# Polynomial equations parametrized by binary forms

Consider the equation $$\displaystyle Ax^p + By^q = Cz^r, A,B,C \in \mathbb{Z}, \gcd(x,y,z) = 1, p,q,r \geq 2.$$

When $$p^{-1} + q^{-1} + r^{-1} > 1$$, the above equation is called spherical and satisfies an identity of the form

$$\displaystyle x = f(u,v), y = g(u,v), z = h(u,v)$$

where $$f,g,h$$ are binary forms with complex coefficients. Beukers showed that all solutions to the equation are parametrized by a finite family of such binary forms with integer coefficients.

We now consider the equation

$$\displaystyle x^r + y^2 = Cz^2, \gcd(x,y,z) = 1, C \in \mathbb{Z}, r \geq 3.$$

This equation is spherical, and if it has one integer solution then it will have infinitely many due to the existence of a polynomial parametrization (by binary forms). Suppose that it does indeed have a solution and we have a parametrization $$x = f, y = g, z = h$$ as above. What can we say about the discriminants of $$f,g,h$$?

• I get it, binary forms but not quadratic... $$\left(x^3 - 3 x y^2 \right)^2 + \left(3x^2 y - y^3 \right)^2 = \left(x^2 + y^2 \right)^3$$ – Will Jagy Jun 14 at 3:16
• so, $C \neq k^2, \;$ $f^2 - C g^2 = \left( x^2 - C y^2 \right)^r$ using Gauss composition – Will Jagy Jun 14 at 3:42
• @WillJagy I don't think it's quite that simple, since if you re-arrange the equation you put down you don't recover the original equation... you get $h^r - f^2 = -Cg^2$, which is easier since for even $r$ the LHS is reducible. – Stanley Yao Xiao Jun 14 at 14:38