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I understand, on a layman's level, Fray's motivation to write an elliptic equation corresponding to an assumed solution to FLT. My question is, how technically is Frey's equation derived?

$1.$ FLT : Contradiction. There exist integers $A,B,C,n>2$ such that $A^n + B^n = C^n$

$2.$ Frey: $y^2 = x(x-A^n)(x+B^n)$

$3.$ Elliptic curve $y^2= x^3 + ax + b$

$4.$ The curve is not modular. => not elliptic => FLT proved.

I need is the algebra to go from 1. to 3. above.

Assume Numbers A,B,C such that A^n + B^n = C^n contradicting FLT.

$y^2 = x^3 + ax + b$ an elliptic curve in Weirstrass form. I am guessing A and B go unchanged into x(x-A^n)(x+B^n)

Is that correct ?

Expanding 2. I get a term in x^2 , there is none in 3. This is my problem.

What does a $2$ dimensional graph of the Frey curve look like ?

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    $\begingroup$ My understanding is that if $a^n+b^n=c^n$ then the Frey curve $y^2=x(x-a^n)(x+b^n)$ is not modular. I think you'd have to dig into the details of the proof of that statement in order to understand where the Frey curve comes from. As to where $c$ goes, just make a linear change of variables $x\mapsto x+a^n$ and the curve becomes $y^2=x(x+a^n)(x+c^n)$, and now you can worry about where the $b^n$ went. $\endgroup$ – Gerry Myerson Jun 24 '16 at 13:06
  • $\begingroup$ Or could we make this replacement: A = a^n/c^n, B = b^n/c^n ? $\endgroup$ – xatabay Jun 24 '16 at 14:01
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    $\begingroup$ No, that they are integers is crutial. It allows you to define the curve over the integers, and by extension, even make sense of all the conjecture you need to use. $\endgroup$ – max Jun 24 '16 at 21:32
  • $\begingroup$ I have edited question, no one is answering, can you look again at my question please. $\endgroup$ – xatabay Jun 30 '16 at 10:27
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    $\begingroup$ An elliptic curve over $\mathbb Q$ can be given in Weierstrass form as $y^2=x^3+ax+b$. But just because it's not given in Weierstrass form, doesn't stop it being an elliptic curve! Specifically, Frey's curve is an elliptic curve not in Weierstrass form. It will be isogenous to a Weierstrass curve, but that's not necessary. $\endgroup$ – Mathmo123 Jul 11 '16 at 22:55
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The algebra to go from 1) to 3) is given in Cremona Chapter III

For clarity here, I will go from 3) to 1) for the general case first.

Basically, $y^2=x^3-27c_4x-54c_6$ following Cremona's change of variables.

$c_4=b_2^2-24b_4$

$c_6=b_2^3+36b_2b_4-216b_6$

$b_2=a_1^2+4a_2$

$b_4=2a_4+a_1a_3$

$b_6=a_3^2+4a_6$

for the Weierstrass equation:

$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$

Next, we write the Frey equation:

$y^2=x(x-a^n)(x+b^n)$ which when expanded is

$y^2=x^3-a^px^2+b^px_2-a^pb^px$

Now you can easily see that for the Frey curve:

$a_1=0$

$a_3=0$

$a_2=b^p-a^p$

$a_4=-a^pb^p$

$a_6=0$

Plug these values in to get $c_4$ and $c_6$ for this Frey curve:

$c_4=16(a^{2p}+b^pa^p+b^{2p})$

$c_6=-32(a^p-b^p)(2a^{2p}-13b^pa^p+2b^{2p})$

Plug these into Cremona's equation and we get the rather unwieldy:

$y^2=x^3-432(a^{2p}+b^pa^p+b^{2p})x+1728(a^p-b^p)(2a^{2p}-13b^pa^p+2b^{2p})$

which is the original Frey curve in 1) in the form you asked for in 3).

Of course, you don't need the curve in this form to prove FLT. Ribet's level lowering of the conductor of the curve shows that the conductor is $2$ and there is no modular elliptic curve of level $2$. $11$ is the first level for a modular elliptic curve. The key is not the Weierstrass manipulation you were seeking. The key is that the discriminant of the original equation in 1) has an exponent due to FLT's signature $(p,p,p)$.

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