This is a refinement of my [question asked earlier](https://mathoverflow.net/questions/435241/ruling-out-the-existence-of-a-strange-polynomial), which is answered beautifully in the negative by Thomas Browning. The example he gave was geometrically reducible. Now I want to ask the same question, but with the extra assumptions that $f(0,0) = 0$ and $f$ is geometrically irreducible for all $c \in \mathbb{R}$.


For convenience, the question is: does there exist a polynomial $f \in \mathbb{Z}[x,y]$ such that $f(0,0) = 0$ and $f$ is geometrically irreducible, and such that


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


and


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



Ruling out the existence of a strange polynomial II
Just a slight modification of the previous [example](https://mathoverflow.net/a/435244/36721):
$$f(x,y):=\left(5 y^2+5 y+1\right) \left(x^2+y^2\right)+\left(10 y^2+10 y+1\right) y^2.$$




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Your conditions $f(0,0) = 0$ and
$$f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$
contradict each other. So, I was assuming that you meant
$f(0,0) = 0$ and
$$f(a,b) > 0 \text{ for all } (a,b) \in \mathbb{Z}^2\setminus\{(0,0)\}.$$


