The following question arose while trying to generalize some combinatorial statements from $\mathbb{Z}$ to $\mathbb{R}$.
Suppose I have a multivariate homogenous polynomial $f$ with coefficients in $\mathbb{Z}$, and its integral zeroes lie only on the axes, i.e. $f(\vec x) = 0 \implies$ some coordinate of $\vec x$ is 0.
I want to prove that if I look at $f$ as a polynomial over $\mathbb{R}$ then it satisfies there the same property - its zeroes (this time, real zeroes) must line on the axes. I am not sure that it is true, but I am sure that it is correct in many cases and I was told that it might have a proof that uses model theory (though I prefer an "explicit" proof). If this is incorrect, a counterexample would be nice.
Note: because $f$ is homogenous, it can be seen that its main property carries on to $\mathbb{Q}$.
EDIT: Chris found a nice counterexample. Can all the counterexamples be characterized some how? Chris also showed that a counterexample which is a sum of squares can be found.
What about the following case: $f$ is a sum of squares of product of hyperplanes, i.e. $f = \sum P_i^2$ where $P_i$ is a product of linear forms.