Irreducibility of a class of polynomials This question is directly inspired by this question. Consider polynomials of the form 
$$p(x) = \prod_{i=1}^n(x-i)^2 - d.$$ For which values of $n$ and $d$ is $p(x)$ irreducible? There is a theorem of Polya, cited in the original question, which states that:
If for $n$ integral values of $x,$ a polynomial $q(x)$ of degree $n$ has values which are different from $0,$ are smaller in absolute value than
$$\frac{\lceil n/2\rceil!}{2^{\lceil n/2\rceil}},$$ then $q(x)$ is irreducible over $\mathbb{Q}.$ 
This seems to not quite apply to the given type of polynomial (though it would, if there were no squares).
In this paper, there is a reference to a paper of Brauer, Brauer, and Hopf(!) from 1926 which addresses the question (of Schur) on irreducibility of polynomials of the form $g(f(x)),$ but it seems that their $g$ always has constant coefficient $1,$ so this is not directly applicable.
Surely, technology has advanced since those days, and even the Galois groups can be computed (I assume the Galois group is the full symmetric group for most choices of $n$ and $d.$)
What makes the whole thing strange is that the original question (with $n=2013$ and $d=2014$) was supposedly an entrance exam question at Peking university. So this is either child abuse, or there is some simple trick.
 A: I prove here that $\prod_{i=1}^n (x-a_i)^2 + d$ is irreducible over $\mathbf{Q}$ whenever the $a_i$ are distinct integers, $n\ge 6$, and $d$ is squarefree, $d>1$, and $d\not\equiv 3\pmod{4}$.
The hypotheses on $d$ ensure that the ring of algebraic integers in $\mathbf{Q}(\sqrt{-d})$ is $\mathbf{Z}[\sqrt{-d}]$, and that the only units in this ring are $\pm 1$.  Write $f(x):=\prod_{i=1}^n (x-a_i)$, and let $\alpha$ be a root of $f(x)^2+d$, so that $\beta:=f(\alpha)$ satisfies $\beta^2=-d$.  Suppose that $f(x)^2+d$ is reducible over $\mathbf{Q}$, so that $[\mathbf{Q}(\alpha):\mathbf{Q}]<\deg(f(x)^2+d)=2n$.  Since $$[\mathbf{Q}(\alpha):\mathbf{Q}]=[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)]\cdot [\mathbf{Q}(\beta):\mathbf{Q}] = 2[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)],$$
it follows that $[\mathbf{Q}(\alpha):\mathbf{Q}(\beta)]<n$, so that $f(x)-\beta$ is reducible over $\mathbf{Q}(\sqrt{-d})$.  Write $f(x)-\beta=g(x)h(x)$ with $g,h$ nonconstant polynomials in $\mathbf{Q}(\sqrt{-d})[x]$.  Since $f(x)-\beta$ is monic, we may assume that both $g$ and $h$ are monic.
Since $f(x)-\beta$ is a monic polynomial whose coefficients are algebraic integers, all of its roots are algebraic integers.  Therefore both $g$ and $h$ are monic polynomials all of whose roots are algebraic integers, so the coefficients of $g$ and $h$ are algebraic integers; since these coefficients are also in $\mathbf{Q}(\sqrt{-d})$, it follows that $g$ and $h$ are in $\mathbf{Z}[\sqrt{-d}][x]$.
Now substitute $x=a_i$ into the identity $f(x)-\beta=g(x)h(x)$, to get $-\beta=g(a_i)h(a_i)$.  Hence $g(a_i)$ is an element of $\mathbf{Z}[\sqrt{-d}]$ which divides $\beta$, so taking norms gives $N(g(a_i))\mid N(\beta)=d$.  Since $d$ is squarefree, the only elements of $\mathbf{Z}[\sqrt{-d}]$ whose norm divides $d$ are $\pm 1$ and $\pm\sqrt{-d}$.  Thus $g(a_i)\in\{\pm 1,\pm\sqrt{-d}\}$.
First suppose that $g(a_i)$ takes the same value (call it $c$) for every $i$.  Then $g(x)-c=f(x)G(x)$ for some $G(x)\in\mathbf{Z}[\sqrt{-d}][x]$, which is impossible since $0<\deg(g)<\deg(f)$.
Hence there exist $i,j$ with $g(a_i)\ne g(a_j)$.  Since $n\ge 6$, it follows that there exist $r,s$ for which $g(a_r)\ne g(a_s)$ and $a_r\ge a_s+3$: for, this is clear if $\lvert a_i-a_j\rvert\ge 3$, and if $\lvert a_i-a_j\rvert<3$ then there exists $k$ such that $\lvert a_i-a_k\rvert\ge 3$ and $\lvert a_j-a_k\rvert\ge 3$, and we must have either $g(a_i)\ne g(a_k)$ or $g(a_j)\ne g(a_k)$.
Finally, $a_r-a_s$ divides $g(a_r)-g(a_s)$ in $\mathbf{Z}[\sqrt{-d}]$.  Since $a_r-a_s$ is an integer $\ge 3$, and $g(a_r),g(a_s)$ are distinct elements of $\{\pm 1, \pm\sqrt{-d}\}$, this is impossible.
