Lemmermeyer May 14 '11 at 12:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.