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We have $F(x,y)\in\mathbb{Z}[X,Y]$ a positive definite binary form of degree $\geq 3$. I have to prove, without using lower bounds on linear forms in logarithms (we were working with Baker's theorems), that for each positive integer $m$ the equation $F(x,y)=m$ has only finitely many solutions in $x,y\in\mathbb{Z}$. I also have to describe a method to find these.

We know that the coefficient of $X^d$ of $F$ is positive, and that all zeros of $F(X,1)$ are in $\mathbb{C}\backslash \mathbb{R}$. We also know the discriminant of $F$ is negative.

I have trouble finding this out by myself. I also looked up some number theory books, but most literature is about binary quadratic forms, not the binary form I have to prove this for. Hope someone can help me!

jbuser430
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1 Answers1

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I'd say you mean a degree $d$ homogeneous polynomial $f\in \Bbb{Z}[X,Y]$ having no real zeros other than $(0,0)$.

With $c(X^d),c(Y^d)$ the coefficients of $X^d,Y^d$ and $$A= \min(|c(X^d)|,|c(Y^d)|,\inf_{t\in \Bbb{R}}|f(1,t)|,\inf_{t\in \Bbb{R}}|f(t,1)|)$$ you get

  • $|f(x,0)|\ge A |x|^d, |f(0,y)| \ge A|y|^d$

  • if $y\ne 0$, $ |f(x,y)|= |y|^d |f(x/y,1)| \ge A|y|^d$,

  • if $x\ne 0$, $|f(x,y)|=|x|^d f(1,y/x)| \ge A|x|^d $ ie. $$|f(x,y)|\ge \max(A|x|^d,A|y|^d)$$ which implies that $f(X,Y)-m$ has finitely many zeros $\in \Bbb{Z}^2$.

reuns
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