Does there exist a bivariate polynomial that is positive exclusively in the 1st quadrant?

Does there exist a bivariate polynomial $$p \in \Bbb R[x,y]$$ that is positive iff $$x, y > 0$$?

My motivation was originally to state multiple positivity conditions with one expression but now I'm just curious and unable to find the answer via search.

Consider such a polynomial $$P$$.
$$P$$ must be $$0$$ on the positive real $$x$$ and $$y$$ axes, as it must cross from being positive to negative there. As a result, both $$x$$ and $$y$$ divide such a polynomial. Write $$P=xyQ$$.
Now, $$Q$$ is positive everywhere except the third quadrant, and so $$R(x,y)=-Q(-x,-y)$$ is positive only on the first quadrant. As a result, we have a polynomial $$R$$ that satisfies the same property, but has lower degree than $$P$$. This process cannot continue infinitely, giving a contradiction.
• I think you have to be a bit more careful. $Q$ is certainly positive in the first quadrant, but in the second, third and fourth quadrants, it might be zero. Importantly, I don't see how we can assert that $Q$ is nonzero everywhere in the third quadrant? Aug 11, 2020 at 9:35
• @Vincent You're right that there are some issues with $0$, but the proof works verbatim to show that "there is no nonzero polynomial that is nonnegative on the first quadrant and nonpositive on the other three quadrants." Aug 11, 2020 at 21:02