Is the following polynomial irreductible over $\mathbb{Z}[X]$? Is the following polynomial irreductible over $\mathbb{Z}[X]$?
$f(x) = (x-11)(x-8)(x-2017)(x-17)(x+5)-1$  What I have tryied: Assume that there is a root a in $\mathbb{Z}$ then $f(a) = 0$ which implies that the polynomial $(a-11)(a-8)(a-2017)(a-17)(a+5) = 1$ And is obvious that there is not such number $a \in \mathbb{Z}$. Therefore $f$ does not have a root in $\mathbb{Z}$. But that is not a proof that f is irreductible. 
 A: I found a solution for those who might encounter this exercise. It's a general case. Let's say we have the polynomial $f(x) \in \mathbb{Z}[X]$, and $f(x) = (x-a_1)...(x-a_n) - 1$, where $a_1,...,a_n \in \mathbb{Z}$, and $a_i \neq a_j, \forall i \neq j$. Assume that $f$ is not irreductible, then $\exists f,g \in  \mathbb{Z}[X]$ such that $f=gh$, with $deg(g), deg(h) \leq deg(f)$. We have $f(a_i) = -1, \forall i \in \{1,2,..,n\} \Rightarrow g(a_i)h(a_i) = -1 \in \mathbb{Z} \Rightarrow g(a_i), h(a_i)\in \{-1,1\}$ So this implies that we have two cases: $g(a_i) =1, h(a_i) = -1$ or $g(a_i) = -1, h(a_i) = 1 \Rightarrow g(a_i)+h(a_i) = 0\Rightarrow (g+h)(a_i) = 0, \forall i \in \{1,2,...,n\}$, with $deg(g+h) \leq n \Rightarrow g+h = 0\Rightarrow h = -g$. So $f = gh = -g^2\Rightarrow f(x) = -g^2(x)$. Let assume that leading coefficient for $g$ is $b$, and leading coeffictient for $f$ is $-1$. So we have $1 = -b^2 \Rightarrow b^2 = -1$. But we assumped that $b\in\mathbb{Z}$. So we've arrived at a contradiction. Therefore $f$ is irreductible over $\mathbb{Z}[X]$
