Reducible polynomials in $\mathbb{Z}[X]$ 
Let $(a_n)_{n\geq 1}$ be a strictly increasing sequence of integers and $k$ an integer different from $0$. There exists among the polynomials 
  $$
(X-a_1)(X-a_2)\cdots(X-a_n)+k,\ n\geq 1
$$
  only a finite number of reducible polynomials in $\mathbb{Z}[X]$? 

I think this must be true, but I have yet no proof. Any idea / solution? 
Thank you!
Nicholas Tatsis  
 A: If $n$ is sufficiently big, then $f(x)=k+\prod_{i=1}^n (x-a_i)$ is irreducible in
$\Bbb{Z}[x]$.
For $n=2R$ sufficiently big means
$$
|k|<\frac 14 (R-1)!\quad\text{and}\quad 2|k| R^3< (\lfloor R/2\rfloor-1)!.
$$
In the odd case one should obtain similar bounds.
$\bf{First\ Step:}$ There exist $b_i\in\Bbb{R}$ for $i=1,\dots,n$ such that
$$
f(x)=\prod_{i=1}^n(x-b_i)\quad\text{and}\quad \Delta_i:=|a_i-b_i|<\frac 12.
$$
${\bf{Proof\ of\ the\ First\ Step:}}$ We will do only the case $k>0$ and $n$ even, since the other cases are similar. Set $c_{2i+1}:=a_{2i+1}+\frac 12$. Then
$f(c_{2i+1})<0$. In fact, by elementary considerations on the sign of $f_0(x):=\prod_{i=1}^n (x-a_i)$ we have $f_0(c_{2i+1})<0$.
 There are at most two $a_j$'s with $|c_{2i+1}-a_j|=1/2$, for all the other $a_j$'s we have $|a_j-c_j|>1$. Moreover
either we have at least $R$ of the $a_j$'s to the right or at least $R$ to the left of $c_{2i+1}$, and the differences
$|a_j-c_{2i+1}|$ are different for all of them, so one obtains that
$$
|f_0(c_{2i+1})|=\prod_{j=1}^n|c_{2i+1}-a_j|\ge \frac 14 (R-1)!>k,
$$
so $f(c_{2i+1})=f_0(c_{2i+1})+k<0$, as desired. Since $f(a_{2i+1})=k>0$, we obtain
 $b_{2i+1}\in\ ]a_{2i+1},c_{2i+1}[$ with $f(b_{2i+1})=0$.
Similarly one obtains $f(c_{2i})<0$ for $c_{2i}:=a_{2i}-\frac 12$, and so there
exists $b_{2i}\in\ ]c_{2i},a_{2i}[$ with $f(b_{2i})=0$.
So we have found $b_i$ for $i=1,\dots,n$ with $f(b_i)=0$ and $|a_i-b_i|<\frac 12$. Since the intervals $]a_{2i+1},c_{2i+1}[$,
$]c_{2i},a_{2i}[$, are pairwise disjoint, the $b_i$'s are all different, hence, since the degree of $f$ is $n$ and $f$ is monic, we must have
$$
f(x)=\prod_{i=1}^n(x-b_i).
$$
$\bf{Second\ Step:}$ We have
$$
\Delta_i=\frac{|k|}{\prod_{j:j\ne i}|b_j-a_i|}
$$
$\bf{Proof\ of\ the\ Second\ Step:}$
Set $\bar f(x):=f(x+a_i)$. On one hand $\bar f(0)=f(a_i)=k$, and on the other hand $\bar f(x)=\prod_{j}(x+a_i-b_j)$ and so
$\bar f(0)=\prod_j(a_i-b_j)=(a_i-b_i)\prod_{j:j\ne i}(a_i-b_j)$. It follows that
$$
|k|=|a_i-b_i|\prod_{j:j\ne i}|a_i-b_j|,
$$
and so $\Delta_i=|a_i-b_i|=\frac{|k|}{\prod_{j:j\ne i}|b_j-a_i|}$, as desired.
