“Close” roots of polynomials and Pillai property

A sequence of integers $a(n)_{n \geq 0}$ has the Pillai property if there exists an integer $G \geq 2$ such that for all integers $k \geq G$ there exists an integer $n \geq 0$ such that none of the $k$ consecutive terms $a(n + 1), a(n + 2), \ldots, a(n + k)$ is relatively prime with all the others.

In [SS] it is proved that the sequence given by $a_n := f(n)$ has the Pillai property, where $f \in \mathbb{Z}[x]$ is any polynomial of degree at most $3$. (Actually, it is proved for degrees $2$ and $3$, the linear case was already proved by Evans [Eva].)

For $f(X) = a_k X^k + \cdots + a_1 X + a_0 \in \mathbb{Z}[X]$, $a_k \neq 0$, define $$\widetilde{f}(X) := a_k^{2k-2} \prod_{\substack{1 \leq i, j \leq k \\ i \neq j}}(X - (\alpha_i - \alpha_j)) ,$$ where $\alpha_1, \ldots, \alpha_j \in \mathbb{C}$ are the roots of $f$.

It seems that the only part of the proof where the hypothesis $\deg f = 2,3$ is used is the following elementary lemma:

Lemma 2.2. Suppose $f \in \mathbb{Z}[X]$ has degree $2$ or $3$, and that $p \mid \widetilde{f}(r)$ for some prime number $p \nmid 6a_k$ and some positive integer $r$. Then there exists an integer $n$ such that $$f(n) \equiv f(n + r) \equiv 0 \pmod p.$$

The authors showed that Lemma 2.2 is indeed false for quartic polynomials, by giving the examples:

• $f(X) = X^4 + 1$, since $7 \mid \widetilde{f}(3)$ but $f(X) \equiv 0 \pmod 7$ has no solutions.

• $f(X) = g(X)g(X + r)$ for a quadratic irreducible $g \in \mathbb{Z}[X]$ and a positive integer $r$, since $\widetilde{f}(r) = 0$ is divisible by all primes, but $f(X) \equiv 0 \pmod p$ has no solutions for infinitely many primes $p$.

My questions are:

1. Do the polynomials $f$ for which Lemma 2.2 fails (that is, $p \mid \widetilde{f}(r)$ for some prime $p \nmid 6a_k$ and some positive integers $r$, but $f(X) \equiv f(X + r) \equiv 0 \pmod p$ has no solutions) have a nice description? In particular, I am interested in irreducible $f$, since the second example $f(X)=g(X)g(X+r)$ is quite disappointing for the problem of Pillai property, being clear that one can always assume $f$ irreducible.

2. Can Lemma 2.2 be extended to higher degree polynomial by assuming further conditions than $p \mid \widetilde{f}(r)$? The hope of course is showing that also polynomials of degree greater than $3$ have the Pillai property.

[SS] Sanna and Szikszai, A coprimality condition on consecutive values of polynomials, Bull. London Math. Soc. (2017) (https://arxiv.org/abs/1704.01738)

[Eva] Evans, On $N$ consecutive integers in an arithmetic progression, Acta Sci. Math. 33 (1972), 295-296.