0
$\begingroup$

Major rewrite due to comments.

Let $f(x,y) \in \mathbb{Q}[x,y]$ and $f$ depends on both $x,y$.

Q1 Is it possible the number of rational solutions to $f(x,y)=n$ to be uniformly bounded for all rational $n$?

Looking for unconditional answer.

It is conjectured that polynomial injection $\mathbb{Q}^2 \to \mathbb{Q}$ exists and proving this will give bound $1$ (there are suspected injections).

Improve this question
$\endgroup$
8
  • $\begingroup$ Could you please make your question more precise? For every integer $d$, for every rational number $n$, the polynomial $x^d + y = n$ has infinitely many rational solutions. Are you asking about rational points on irrational curves? $\endgroup$
    – Jason Starr
    Nov 20, 2015 at 15:05
  • $\begingroup$ @JasonStarr Over the rationals. Your example has infinitely many points, but I am asking about bounded finitely many. Maybe will edit, thanks. $\endgroup$
    – joro
    Nov 20, 2015 at 15:18
  • $\begingroup$ joro -- your question is uncharacteristically imprecise. What order do the quantifiers go in? And where does f live? I'm assuming the answer isn't "Faltings' theorem". Note also Tom Leinsters comment on mathoverflow.net/questions/21003/… , so one answer to one interpretation of your question might be "bound = 1, answer is unknown" $\endgroup$
    – eric
    Nov 21, 2015 at 9:51
  • $\begingroup$ @eric Maybe will edit, thanks. $f$ is polynomial with rational coefficients and the rational points must be finite. I am asking are the bounded by $\deg{f}$ for all rational $n$ (again, when they are finite). Maybe this is open, since it is not known for Thue equations (in this case they depend on $n$). I know several conjectures imply even constant bound, but am asking about unconditional results. $\endgroup$
    – joro
    Nov 21, 2015 at 10:54
  • $\begingroup$ @eric Injection Q^2 to Q indeed will answer the question with bound one, but AFAICT it is not known. $\endgroup$
    – joro
    Nov 21, 2015 at 16:06

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.