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

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Is it known (or obvious) that there is an injective f? – Tom Leinster Apr 11 '10 at 17:56
Quote from, 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 well-known conjecture." – Jonas Meyer Apr 11 '10 at 19:47
Shouldn't Jonas repost his comment as an answer? – David Corwin Jul 27 '10 at 22:19
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. – Peter Hegarty Sep 22 '11 at 13:52
Now 16 answers, all deleted. – Gerry Myerson Oct 7 '14 at 5:09

protected by Andrés Caicedo Dec 5 '13 at 14:23

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