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.

  • $\begingroup$ $x^4+y^4=z^4$ is not actually an elliptic curve, you have given part of a degree $2$ map from $C$ to some elliptic curve $E$, with $4$ branch points. $\endgroup$ – mercio Jul 2 '15 at 12:16
  • $\begingroup$ Maybe I did not make it clear. I am not saying it is an elliptic curve. All I am saying is that we can relate it to an elliptic curve using a clever substitution and I am seeking for the motivation behind the substitution. $\endgroup$ – Zilin J. Jul 2 '15 at 13:34
  • $\begingroup$ The statement of the post did not say $x^4+y^4=z^4$ is an elliptic curve. $\endgroup$ – Piquito Jul 4 '15 at 3:29
  • $\begingroup$ Are there similar substitutions for any Fermat curve, or only $n=3,4$? $\endgroup$ – Kieren MacMillan Mar 9 '16 at 16:28

@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.

  • $\begingroup$ Thanks. Looking forward to your upcoming post. $\endgroup$ – Zilin J. Jul 4 '15 at 17:39

After reading the following chapters of the book, here are some explanations that can help clarify why one can always transform those two Fermat equations to elliptic curves. Indeed, one can always transform

  1. a cubic equation in $P_K^2$ (2 dimensional projective space over $K$, $\operatorname{char}K \neq 2,3$) or
  2. a quartic equation of the form $v^2 = au^4+bu^3+cu^2+du+e$ ($\operatorname{char}K\neq 2$)

into an elliptic curve (in Weierstrass form $y^2=x^3+Ax+B$) provided a point satisfying the equation.

For the first case, one can refer to Elliptic curve handbook by I. Connell. Here I present another way to deduce the elliptic curve related to the Fermat equation $x^3+y^3+z^3=0$.

Set $x/z = u+v$, $y/z=u-v$. We have $(u+v)^3+(u-v)^3+1=0$, and so $2u^3+6uv^2+1=0$, hence $6(v/u)^2 = (-1/u)^3-2$. Let $x_1 = -6/u$, $y_1 = 36v/u$. We get $y_1^2 = x_1^3 - 432$.

For the second case, suppose $(p,q)$ satisfies $v^2=au^4+bu^3+cu^2+du+e$. By translation $u\mapsto u+p$, we may assume $p=0$. We have two cases:

  1. if $q=0$, then $e=0$ and so $(v/u^2)^2 = d(1/u)^3+c(1/u)^2+b(1/u)+a$.
  2. if $q\neq 0$, set $$x = \frac{2q(v+q)+du}{u^2},\quad y=\frac{4q^2(v+q)+2q(du+cu^2)-(d^2u^2/2q)}{u^3},$$ $$a_1 = d/q, a_2 = c-(d^2/4q^2), a_3 = 2qb, a_4 = -4q^2a, a_6 = a_2a_4.$$ By straightforward calculation, $$y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6.$$ One can deduce Weierstrass form by completing squares for $y$ and translating $x$.

Although I still have no idea how one would stumble upon the substitution above.


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