Can anyone give an overview showing how the elliptic curve $E (a, b) := y^2=x(x-a)(x-b)$ is associated with the solution of $a^n+b^n=c^n$?
In 1969 Hellegouarch performed the elliptic curves $E (a, b)$, which were later named after Gerhard Frey and were constructed from the solutions of the Fermat equation, he reported on the curves in an article in Acta Arithmetica. I have found the article but it is not in English.
I was looking for a document (lecture note, book, blog) that elaborately explains how the elliptic curve $E (a, b) := y^2=x(x-a)(x-b)$ is associated with the solution of $a^n+b^n=c^n$ and asked it on mathoveflow, I got some documents but I could not understand the whole thing. So, I am asking for help of this forum.
What I got so far:
Let $y^2=x^3+ax^2+bx+c$, $\Delta=(ab)^2+4a^3c-4b^3+18abc-27c^3.$
If $c=0$, then $y^2=x^3+ax^2+bx$, $\Delta=(ab)^2-4b^3.$
Now the Frey's curve is , $y^2=x(x-A)(x+B)=(x^2-Ax)(x+B)=x^3-Ax^2+Bx^2-ABx.$
Let, $a=(B-A), b=-AB,$ then $\Delta= \{(B-A)(-AB)\}^2-4(-AB)^3= \{(B-A)^2-4(-AB)\}(-AB)^2=(A^2+B^2-2AB+4Ab)(AB)^2=(A+B)^2(AB)^2.$
Now, let, $A=a^q, B=b^q,$ then, $\Delta=(a^q+b^q)^2(ab)^{2q},$ if $a^q+b^q=c^q$ for $q>2$ (this is Fermat's last theorem) then $\Delta=(c^q)^2(ab)^{2q}=(abc)^{2q}.$
Now whats wrong with $\Delta=(abc)^{2q} ?$