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

my try: since
I know this
$$g(x)=(x-1)(x-2)(x-3)\cdots(x-n)+1 \qquad(n\in N^{+},n\neq 4)$$is reducible on the field of rational numbers $\mathbb Q$
because when $n=4$ we have
$$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+4)(x^2-5x+6)+1=(x^2-5x+5)^2$$
this simaler problem:
$[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ is irreducible over $\mathbb Q$
$(1)$ problem is from Beijing university mathematics postgraduate entrance exams today,and this constant is $2014$ follow I can't works,
 A: As Michael Zieve said my previous answer had a wrong quote, which I am terribly sorry for.
However, the following is true: 
if $ax^2+bx+c$ is irreducible $a,b,c \in \mathbb{Z}$ then $Q(x)=aP(x)^2 + bP(x) + c$ is irreducible where $P(x)=(x-a_1)...(x-a_n), n>2\tau(c)(2+[\log_2|c|]), a_1, ... , a_n \in \mathbb{Z}$ see theorem 7.1 in K. Gyoury,L.Hajdu and R. Tijdeman ( http://www.math.unideb.hu/~hajdul/ght20.pdf‎)
Below is the previous version of the answer. Dorwart and Ore proved the result only for $c=1$.
There is a result by Dorwart and Ore (see http://www.jstor.org/discover/10.2307/1968341?uid=3739232&uid=2&uid=4&sid=21103238485287) stating that
if $ax^2+bx+c$ is irreducible $a,b,c \in \mathbb{Z}$ then $Q(x)=aP(x)^2 + bP(x) + c$ is irreducible where $P(x)=(x-a_1)...(x-a_n), n \geq 5, a_1, ... , a_n \in \mathbb{Z}$
This answers both your questions
A: Since $f(x)$ has integer coefficients and the gcd of these coefficients is $1$, by Gauss's lemma it suffices to show that $f(x)$ is irreducible in $\mathbf{Z}[x]$.  Suppose that $f(x)=g(x)h(x)$ where $g,h\in\mathbf{Z}[x]$ are nonconstant.  For $1\le c\le 2013$, we have $2014=f(c)=g(c)\cdot h(c)$, so both $g(c)$ and $h(c)$ are (positive or negative) divisors of $2014$.  Since $2014=2\cdot 19\cdot 53$, we see that $2014$ is divisible by exactly $16$ integers, namely $\pm 2^i 19^j 53^k$ with $i,j,k\in\{0,1\}$.  Thus there must be some divisor $d$ of $2014$ which occurs as $g(c)$ for at least $\lceil 2014/16\rceil=126$ values $c\in\{1,2,\dots,2013\}$.
I claim that $g(x)-d$ is divisible by $(x-1)(x-2)\dots (x-2013)$ in $\mathbf{Z}[x]$.  To show this, let $S$ 
 be the set of all $c\in\{1,2,\dots,2013\}$ for which $g(c)=d$.  Since $p(x):=\prod_{s\in S} (x-s)$ is monic, we can write $g(x)-d=p(x)q(x)+r(x)$ with $q,r\in\mathbf{Z}[x]$ and $\deg(r)<\deg(p)$.  But every element of $S$ is a root of both $p(x)$ and $g(x)-d$, and hence is a root of $r(x)$, so the number of roots of $r(x)$ is at least $\#S=\deg(p)$, and hence is bigger than $\deg(r)$, so we must have $r(x)=0$.  Now suppose that $S\ne\{1,2,\dots,2013\}$, and let $c$ be an integer with $1\le c\le 2013$ and $c\notin S$. Since $g(c)-d=p(c)q(c)$ where $d$ and $g(c)$ are distinct divisors of $2014$ (and where $p(c)$ and $q(c)$ are integers), it follows that $\lvert p(c)\rvert\le 4018$.  But $p(c)=\prod_{s\in S}(c-s)$ is the product of at least $126$ distinct nonzero integers, hence has absolute value at least $(63!)^2$, which is bigger than $4018$, a contradiction.  Therefore $S=\{1,2,\dots,2013\}$, so that $p(x)=(x-1)(x-2)\dots(x-2013)$ and $q(x)\in\mathbf{Z}[x]$ satisfy $g(x)-d=p(x)q(x)$.
The same argument shows that there is a divisor $D$ of $2014$ for which $h(x)-D=p(x)Q(x)$ with $Q(x)\in\mathbf{Z}[x]$ and $p(x)=(x-1)(x-2)\dots (x-2013)$ as above.  In particular, since we assumed that both $g(x)$ and $h(x)$ are nonconstant, it follows that both of them have degree at least $2013$.  But their product has degree $4026$, so we must have $\deg(g)=\deg(h)=2013$, whence both $q(x)$ and $Q(x)$ are nonzero constants (which we will call $q$ and $Q$).  Hence we have
$$
p(x)^2 + 2014 = f(x) = g(x)\cdot h(x) = (d+p(x)\cdot q)\cdot (D+p(x)\cdot Q).
$$
It follows that $y^2+2014=(d+yq)(D+yQ)$ (since this is an equality of degree-$2$ polynomials which is valid for more than two values $y$, namely for any value $y=p(x)$).  But this is impossible, since $\sqrt{-2014}$ is irrational.  This contradiction shows that $f(x)$ is irreducible in $\mathbf{Z}[x]$, and hence (by Gauss's lemma) in $\mathbf{Q}[x]$.
