Consider in $\mathbb{F}_q[x_1,\dots,x_n]$, where $r$ is a positive integer dividing $n$, the polynomial
$$
f(x_1,\dots,x_n)=x_1x_2\dots x_r+x_{r+1}x_{r+2}\dots x_{2r}+\dots+x_{n-r+1}x_{n-r+2}x_{n}.
$$
What is the cardinality of the set of the solutions to $f(x_1,\dots,x_n)=0$?
Bonus question:
and if $n\neq 0\mod{r}$, so that the last monomial of $f$ has less than $r$ variables multiplied?
Edit:
$f(x_1,\dots,x_n)$ is the product of the first $r$ variables plus the product of the second r variables and so on (so it is of degree one in each variable, with total degree $r$).
It is clear that imposing a variable in each monomial to be zero then the set of the solutions has at least size around $q^{(r-1)n/r}r-n/r+1$, but my question arises from the fact that I don't know what is the actual size of this set (it is not difficult to find some solutions having $x_i\ne0$, for all $i\in\{1,\dots,n\}$).
Best Answer
My previous attempt having failed due to dependence between the terms (see edit history if you want to laugh and point), I think the best approach is to see this as a Markov process over prefixes of the sum, with two states: zero and non-zero.
Let $n = rs$.
There are $q^r$ possibilities for each term, of which $(q-1)^r$ are non-zero, distributed evenly over the $q-1$ non-zero field elements. Therefore the number you want is $$\begin{bmatrix}1 & 0 \end{bmatrix} \begin{bmatrix}q^r - (q-1)^r & (q-1)^{r-1} \\ (q-1)^r & q^r - (q-1)^{r-1}\end{bmatrix}^s \begin{bmatrix}1 \\ 0\end{bmatrix}$$ The transition matrix diagonalises (I used a CAS) as $$\begin{bmatrix}\frac{1}{q-1} & -1 \\ 1 & 1\end{bmatrix} \begin{bmatrix}q^r & 0 \\ 0 & q^r - q(q-1)^{r-1}\end{bmatrix} \begin{bmatrix}\frac{q-1}{q} & \frac{q-1}{q} \\ \frac{1-q}{q} & \frac{1}{q} \end{bmatrix} $$ so we get $$\begin{bmatrix}1 & 0 \end{bmatrix} \begin{bmatrix}\frac{1}{q-1} & -1 \\ 1 & 1\end{bmatrix} \begin{bmatrix}q^r & 0 \\ 0 & q^r - q(q-1)^{r-1}\end{bmatrix}^s \begin{bmatrix}\frac{q-1}{q} & \frac{q-1}{q} \\ \frac{1-q}{q} & \frac{1}{q} \end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} \\ = \begin{bmatrix}\frac{1}{q-1} & -1 \end{bmatrix} \begin{bmatrix}q^n & 0 \\ 0 & [q^r - q(q-1)^{r-1}]^s\end{bmatrix} \begin{bmatrix}\frac{q-1}{q} \\ \frac{1-q}{q} \end{bmatrix} \\ = q^{n-1} + q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s $$
For the bonus question, let $n = rs + t$ with $0 < t < r$. Then we take the same transition matrix, substituting $t$ for $r$, and apply to the previous result to get $$ \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}q^t - (q-1)^t & (q-1)^{t-1} \\ (q-1)^t & q^t - (q-1)^{t-1}\end{bmatrix} \begin{bmatrix}q^{rs-1} + q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s \\ q^{rs} - q^{rs-1} - q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s\end{bmatrix} \\ = [q^t - (q-1)^t][q^{rs-1} + q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s] + [(q-1)^{t-1}][q^{rs} - q^{rs-1} - q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s] \\ = q^{rs} (q-1)^{t-1} + [q^t - q(q-1)^{t-1}][q^{rs-1} + q^{s-1}(q-1)[q^{r-1} - (q-1)^{r-1}]^s] \\ = q^{n-1} + q^s(q-1)[q^{t-1} - (q-1)^{t-1}][q^{r-1} - (q-1)^{r-1}]^s $$
In fact, there's enough structure there to conjecture and prove that for a general composition of $n$, $\lambda_1 + \cdots + \lambda_k = n$, the number of solutions of $$x_1x_2\cdots x_{\lambda_1}+x_{\lambda_1+1}x_{\lambda_1+2}\cdots x_{\lambda_1 + \lambda_2}+\cdots = 0$$ is given by $$q^{n-1} + q^{k-1}(q-1)\prod_{i=1}^k [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}]$$
By induction on $k$.
If $k=1$ we have the original problem with $r=\lambda_1, s=k=1$.
If it holds for $k$ then for $k+1$ we have $$ \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}q^{\lambda_{k+1}} - (q-1)^{\lambda_{k+1}} & (q-1)^{\lambda_{k+1}-1} \\ (q-1)^{\lambda_{k+1}} & q^{\lambda_{k+1}} - (q-1)^{\lambda_{k+1}-1}\end{bmatrix} \begin{bmatrix}q^{n - \lambda_{k+1}-1} + q^{k-1}(q-1)\prod_{i=1}^k [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}] \\ q^{n - \lambda_{k+1}} - q^{n - \lambda_{k+1}-1} - q^{k-1}(q-1)\prod_{i=1}^k [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}]\end{bmatrix} \\ = [q^{\lambda_{k+1}} - (q-1)^{\lambda_{k+1}}][q^{n - \lambda_{k+1}-1} + q^{k-1}(q-1)\prod_{i=1}^k [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}]] + [(q-1)^{\lambda_{k+1}-1}][q^{n - \lambda_{k+1}} - q^{n - \lambda_{k+1}-1} - q^{k-1}(q-1)\prod_{i=1}^k [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}]] \\ = q^{n-1} + q^k (q-1) \prod_{i=1}^{k+1} [q^{\lambda_i-1} - (q-1)^{\lambda_i-1}] $$ as desired.