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Does there exist a polynomial $f \in \mathbb{Z}[x,y]$ such that

$$\displaystyle f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$

and

$$\displaystyle \liminf_{(x,y) \in \mathbb{R}^2} f(x,y) = -\infty?$$

In other words, does there exist a polynomial $f$ which takes on positive values at every integer point, but still there exists a sequence $(x_k, y_k)$ of real pairs such that $\lim_{k \rightarrow \infty} f(x_k, y_k) = -\infty$?

Note that if such a sequence exists, the norm of its elements must tend to infinity. This is because $f$ is continuous, and therefore the image of any compact set under $f$ is necessarily compact, and thus in particular must be bounded.

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  • $\begingroup$ What about $(x^2+1)(5y(y-1)+1)$? $\endgroup$ Nov 24 at 18:47
  • $\begingroup$ Ah yes, I was contemplating a similar example but it fell short of producing a counter-example. I'll accept this as an answer if you write it. $\endgroup$ Nov 24 at 18:53
  • $\begingroup$ Is this a Math Olympiad problem? $\endgroup$
    – Pablo H
    Nov 25 at 17:22
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    $\begingroup$ @PabloH No, it is not, at least not one that I am aware of. This arose from a research problem I am thinking about. $\endgroup$ Nov 25 at 17:26

1 Answer 1

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The polynomial $f(x,y)=(x^2+1)(5y^2+5y+1)\in\mathbb{Z}[x,y]$ is an example. Note that $5y^2+5y+1>0$ for $y\in\mathbb{Z}$, but $5y^2+5y+1<0$ at $y=-\frac{1}{2}$.

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    $\begingroup$ Your comment has a minus sign but your answer doesn't... did you intend them to match? $\endgroup$
    – user541686
    Nov 25 at 8:46
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    $\begingroup$ Either works. I just thought that $5y^2+5y+1$ looked a bit more aesthetically pleasing that $5y(y-1)+1$. $\endgroup$ Nov 25 at 13:22

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