TITLE: Chebyshev-like polynomials with integral roots
QUESTION [7 upvotes]: Chebyshev polynomials have real roots and satisfy a recurrence relation. I was wondering if one can find a sequence of polynomials with integral or rational roots with similar properties. More precisely, one is looking for a sequence of polynomials $(f_n),f_n\in\mathbf{Q}[t]$ such that

$\deg f_n\to\infty$ as $n\to\infty$;
$\sum_{n=0}^\infty f_n(t) x^n$ is (the Taylor series of) a rational function $F$ in $x$ and $t$.
All roots of any $f_n$ are integer and have multiplicity 1. (A weaker version: the roots are allowed to be rational and are allowed to have multiplicity $>1$ but there should be an $a>0$ such that the number of distinct roots of $f_n$ is at least $a\deg f_n$.)

REPLY [3 votes]: Here's one thought.  For each integer k, f_n(k) satisfies a recurrence relation.  If the roots of f_n are all integers, then f_n(k) and f_n(k+1) have to be "in sync" in the sense that they never have opposite sign.  This is a strong condition!  For instance, suppose the sequences f_n(k) and f_n(k+1) each have unique largest eigenvalue:  then these eigenvalues would have to have the same argument.
Update:  Qiaochu's answer suggests that in fact working mod p would be just better than the "archimedean" picture sketched above, since it is really F_q[t], not Z[t] or Q[t], that is analogous to the integers.  Let F_n(t) be the reduction of f_n(t) to F_p[t].  If f_n(t) has all roots rational for every n, then the reduction of f_n(t) mod p has the same property.  But now we are saying something quite strong; that f_n(t) lies in a finitely generated subgroup of F_q(t)^*!  This is presumably ruled out by Mason's theorem (ABC for function fields.)  Indeed, you could probably prove in this way that not only are the roots of f_n(t) not rational, but f_n(t) has irreducible factors of arbitrarily large degree.
I don't think this approach would touch a harder question along the same lines like this one.