We have F ( x , y ) &#x2208;<!-- ∈ --> <mrow class="MJX-TeXAtom-ORD">

babajijwerz

babajijwerz

Answered question

2022-05-30

We have F ( x , y ) Z [ X , Y ] a positive definite binary form of degree 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 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 C 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!

Answer & Explanation

Gloletheods6g

Gloletheods6g

Beginner2022-05-31Added 6 answers

I'd say you mean a degree d homogeneous polynomial f 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 R | f ( 1 , t ) | , inf t R | f ( t , 1 ) | )you get
1. | f ( x , 0 ) | A | x | d , | f ( 0 , y ) | A | y | d
2. if y 0, | f ( x , y ) | = | y | d | f ( x / y , 1 ) | A | y | d ,
3. if x 0, | f ( x , y ) | = | x | d f ( 1 , y / x ) | A | x | d ie.
| f ( x , y ) | max ( A | x | d , A | y | d )
which implies that f ( X , Y ) m has finitely many zeros Z 2 .

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Linear algebra

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?