Chebyshev-like polynomials with integral roots 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 $a\deg f_n$.)
 A: I can satisfy conditions 1, 3 and almost satisfy condition 2.  Letting $f_n(t) = {t+n-1 \choose n}$ we have the well-known generating function
$\displaystyle \sum_{n \ge 0} f_n(t) x^n = \frac{1}{(1 - x)^{-t}}$
which is rational in $x$ for any fixed integer value of $t$.  I think condition 2 will end up being the hardest to satisfy because rational functions are quite rigid.

Edit 1:  My current opinion is that the conditions are not satisfiable.  Based on the analogous situation with linear homogeneous recurrences on the integers I am going to conjecture that any polynomial sequence which obeys a polynomial linear recurrence and is not essentially a geometric series has terms divisible by irreducible polynomials of arbitrarily high order.  
Edit 2:  A very strong result available in the integer case is Zsigmondy's theorem, but we don't need a result this strong.  Here's a nice result in the integer case.  Suppose an integer sequence $a_n$ satisfies a linear homogeneous recurrence with integer coefficients, and let $p$ be a prime not dividing those coefficients.  Then the sequence $a_n \bmod p$ is periodic (not just eventually periodic) $\bmod p$ by Pigeonhole.  If in addition there exists $n$ such that $a_n = 0$ and $a_n$ is unbounded, then it follows that there is a nonzero term of the sequence divisible by $p$.  For example, this is true of the Fibonacci sequence; in fact we have the much stronger result that for $p > 5$, either $p | F_{p+1}$ or $p | F_{p-1}$.
My guess is that a result like this holds in the polynomial case with $p$ replaced by a monic irreducible polynomial (say, of degree $2$), although the argument above breaks down as written.
A: 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.  
