$y^2 = x^3 + mx + c$ An elliptic curve in the form defined in Wikipedia
$y^2 = x(x-A)(x+B) = x^3 +(B-A)x^2 + ABx$ Frey's curve
has no term in $x^2$, but $2$. does because from Fermat, $A=a^n$ not equal to $B=b^n$
Question: Is there a simple way to transform Frey's curve with given $A,B$ into form $1$.
(Has it got anything to do with Weierstrass ?)