TITLE: Fully factored integer polynomials with constant differences
QUESTION [12 upvotes]: 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 $d>2$ ?

REPLY [2 votes]: I couldn't resist seeing the degree-$12$ solution with my own eyes (from the website linked in Peter Košinár's answer).  It's $$F(X)=X^{12} - 1812X^{11} + 1434735X^{10} - 651551410X^9 + 187228503603X^8 - 35425220352696X^7 + 
    4450555420480105X^6 - 365361907449541290X^5 + 18776316466170261396X^4 - 
    556138149602905800792X^3 + 8118307377660639252960X^2 - 42308199268401215635200X$$ where $$\delta=67440294559676054016000.$$
Here both $F(X)$ and $F(X)+\delta$ are monic and squarefree, and the roots of $F(X)$ are
$$0, 11, 24, 65, 90, 129, 173, 212, 237, 278, 291, 302$$
while the roots of $F(X)+\delta$ are
$$3, 5, 30, 57, 104, 116, 186, 198, 245, 272, 297, 299.$$
After writing this out, I can see why the PTE community focuses on the roots rather than the polynomials...