Fully factored integer polynomials with constant differences Given a degree $d$, it is possible to construct a pair $(F,\delta),$ where $F$ is a polynomial in $\mathbb{Z}[X]$ and $\delta$ a non-zero integer, such that $F(X)$ and $F(X)+\delta$ both split into linear factors over $\mathbb{Z}[X]$ ?
This is easy for $d=2$, as shown by the pair $F(X)=X^2$ and $\delta=-a^2$ (for an arbitrary integer $a$). Indeed, $F(X)=X \cdot X$ and $F(X)+\delta=(X-a)\cdot (X+a).$ What can be said for $d>2$ ?
 A: This question is closely related to the Prouhet-Tarry-Escott problem.
The so-called ideal solution to PTE asks for two distinct sets of integers $A$ and $B$ with $|A|=|B|=d$, such that for all $k$ with $1\leq k\leq (d-1)$, we have 
$$\sum_{a\in A} a^k = \sum_{b\in B} b^k$$
It's easy to see that these equalities imply that all the symmetric polynomials of degree up to $(d-1)$ applied to members of $A$ and $B$ have identical values. Therefore, the difference of $$P_A=\prod_{a\in A}(x-a)$$ and $$P_B=\prod_{b\in B}(x-b)$$
must be a (non-zero) constant, making the pair $(P_A, P_B-P_A)$ a solution to this question for degree $d$.
At least one ideal PTE solution is known for $d\leq 10$ and for $d=12$; examples can be found at this page. Although the page is somewhat aged, I believe no solution of higher degree has been discovered subsequently, nor has the $n=11$ case been resolved. Also, as far as I know, there is no reason to believe that there is some upper bound on $d$ after which no solutions can be found; although the search is certainly getting considerably more difficult with increasing $d$.
The main difference between PTE and this problem is that this one allows non-monic polynomials and also repeated roots of the polynomials. Thus this question admits solutions which wouldn't be solutions of PTE; e.g. the one listed in the problem statement.
