Ruling out the existence of a strange polynomial

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.

• What about $(x^2+1)(5y(y-1)+1)$? Nov 24, 2022 at 18:47
• 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. Nov 24, 2022 at 18:53
• Is this a Math Olympiad problem? Nov 25, 2022 at 17:22
• @PabloH No, it is not, at least not one that I am aware of. This arose from a research problem I am thinking about. Nov 25, 2022 at 17:26

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 $$y=-\frac{1}{2}$$.
• Either works. I just thought that $5y^2+5y+1$ looked a bit more aesthetically pleasing that $5y(y-1)+1$. Nov 25, 2022 at 13:22