I am working on the following proof:
Prove that $f_n(x)$ is irreducible over $\mathbb Z$ for all $n\in$ Z, with $n\not=4$, where $f_n(x)=1+(x-1)(x-2)\cdots(x-n)$.
I found that $f_4(x)$ can be factored into $(x^2-5x+5)^2$.
I was thinking of proving cases, 1, 2, 3, and 5, and then performing induction to get the rest, but I have come to a roadblock. After assuming irreducibility in the $k^{th}$ case, I arrive at:
$f_{k+1}(x)=1+(x-1)(x-2)...(x-k)(x-k-1)$.
I honestly have no idea how to proceed from here, or if induction will work for this proof.
Thanks for any ideas.
-MathUser3
Assume on the contrary that $f_n$ is reducible, then write $$1+(x-1)(x-2) \ldots (x-n)=f_n(x)=g(x)h(x)$$ where $1 \leq \deg(g), \deg(h) \leq n-1$. Then $g(i)h(i)=1$ for $1 \leq i \leq n$, so $g(i)=h(i)= \pm 1$. This implies that $g(x)-h(x)$ is a polynomial of degree at most $n-1$, with $1, 2, \ldots , n$ as roots. Thus $g(x)-h(x)$ is identically $0$, so $f_n(x)=g(x)h(x)=g(x)^2$.
In my answer here, I show (after the first part about $P(n)$ always perfect square implying that $P(x)$ is a square of a polynomial) that $m+x(x+1) \ldots (x+k-1)$ is not a square of a polynomial, unless $k=4$ and $m=1$. This essentially finishes off the problem, since here we have $m=1, k=n$ and we may take $x \to x-n$.
Andreas Caranti's answer there also links to a proof of irreducibility of $f_n$ for $n>4$; I take the liberty of quoting him: