If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers. 
If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers.

This question is from Pathfinder for Olympiad Mathematics by Vikash Tiwari and V. Seshan. I tried using Eisenstein's Irreducibility Criterion theorem or using contradiction method. I tried to equate it with $f(x)\cdot g(x)$ and prove that either $f(x)$ or $g(x)$ is equal to $1$ but I'm unable to prove that. Please help me with this.
 A: Hint: Write $PQ=-1+\prod\limits_{i=1}^n{(x-a_i)}$ with $P,Q$ monic with integer coefficients of degree less than $n$. Show that each $a_i$ is a root of $P+Q$.
A: 
If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1$ is irreducible over integers.

If that polynomial were reducible over integers, then there would exist two monic polynomial $\;P(x)\;$ and $\;Q(x)\;$ with integer coefficients of degree less than $\;n\;$ such that
$P(x)Q(x)=(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1\;.$
Consequently, it results that
$P(a_i)Q(a_i)=-1\quad\forall i\in\{1,2,\ldots,n\}\;,$
but $\;P(a_i)\;$ and $\;Q(a_i)\;$ are integers, so there are only two possibilities:

*

*$\;P(a_i)=1\;$ and $\;Q(a_i)=-1\;,$


*$\;P(a_i)=-1\;$ and $\;Q(a_i)=1\;.$
In any case we get that
$P(a_i)+Q(a_i)=0\quad\forall i\in\{1,2,\ldots,n\}\;,$
but it is impossible because $\;P(x)+Q(x)\not\equiv0\;$ (sum of monic polynomials) has degree less than $\;n\;,\;$ so according to the fundamental theorem of algebra, it cannot have $\;n\;$ different roots $\;a_1, a_2, \ldots, a_n\;.$
Hence, there do not exist such polynomials $\;P(x)\;$ and $\;Q(x)\;,\;$ consequently
$(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1\;$ is irreducible.
