Study the Irreducibility of polynomial

Question

How could I study the irreducibility of the polynomial below

$((x-1)(x-2) \cdots (x-n))+1$ in $\mathbb{Q}[x]$

Answer 1

Proposition:
a) The polynomial $P_n(x)=(x-1)(x-2) \cdots (x-n)+1 \in \mathbb{Q}[x]$ is irreducible for $n\neq 4$
b) $P_4(x)=(x^2-5x+5)^2$
Proof:
Suppose that $P_n(x)$ is reducible and write $P_n(x)=f(x)g(x)$ with $\operatorname {deg}f(x), \operatorname {deg}g(x)\lt n$ and assume (Gauss) that $f(x),g(x)\in \mathbb Z[x]$.
Then evaluating at the integers $1\leq k \leq n$ we see that $P_n(k)=1=f(k)g(k)$, forcing $$f(k)=g(k)=\epsilon_k=\pm1$$ But then the polynomial $f(x)-g(x)$ has $n$ zeros (namely $1,\cdots,n$) and degree $\lt n$.
This implies $f(x)-g(x)=0$ so that $f(x)=g(x)$ and $$P_n(x)=(x-1)(x-2) \cdots (x-n)+1 =(f(x))^2 \quad (\bigstar)$$ This shows that the degree $n$ of $P_n$ must be even: $n=2m$.
Finally evaluating at $x=\frac 32$, we get from $(\bigstar)$:$$-\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdots \frac{4m-3}{2}+1=f(\frac 32)^2$$ The right-hand side $(f(x))^2$ is always $\geq 0$ but the left-hand side is $\geq 0$ only for $m=1$ or $2$, so that we already see that $P_n(x)$ can be factored as $f(x)g(x)$ only if $n=2m$ equals $2$ or $4$.
$\bullet $ For $m=1$ we get $P_2(x)=(x-1)(x-2)+1=x^2-3x+3$, which is irreducible.
$\bullet \bullet$ For $m=2$ we get $$P_4(x)=(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2$$ which is thus (strangely!) the only reducible polynomial of the family $P_n(x)$ .