3

The following is quoted from "Diophantine Equations (Pure & Applied Mathematics)" by L.J Mordell, p.77

Theorem 2, If the quartic curve $$y^2=ax^4+bx^3+cx^2+dx+e$$ has a rational point, it is equivalent to the cubic curve $$Y^2=4X^3-g_2X-g_3$$ We may assume e is a perfect square. Then on writing $x=\dfrac{1}{X},y=\dfrac{Y}{X^2}$, we may assume that a is a perfect square.

Why we are able to make the assumptions that a and e are perfect squares?

Ameet Sharma
  • 2,957

1 Answers1

2

Let $p(x) = ax^4+bx^3+cx^2+dx+e$.

If the curve $y^2 = p(x)$ has a rational solution $(X,Y)$ then letting $x' = x-X$ you obtain $y^2 = q(x')$ where $q(x') = p(x'+X)$.

What's important is that $q(0) = p(X) = Y^2$ which is a square, so after the translation by $X$ your curve has the form $y^2 = ax'^4 + b'x'^3+ c'x'^2 + d'x' + Y^2$ as you wanted.

mercio
  • 50,180