<s>[INST] Substitutions that transform Fermat Equations to Elliptic Curves I was reading Chapter 1 of Elliptic Curves - Number Theory and Cryptography by Lawrence C Washington. He was considering Fermat equations $$a^4+b^4=c^4\text{ and }a^3+b^3=c^3.$$ For the 1st equation, let $$x = 2\frac{b^2+c^2}{a^2},\quad y=\frac{4b(b^2+c^2)}{a^3};$$ and for the 2nd equation, let $$x=\frac{12c}{a+b},\quad y=\frac{36(a-b)}{a+b}.$$ Straightforward calculation shows that under these substitutions we transform Fermat equations to elliptic curves: $$y^2=x^3-4x\text{ and }y^2=x^3-432.$$ Before I looked at the substitutions, I tried and failed to find the solutions myself. I am wondering if there is a natural way to deduce those substitutions and the motivation behind.
 [/INST] @Zilin J.:I have a method that sends all binomial addition $a+b=c$ (so, for example, $a^4+b^4=c^4$) to an elliptical curve, noted $V_A$, of the shape $$X^3+Y^3=AZ^3$$, where $A$ is a cube-free number, and this curve is birational equivalent to a Wierstrass one, the curve $$y^2z=x^3-432A^2z^2$$ (one can obviously work with affine curve making $z=1$), via $$(X,Y)\to ({\frac {12A}{X+Y}, \frac {36A(X-Y)}{X+Y}})$$. We have $$a+b=c\iff [9ab^2+(b-a)^3]^3+[9a^2b-(b-a)^3]^3=abc[3(a^2+ab+b^2]^3 (*)$$  where $A$ can be taken as the cube-free part of $abc$.
This equivalence can be verified straightforward way, simplifying the right side (a more theoretical way is as follow: 
$a+b=c\iff a^2b+ab^2=abc\iff P=(\sqrt[3]{a^2b},\sqrt[3]{ab^2},1)\in V_{abc}$; 
one can verify that the irrational point $P\in V_{abc}$ becomes rational and gives (*) via the isogeny $P\to3P$).
I will post about this topic with more information soon.
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