6
$\begingroup$

I have read in many places that the Frey curve (if it existed) $y^2=x(x-A)(x+B)$ (or equivalently, $y^2=x(x-A)(x-C)$ corresponds to the solutions of $a^n+b^n=c^n$, where $A=a^n/c^n$ and $B=b^n/c^n$.

However, the connection is not very clear to me. Can someone either explain it or point me to a source that would spell this out step by step?

Thank you!

Edit: Just to clarify, I am asking about the correspondence between the Frey curve and the solutions to Fermat's Last Theorem. I take it there is no simple relation between the two?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ If FLT has a solution, construct the corresponding Frey curve. Wiles proved that every elliptic curve over $\mathbb{Q}$ is modular. Ribet showed that this curve is then modular of level 2, but it's known that there is no nonzero cusp form of level 2, contradiction. Thus FLT cannot have solution. $\endgroup$ – user27126 Oct 31 '13 at 20:12
  • $\begingroup$ I believe $A$ should simply be $a^n$ and likewise $B = b^n$ (see wikipedia). But with this, the correspondence is given in your question. If you give me three integers $a,b,c$ with $a^n + b^n = c^n$, then the corresponding Frey curve is $y^2 = x(x-A)(x+B)$. Likewise, if you have a Frey curve, then by definition there are integers $a$, $b$, $c$, and $n$ with $a^n = A$, $b^n = B$, and $a^n + b^n = c^n$. $\endgroup$ – RghtHndSd Oct 31 '13 at 20:46
3
$\begingroup$

I think the proof is very advanced and done by Ribet. If I have understood correctly, the connection is given in http://math.berkeley.edu/~ribet/Articles/invent_100.pdf

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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