Let $F: \mathbb{R}^{n} \to \mathbb{R}$ be a smooth function. Suppose $B$ be a closed and bounded box. I would like to obtain for fixed $q \in \mathbb{N}$ $$ \# \{ \mathbf{a} \in \mathbb{Z}^n : F(\frac{\mathbf{a}}{q} ) = 0 , \ \frac{\mathbf{a}}{q} \in B \} < C q^{n-1}, $$ where $C$ is independent of $q$.

I think this is not possible if $F$ has a region where it is identically zero inside $B$. So my question is what assumption on $F$ is necessary on $B$ so that this bound can be achieved. Any comments would be appreciated. Thank you.

  • $\begingroup$ Is $F$ a polynomial, or just any $C^\infty$-function? I am sure with the latter one can come up with an example that hits way more rational points than it should $\endgroup$ – Stanley Yao Xiao Jul 8 at 1:55
  • $\begingroup$ $F$ is any $C^{\infty}$-function here. Yes, you are right, for example $F \equiv 0$. That's why I was hoping for some conditions I could put on $F$ such that this bound holds. One condition one could assume is that $F$ is non-singular on $B$. Then by implicit function theorem (and using compactness) the bound follows. Maybe this is as good as one can hope for... $\endgroup$ – Johnny T. Jul 8 at 6:55

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