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In what follows let $p$ be a (odd, if needed) prime, $e$ a positive integer, $q = p^e$; $\mathbb{F}_q$ will denote a finite field with $q$ elements whose prime subfield will be denoted as $\mathbb{F}_p$. Also $f\in\mathbb{F}_q[x,y]$ will denote a bivariate polynomial with coefficients from $\mathbb{F}_q$.

For any such polynomial $f$, any integer $k\ge 2$ and any $k$-tuple $(x_1,\dotso,x_k)\in\mathbb{F}_q^k$ I define \begin{equation} F_f(k) := f(x_{k-1},x_k) - f(x_{k-2},x_{k-1})+\dotsb+(-1)^k f(x_1,x_2)\quad(*) \end{equation} to be an alternating sum of length $k$ of (evidently not independent one from another) values of $f$.


''Describe'' (in whatever terms possible) the class of such polynomials $f$ for which $$ \{F_f(k)\colon k\ge 2,(x_1,\dotso,x_k)\in\mathbb{F}_q^k\} = Span_{\mathbb{F}_p}(Range(f))\quad(**). $$ In other words, find all such polynomials for which these alternating sums represent any element from the span of the range of that polynomial with integer coefficients.

Perhaps anyone could point to me a source where a similar problem is targeted...

Observation and examples

  1. If there exists an $a\in\mathbb{F}_q$ such that $f(x,y) = (x-a)(y-a)\hat{f}(x,y)$ for some $\hat{f}\in\mathbb{F}_q[x,y]$, then $f$ meets (**) the condition of the problem. Informally this is because one can choose the $x_i$'s in such a way as to guarantee that only the 1st, 5th, 9th,... terms are left and all others vanish. The terms left are however independent one from another and their sums may produce every element from the span of the range.
  2. $f(x,y) = x^2$ or $f(x,y) = x^2 + 1$ for $q\equiv 3\pmod 4$ are not of the form mentioned in 1), but they also meet that condition. In both cases the two sides of (**) are simply all of the field. An argument similar to the one above applies here as well. Combine the 1st and the 4th terms in (*) and make all others vanish. Since every element of the field is a difference of two squares, we're done.
  3. Write $\mathbb{F}_{25} = \mathbb{F}_5(\xi)$, where $\xi$ is a root of a quadratic monic irreducible over $\mathbb{F}_5$. Define $f\in\mathbb{F}_{25}[x,y]$ piece-wise $$ f(x,y) = \begin{cases} 1, & x\in\mathbb{F}_5\\ \xi, & x\in\mathbb{F}_{25}\setminus\mathbb{F}_5 \end{cases} $$ so that $Range(f)$ contains a basis of the field viewed as a vector space over its prime subfield. One can check however that condition (**) is not met, i.e. that alternating sums (*) can't produce any given element of the field. In this example every sum (*) may be written as $\alpha\cdot 1 + \beta\xi$, $\alpha,\beta\in\mathbb{F}_5$, but these $\alpha$ and $\beta$ are not independent from each other
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