I am wondering what a $q$-analog of the following should be:
Proposition. If $f(x)\in\Bbb Q[x]$ is a polynomial for which $f(\Bbb Z)\subseteq\mathbb{Z}$ (that is, $f(x)$ is an integer for all integers $x$), then $f(x)=\sum_{k\ge0} a_k\binom{x}{k}$ for some integer coefficients $a_k$.
This can be proved with the forward difference operator $\Delta f(x):=f(x+1)-f(x)$. Since $\Delta\binom{x}{k}=\binom{x-1}{k}$, one can solve for the $a_k$ explicitly. In fact, we get a "discrete Taylor series": $f(x)=\sum (\Delta^nf)(0)\binom{x}{k}$ for arbitrary arithmetic functions $f$.
Defining $b_k(x):=\displaystyle \frac{(x-1)(x-q)\cdots(x-q^{k-1})}{(q^k-1)(q^k-q)\cdots(q^k-q^{k-1})}$, we have $\left[\begin{smallmatrix}n\\k\end{smallmatrix}\right]_q=b_k(q^n)$. What comes to mind is
Candidate. If $f\in\mathbb{F}(q)[x]$ is such that $f(q^n)\in\mathbb{F}[q]$ for all $n=0,1,2,\dots$ then $f(x)=\sum a_k(q)b_k(x)$ for some $a_k\in\mathbb{F}[q]$.
The $q$-derivative $Df(x)=\displaystyle \frac{f(qx)-f(x)}{qx-x}$ seems to give $D b_k(x)=q^{1-k}b_{k-1}(x)$, so the same proof won't work.
