Long time ago there was a [question](https://mathoverflow.net/questions/21003/)
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](http://arxiv.org/abs/0902.3961)
(and in comments to [this question](https://mathoverflow.net/questions/21003/)).
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](https://mathoverflow.net/users/5583/harry-altman) and
[Will Jagy](https://mathoverflow.net/users/3324/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](http://arxiv.org/abs/0902.3961)
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?



Polynomials with rational coefficients
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 $5$, ... 


