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.

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.

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.

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.