This question is inspired by [this](https://mathoverflow.net/q/300946/12218) MO question; indeed it is a special case on which to focus.


An *exotic affine space* is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$.


Say that two varieties are *count equivalent* if they are both [polynomial count varieties](https://arxiv.org/pdf/math/0612668.pdf) with the same counting polynomial.


As shown in the comments [here](https://mathoverflow.net/a/300971/12218), the [Russell Cubic](https://en.wikipedia.org/wiki/Exotic_affine_space) is count equivalent to $\mathbb{A}^3$ although it is not isomorphic to $\mathbb{A}^3$.



> 
> **Question:** Are all exotic affine spaces count equivalent to affine space?
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Are all exotic affine spaces count equivalent to affine space?
For all but finitely many primes, yes. Any such $V$ has $V\_{\mathbb C}$ smooth and has $H^i(V\_{\mathbb C}, \mathbb Q\_\ell)=0$ for $i\neq 0$ and $=\mathbb Q\_\ell$ for $i=0$. Both these properties are known to be constructible, so they hold for $V\_{\mathbb F\_q}$ for all but finitely many $q$.


For any such $q$, the action of $\operatorname{Frob}\_q$ on $H^0(V\_{\overline{\mathbb F}\_q}, \mathbb Q\_\ell)$ must be trivial as it is one-dimensional and contains at least one invariant class. So its cohomology, with Frobenius action, matches affine space. Applying Poincare duality (using smoothness), the same is true for its cohomology with compact supports. Applying the Grothendieck-Lefschetz formula, its number of points matches affine space.


