Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}$ such that $f\colon\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

62$\begingroup$ Is it known (or obvious) that there is an injective f? $\endgroup$– Tom LeinsterCommented Apr 11, 2010 at 17:56

178$\begingroup$ Quote from arxiv.org/abs/0902.3961, Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a wellknown conjecture." $\endgroup$– Jonas MeyerCommented Apr 11, 2010 at 19:47

18$\begingroup$ Shouldn't Jonas repost his comment as an answer? $\endgroup$– David CorwinCommented Jul 27, 2010 at 22:19

7$\begingroup$ If there exists such an $f$, then there does so in any number $n$ of variables, by a simple induction. So does there exist, for some $n \geq 2$, a polynomial $p(x_1,...,x_n)$ in $n$ variables over $\mathbb{Q}$ such that $p : \mathbb{Q}^n \rightarrow \mathbb{Q}$ is bijective ? Replace bijective everywhere by injective if you like. I don't know if this is any easier to answer, but sometimes you can say a lot more about Diophantine equations in many variables. $\endgroup$– Peter HegartyCommented Sep 22, 2011 at 13:52

34$\begingroup$ Now 16 answers, all deleted. $\endgroup$– Gerry MyersonCommented Oct 7, 2014 at 5:09
3 Answers
Jonas Meyer's comment:
Quote from arxiv.org/abs/0902.3961, Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a wellknown conjecture." – Jonas Meyer
Added June 2019 Poonen's paper is published as:
Bjorn Poonen, Multivariable polynomial injections on rational numbers, Acta Arith. 145 (2010), no. 2, pp 123127, doi:10.4064/aa14522, arXiv:0902.3961.

5$\begingroup$ This was posted (by Jonas Meyer) as a comment on 11 April 2010, and got 125 upvotes (so far!) as a comment. Why post it now as an answer? $\endgroup$ Commented Jan 13, 2016 at 5:04

43$\begingroup$ @GerryMyerson Probably on suggestion from David Corwin, and/or on the Stack Exchange principle that important relevant information shouldn't be relegated to comments. I guess Boaz signals, by making it CW, that this isn't for reputation gain, but as a public service. $\endgroup$– Todd Trimble ♦Commented Jan 13, 2016 at 5:16

32$\begingroup$ @GerryMyerson is right. The issue is that the problem is presented as unsolved, drawing unnecessary attention and time, just to find in the comments that it is as answered as it could be. I believe, MO is not expected to provide answers like complete solutions of P=NP. Rather, answers that are within knowledge, or easy reach, of experts. I can stop doing this service if this is against the policies, but then an alternative solution to the issue I raise here better be found. $\endgroup$ Commented Jan 13, 2016 at 12:31
There is a new manuscript on the arXiv by Giulio Bresciani, A higher dimensional Hilbert irreducibility theorem, arXiv:2101.01090, which shows that assuming the weak BombieriLang conjecture, there cannot be a polynomial bijection from $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$.
The author writes that:
Our strategy is essentially the one followed in a "polymath project" led by T. Tao, see [Tao19], hence this result should be credited to the polymath project as a whole.

14$\begingroup$ I should credit Daniel Loughran for pointing me to this question, see mathoverflow.net/q/373221/45660 $\endgroup$ Commented Jan 5, 2021 at 6:25
This is a link to a new, crowdsourced attempt to resolve this question (at least conditional on the assumption of some strong numbertheoretic conjectures) being led by Terry Tao.

13$\begingroup$ While what is outlined there is a good strategy of attack, Terry does not suggest it is a rigorous proof, and posts it in the hopes that specialists will fill in the holes and reduce the problem to a consequence of one or more conjectures in algebraic geometry. Gerhard "That Means It's Conditionally False" Paseman, 2019.06.08. $\endgroup$ Commented Jun 8, 2019 at 16:30