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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).

joro
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  • 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? – Jason Starr Nov 20 '15 at 15:05
  • @JasonStarr Over the rationals. Your example has infinitely many points, but I am asking about bounded finitely many. Maybe will edit, thanks. – joro Nov 20 '15 at 15:18
  • 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 http://mathoverflow.net/questions/21003/polynomial-bijection-from-mathbb-q-times-mathbb-q-to-mathbb-q , so one answer to one interpretation of your question might be "bound = 1, answer is unknown" – eric Nov 21 '15 at 09:51
  • @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. – joro Nov 21 '15 at 10:54
  • @eric Injection Q^2 to Q indeed will answer the question with bound one, but AFAICT it is not known. – joro Nov 21 '15 at 16:06
  • I see. So here's the question: Does there exist $f$ in $Q[x,y]$ such that there's a uniform bound independent of $n$ for number of solutions to $f(x,y)=n$ in rationals $x$ and $y$? That's a nice question because it's weaker than the question which is apparently probably true but open (same question but with bound 1). – eric Nov 23 '15 at 17:11
  • @eric Yes, this is the question. – joro Nov 23 '15 at 17:43
  • Conditionally on the Bombieri-Lang conjecture, this should be true for all $f$ that are not degenerate in the sense that the curves $f(x,y)=n$ are reducible or have genus at most 1, from the work of Caporaso, Harris, and Mazur: http://www.ams.org/mathscinet-getitem?mr=1325796 . But this type of result is unlikely to be made unconditional in the near future. – Terry Tao Nov 25 '15 at 18:40

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