Consider an algebraic manifold whose number of points is $q^n ([n+1]\_q)$. Is there a geometric relation to $A^n (P^n)$? In particular, is there an equivalence in the Grothendieck ring of varieties or could there be a birational equivalence?


If there is no such equivalence in general, might some additional reasonable requirements on a manifold force there to be such an equivalence? 




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**Motivation:** one can see that some examples of identities on the level of enumerating $F\_q$ points can be lifted to geometric relations:


[Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F\_q$?](https://mathoverflow.net/questions/299581/is-there-a-lift-of-the-q-vandermonde-identity-to-some-geometric-motivic-identi)


[Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$](https://mathoverflow.net/questions/299748/can-one-divide-algebraic-manifolds-make-sense-gr2-n-gr2-nm-pn-1-p?noredirect=1&lq=1)



If number of points on a manifold is $q^n ( [n+1]_q )$ does it imply a geometric relation to $A^n (P^n)$?
The Russell Cubic $R:=V(x + x^2 y + z^2 + t^3)\subset \mathbb{A}^4$ is not isomorphic to $\mathbb{A}^3$ although over $\mathbb{C}$ they are both diffeomorphic to $\mathbb{R}^{6}$ (see this [Wikipedia](https://en.wikipedia.org/wiki/Exotic_affine_space) page).


I ran a *Mathematica* program I quickly wrote to compute the number of solutions of $x + x^2 y + z^2 + t^3$ over $\mathbb{F}\_p$. For the first twelve primes $p$ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) I got $p^3$.


Based on this evidence I guessed that although $R$ and $\mathbb{A}^3$ are not isomorphic that they had the same counting polynomial.


In the comments below, Vladimir Dotsenko, provides an elementary proof of my guess:



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> "Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\not=0$ we have the unique value for $y$, so this gives $(p−1)p^2$ points ($p−1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clearly has $p$ zeros by the usual parametrization of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\not=0$, we have $t=−(z/t)^2$ so denoting $z/t=u$, we have $(p−1)$ solutions $(u^3,−u^2)$. "
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