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Long time ago there was a question on whether a polynomial bijection $\mathbb Q^2\to\mathbb Q$ exists. Only one attempt of answering it has been given, highly downvoted by the way. But this answer isn't obviously unsuccessful, because the following problem (for case $n=2$) remains open.

Problem. Let $f$ be a polynomial with rational (or even integer!) coefficients in $n$ variables $x_1,\dots,x_n$. Suppose there exist two distinct points $\boldsymbol a=(a_1,\dots,a_n)$ and $\boldsymbol b=(b_1,\dots,b_n)$ from $\mathbb R^n$ such that $f(\boldsymbol a)=f(\boldsymbol b)$. Does this imply the existence of two points $\boldsymbol a'$ and $\boldsymbol b'$ from $\mathbb Q^n$ satisfying $f(\boldsymbol a')=f(\boldsymbol b')$?

Even case $n=1$ seems to be non-obvious.

EDIT. Just because we have a very nice counter example (immediately highly rated by the MO community) by Hailong Dao in case $n=1$ and because for $n>1$ there are always points $\boldsymbol a,\boldsymbol b\in\mathbb R^n$ with the above property, the problem can be "simplified" as follows.

Is it true for a polynomial $f\in\mathbb Q[\boldsymbol x]$ in $n>1$ variables that there exist two points $\boldsymbol a,\boldsymbol b\in\mathbb Q^n$ such that $f(\boldsymbol a)=f(\boldsymbol b)$?

The existence of injective polynomials $\mathbb Q^2\to\mathbb Q$ is discussed in B. Poonen's preprint (and in comments to this question). What can be said for $n>2$?

FURTHER EDIT. The expected answer to the problem is in negative. In other words, there exist injective polynomials $\mathbb Q^n\to\mathbb Q$ for any $n$.

Thanks to the comments of Harry Altman and Will Jagy, case $n>1$ is now fully reduced to $n=2$. Namely, any injective polynomial $F(x_1,x_2)$ gives rise to the injective polynomial $F(F(x_1,x_2),x_3)$, and so on; in the other direction, any $F(x_1,\dots,x_n)$ in more than 2 variables can be specialized to $F(x_1,x_2,0,\dots,0)$.

In spite of Bjorn Poonen's verdict that case $n=2$ can be resolved by an appeal to the Bombieri--Lang conjecture for $k$-rational points on surfaces of general type (or even to the 4-variable version of the $abc$ conjecture), I remain with a hope that this can be done by simpler means. My vague attempt (for which I search in the literature) is to start with a homogeneous form $F(x,y)=ax^n+by^n$, or any other homogeneous form of odd degree $n$, which has the property that only finitely many integers are represented by $F(x,y)$ with $x,y\in\mathbb Z$ relatively prime. In order to avoid this finite set of "unpleasant" pairs $x,y$, one can replace them by other homogeneous forms $x=AX^m+BY^m$ and $y=CX^m+DY^m$ (again, for $m$ odd and sufficiently large, say), so that $x$ and $y$ escape the unpleasant values. Then the newer homogeneous form $G(X,Y)=F(AX^m+BY^m,CX^m+DY^m)$ will give the desired polynomial injection. So, can one suggest a homogeneous form $F(x,y)$ with the above property?

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Well, it's clear that this won't work if you replace $\mathbb{R}$ with $\mathbb{C}$ from considering $x^n$ when $n$ is odd. That also shows it won't work if you replace it with general $\mathbb{Q}_p$, either... –  Harry Altman Jun 6 '10 at 9:02
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For any polynomial in two variables there exist distinct $a,b$ so that $f(a)=f(b)$, so the condition you put on $f$ is always fulfilled. –  Guy Katriel Jun 6 '10 at 9:18
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As regards the new question, if there's a counterexample $f$ for $n=2$, there's a counterexample for any $n$, as you can just take $f(f(x,y),z)$ when $n=3$, etc. So if we expect there is a counterexample for $n=2$ then we shouldn't be able to prove this at all; I guess considering $n>2$ might still be helpful if that makes finding counterexamples easier? –  Harry Altman Jun 6 '10 at 17:58
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@Harry, on the other hand, if there is an injective example in $ n \geq 3$ variables, by setting $n-2$ of them to $0$ we get an injective example in dimension $2.$ So you have shown that there is an injective polynomial in dimension 2 if and only if there is an example for every $n \geq 2.$ –  Will Jagy Jun 6 '10 at 18:55
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Harry and Will, thank you for these comments. So, the problem is reduced to finding just one counter example for some $n>1$. (On the other hand, I guess that your comments have resulted in somebody's downvote.) –  Wadim Zudilin Jun 6 '10 at 21:53
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1 Answer

up vote 27 down vote accepted

Let $f(x)=x^3-5x/4$. Then for $x\neq y$, $f(x)=f(y)$ iff $x^2+xy+y^2=5/4$ or $(2x+y)^2+3y^2=5$. The last equation clealy have real solutions. But if there are rational solutions, then there are integers $X,Y,N$ such that $(2X+Y)^2+3Y^2=5N^2$. This shows $X,Y,N$ all divisible by $5$, ...

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This is a very nice counter example for $n=1$! –  Wadim Zudilin Jun 6 '10 at 9:52
    
I'm surprised this was possible with just a cubic. –  Harry Altman Jun 6 '10 at 10:15
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Also, this solution can be quickly tweaked to have integer coefficients; if there are no rational solutions to $x^2+xy+y^2=5/4$, then there can't be any to $x^2+xy+y^2=5$, either. Hence $x^3-5x$ also works. –  Harry Altman Jun 6 '10 at 10:35
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Hailong, I used your solution as the hardest problem in my number theory class (to show that your $f(x)$ is injective). Two students (of 16) could do it. –  Wadim Zudilin Dec 9 '10 at 4:40
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@Wadim: (-: That's a good use of MO. –  Hailong Dao Dec 9 '10 at 22:15
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