You are currently browsing the tag archive for the ‘elliptic curves’ tag.

I have been reading “Number Theory I: Fermat’s Dream” by Kazuya Kato, Nobushige Kurokawa, and Takeshi Saito. In Chapter 1: Rational Points on Elliptic Curves, the authors state two propositions:

Proposition 1.1. There is no integral solution $(x,y,z)$ to $x^4 + y^4 = z^4$ satisfying $xyz \neq 0.$

and

Proposition 1.2. The only rational solutions to $y^2=x^3-x$ are
$(x,y)=(0,0)$ and $(\pm1,0).$

They then show how 1.2 implies 1.1:

Suppose $(x,y,z)$ is an integral solution to $x^4 + y^4 = z^4$ satisfying $xyz \neq 0.$ We rewrite the equation as

$x^4 = z^4 - y^4$

We then multiply each side by $\dfrac{z^2}{y^6}$ to get

$\dfrac{x^4z^2}{y^6} = \dfrac{z^6}{y^6} - \dfrac{z^2}{y^2}$
$\left(\dfrac{x^2z}{y^3}\right)^2=\left(\dfrac{z^2}{y^2}\right)^3-\left(\dfrac{z^2}{y^2}\right).$

Setting $Y=\dfrac{x^2z}{y^3},\, X=\dfrac{z^2}{y^2}$, we have
$Y^2=X^3-X.$

Note that X,Y are rational numbers. The only rational points of this elliptic curve are $(0,0), (\pm1,0)$. In either case, $0=Y=\dfrac{x^2z}{y^3}$ implies either $x=0$ or $z=0$, contradicting our assumption. So there are only trivial integral solutions $(x,y,z)$ to $x^4 + y^4 = z^4$.

We can extend this argument to prove that there are no non-trivial integral solutions to $x^n + y^n = z^n$, for $n\equiv 0\, ({\rm mod}\, 4)$.