 # We have F ( x , y ) &#x2208;<!-- ∈ --> <mrow class="MJX-TeXAtom-ORD"> babajijwerz 2022-05-30 Answered
We have $F\left(x,y\right)\in \mathbb{Z}\left[X,Y\right]$ a positive definite binary form of degree $\ge 3$. 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 $F\left(x,y\right)=m$ 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 $F\left(X,1\right)$ are in $\mathbb{C}\mathrm{\setminus }\mathbb{R}$. 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!
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I'd say you mean a degree $d$ homogeneous polynomial $f\in \mathbb{Z}\left[X,Y\right]$ having no real zeros other than (0,0).
With $c\left({X}^{d}\right),c\left({Y}^{d}\right)$ the coefficients of ${X}^{d},{Y}^{d}$ and
$A=min\left(|c\left({X}^{d}\right)|,|c\left({Y}^{d}\right)|,\underset{t\in \mathbb{R}}{inf}|f\left(1,t\right)|,\underset{t\in \mathbb{R}}{inf}|f\left(t,1\right)|\right)$you get
1. $|f\left(x,0\right)|\ge A|x{|}^{d},|f\left(0,y\right)|\ge A|y{|}^{d}$
2. if $y\ne 0$, $|f\left(x,y\right)|=|y{|}^{d}|f\left(x/y,1\right)|\ge A|y{|}^{d}$,
3. if $x\ne 0$, $|f\left(x,y\right)|=|x{|}^{d}f\left(1,y/x\right)|\ge A|x{|}^{d}$ ie.
$|f\left(x,y\right)|\ge max\left(A|x{|}^{d},A|y{|}^{d}\right)$
which implies that $f\left(X,Y\right)-m$ has finitely many zeros $\in {\mathbb{Z}}^{2}$.