Low degree polynomials over finite fields with no roots Let $q$ be some prime. Is there a low* (preferably constant) degree multivariate polynomial over $GF(q)$ that does not have 0,1 roots?
In other words, is there a low degree $f(X)\in \mathbb{F}_q[X]$, for $X:=\{x_1,\ldots x_n\}$, such that for all $X\in\{0,1\}^n$, $f(X)\neq 0$?
Furthermore, is there a symmetric such $f(X)$ ? 

(*) low degree here means logarithmic in $n$, i.e., $O(\log n)$ ($q$ is constant, so the constant hidden in the $O$ notation may depend on $q$). Constant degree (i.e., a degree independent of $n$) is even better. Also, $O(\log^c n)$ degree for a constant $c$ independent of $n$ is interesting.

Comment: Other simple examples of such $f(X)$, even if they are of high degree are also interesting to me. For instance, a "simple" polynomial would be one with few monomials. I could only find high degree, but not extremely simple, such polynomials $f$. For instance, based on exclusion-inclusion principle one can show that $1+x_1+x_2+x_1x_2$ has no 0,1 roots over $GF(3)$; this example generalizes to a degree $n$ polynomial when the number of variables is $n$, and thus it is not a low-degree polynomial (and not very simple).

Example, where this exists: Let $q=3$ and $n=1$, then $f(x) = (x-2)$ is a polynom with one root ($x_0 = 2$) with $f(0) \neq 0$ and $f(1) \neq 0$. $f$ is also symmetric (since there is only 1 entry).
