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.

  • 1
    $\begingroup$ It is indeed irreducible. So far, I only found this question in this direction. $\endgroup$ Jan 14, 2018 at 21:12
  • $\begingroup$ Not sure whether there is an easy method to show it by hand, but what might help : The polynomial is irreducible in $\mathbb Z_5[X]$. If you show that, you are done. $\endgroup$
    – Peter
    Jan 14, 2018 at 21:19
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    $\begingroup$ where did you get the question ? $\endgroup$
    – Will Jagy
    Jan 14, 2018 at 21:43
  • $\begingroup$ From a friend of mine. I think he took it from a textbook. $\endgroup$ Jan 15, 2018 at 14:53

1 Answer 1


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 $\mathbb{Z}[X]$


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