How prove $(x-1)^4(x-2)^4\cdots(x-2013)^4+2014$ is reducible? 
Let $$f(x)=(x-1)^4(x-2)^4(x-3)^4\cdots(x-2013)^4+2014\tag{1}$$
Prove or disprove: $f(x)$ is reducible on the field of rational numbers $Q$.

this problem is background from this ：
How to prove that $f(x)=(x-1)^2(x-2)^2(x-3)^2\cdots(x-2013)^2+2014$ is reducible?
and  Dear Michael Zieve use nice methods to solve it.
then I find this similar
problem is from :(George Polya,《Problems and Theorems in Analysis II》)
show that: $$(x-1)^4(x-2)^4\cdots(x-2013)^2+1$$ is also irreducible.
He post this solution on $page 327$.
let $$p_{0}(x)=(x-1)(x-2)\cdots(x-2013)$$
then
$$p^4_{0}(x)+1=[1-p_{0}P_{-1}(x)][1-p_{0}(x)P_{1}(x)]$$
where $p_{-1}(x)$ and $p_{1}(x)$ are polynomial with integral coefficients and with leading cofficient $-1$.and so on,you can download this book see full solution,
But my problem I can't,Thank you.
my try:
$$p^4_{0}(x)+2014=[A+p_{0}(x)P_{-1}(x)][B+p_{0}(x)P_{1}(x)]$$
 A: This polynomial is irreducible.  This follows from the argument in my answer to this MO question.  Namely, write
$$p(x):=(x-1)(x-2)\dots (x-2013).$$  Exactly as in my MO answer, if $p(x)^4+2014$ is reducible over $\mathbf{Q}$ then $p(x)^2-\sqrt{-2014}$ is reducible over $\mathbf{Q}(\sqrt{-2014})$.  Say the factorization is $p(x)^2-\sqrt{-2014}=g(x)h(x)$ where we may assume that $g(x)$ and $h(x)$ are monic.  Then, as in my MO answer, the roots and hence the coefficients of $g$ and $h$ must be algebraic integers, so $g,h\in\mathbf{Z}[\sqrt{-2014}][x]$.  Thus, for $1\le i\le 2013$, we have
$$
-\sqrt{-2014} = p(i)^2-\sqrt{-2014} = g(i)h(i),
$$
so that $g(i)$ divides $\sqrt{-2014}$ and hence (again as in the MO answer) $g(i)\in\{\pm 1,\pm\sqrt{-2014}\}$.
If $g(i)=g(1)$ for every $i=1,2,\dots,2013$, then $g(x)-g(1)=p(x)G(x)$ for some $G\in\mathbf{Z}[\sqrt{-2014}][x]$.  Since $g(i)h(i)=\sqrt{-2014}$, it follows that $h(i)=h(1)$ for every $i=1,2,\dots,2013$, so that $h(i)-h(1)=p(x)H(x)$ for some $H\in\mathbf{Z}[\sqrt{-2014}][x]$.  Now $$p(x)^2-\sqrt{-2014}=g(x)h(x)=(g(1)+p(x)G(x))\cdot (h(1)+p(x)H(x)).$$
Equating leading terms on both sides (and recalling that $g$ and $h$ are nonconstant) implies that $G=H=\pm 1$.  Subtracting $p(x)^2$ from both sides gives
$$
-\sqrt{-2014} = g(1)h(1) + p(x)\cdot G\cdot (g(1)+h(1)),
$$
so we must have $g(1)=-h(1)$ and $g(1)^2=\sqrt{-2014}$, which is impossible.
Hence we have $g(i)\ne g(j)$ for some $i,j$ with $1\le i<j\le 2013$.  This gives a contradiction in exactly the same way as in my MO answer: first, it follows that $g(r)\ne g(s)$ for some $r,s$ with $1\le r < s\le 2013$ and $s>r+2$.  But $s-r$ divides $g(s)-g(r)$ in $\mathbf{Z}[\sqrt{-2014}]$, where $s-r$ is an integer $\ge 3$ and $g(s)-g(r)$ is the difference between two distinct elements in $\{\pm 1,\pm\sqrt{-2014}\}$, which is impossible.
Note: the same argument yields irreducibility of $\prod_{i=1}^n (x-a_i)^4 + d$ whenever the $a_i$ are distinct integers, $n\ge 6$, and $d$ is a squarefree integer such that $d>1$ and $d\not\equiv 3\pmod{4}$.  Presumably this is the result that Wegner proved (as in user68061's comment to the present question).
