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Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters.

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There have been a number of questions related to this, including one of the highest-voted ones by Bjorn Poonen. You might serch through existing questions. –  Will Jagy Nov 7 '10 at 15:59
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In particular, the first link tells us that this question is an open problem. –  Martin Brandenburg Nov 7 '10 at 16:15
    
@Martin actually that is only asking about surjectivity when the domain is $\mathbb{Z}\times\mathbb{Z}$, but I agree that it has some bearing here. –  David Roberts Nov 8 '10 at 2:49

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