$\bf{Third\ Step:}$
In the case $n=2R$, we can assume $a_R=0$ ($x\mapsto x+a_R$), and then
$$
\prod_{j:j\ne i}|b_j-a_i|\ge \frac 12 |a_i|^{R-1}(\lfloor R/2\rfloor-1)!,\quad\text{for $j\ne R$}
$$
and $\prod_{j\ne R}|b_j|\ge \frac{(\lfloor R/2\rfloor-1)!}{2R^2}$. Hence
$$
\Delta_i<\frac{2|k|}{|a_i|^{R-1}(\lfloor R/2\rfloor-1)!}\text{, for $i\ne R$, and}\quad
\Delta_R<\frac{2|k|R^2}{(\lfloor R/2\rfloor-1)!}.
$$
$\bf{Proof\ of\ the\ Third\ Step:}$
We detail the proof for $i>R$: There are $R-1$ numbers $a_j$ with $a_j<a_R=0$ and hence we also have $b_j<0$ for that $j$'s.
Hence $\prod_{j<R}|b_j-a_i|>|a_i|^{R-1}$. There are now $R$ numbers $a_j$ with $j\ne i$ and $j\ge R$. For these numbers, at
most two of them yield equal differences $|a_j-a_i|$, hence there are at least $\lfloor R/2\rfloor $ different positive integers $|a_j-a_i|$ among them.
But there is at most one $j\ne i$ with $\frac 12 <|b_j-a_i|<1$, and for all others $|b_j-a_i|>1$. Moreover, $|a_j-a_i|-1<|b_j-a_i|<|a_j-a_i|+1$, and  hence $\prod_{j\ge R}|b_j-a_i|>\frac 12(\lfloor R/2\rfloor-1)!$. Together with the previous result we obtain
$$
\prod_{j,j\ne i}|b_j-a_i|\ge \frac 12 |a_i|^{R-1}(\lfloor R/2\rfloor-1)!.
$$
For $i<R$ the proof is similar, and for $j=R$ we use $\prod_{j\ne R}|b_j|\ge \frac 12 (R-1)!$ and the fact that $(R-1)!\ge \frac{(\lfloor R/2\rfloor-1)!}{R^2}$.
$\bf{Fourth\ Step:}$
Assume by contradiction that $f$ is reducible and write $f(x)=g(x)h(x)$ with $g(x),h(x)\in \Bbb{Z}[x]$ and $1\le \deg(g(x))\le R$.
Then $g(x)=\prod_{i\in J}(x-b_i)$ for some subset $J\subset\{1,2,...,n\}$ with $|J|=r\le R$. But then all symmetric elementary polynomials
$e_m(b_i,\ i\in J)$, for $m\le r$, are in $\Bbb{Z}$, since they are the coefficients of $g(x)$. Hence all power sums
$p_m(b_i,\ i\in J)=\sum_{i\in J}b_i^m\in\Bbb{Z}$, for $m\le r$. Now for $i\ne R$ we have
$$
|b_i^m-a_i^m|=|(a_i\pm \Delta_i)^m-a_i^m|\le \sum_{j=0}^{m-1}\left| \binom mj a_i^j\Delta_i^{m-j}\right|\le m\left| \binom m{m-1} a_i^{m-1}\Delta_i\right|
\le R^2|a_i|^{R-1}\Delta_i,
$$
where the inequality in the middle follows from the fact that $| \binom mj a_i^j\Delta_i^{m-j}|\le |\binom m{j+1} a_i^{j+1}\Delta_i^{m-j-1}|$
(a straightforward computation shows that $\Delta_i<1/R\le 1/m$).
Hence
$$
\sum_{i\in J\setminus\{R\}}|b_i^m-a_i^m|\le R^2\sum_{i\in J\setminus\{R\}}|a_i|^{R-1}\Delta_i,
$$
and since $|a_R^m-b_R^m|=\Delta_R^m\le \Delta_R$, we obtain from the third step
$$
\sum_{i\in J}|b_i^m-a_i^m|\le \frac{2|k| R^3}{(\lfloor R/2\rfloor-1)!}<1.
$$
Since $\sum_{i\in J}a_i^m,\sum_{i\in J}b_i^m\in\Bbb{Z}$, it follows that $\sum_{i\in J}a_i^m=\sum_{i\in J}b_i^m$, for all $m=1,\dots,r$. Hence
$e_m(b_i,\ i\in J)=e_m(a_i,\ i\in J)$
 and so
$g(x)=\prod_{i\in J}(x-a_i)$, which is a contradiction.
