Let $n>1$ be an integer. For $a_1,a_2,\ldots,a_n\in\mathbb{Z}$ with $a_1< a_2< a_3 < \dots < a_n$, prove that the polynomial $$f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)-1\,.$$ is irreducible in $\mathbb{Z}[x]$.
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Batominovski
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Suppose for the sake of contradiction that $f(x)=p(x)\,q(x)$ for some nonconstant monic polynomials $p(x),q(x)\in\mathbb{Z}[x]$. Without loss of generality, we assume that $\deg\big(p(x)\big)\leq \frac{n}{2}$. Observe that $p\left(a_i\right)\in\{-1,+1\}$ for all $i=1,2,\ldots,n$. Let $$X:=\big\{i\in\{1,2,\ldots,n\}\,\big|\,p\left(a_i\right)=-1\big\}\text{ and }Y:=\{1,2,\ldots,n\}\setminus X\,.$$ Then, either $|X|\geq \frac{n}{2}$ or $|Y|\geq \frac{n}{2}$.
If $|X|>\frac{n}{2}$ or $|Y|>\frac{n}{2}$, then $p(x)$ is constant at more than $\frac{n}{2}\geq \deg\big(p(x)\big)$ values of the inputs $x$, but this contradicts the assumption that $p(x)$ is not a constant polynomial. That is, $$|X|=|Y|=\frac{n}{2}$$ (and $n$ is even). Therefore, $$p(x)=\prod_{i\in X}\,\left(x-a_i\right)-1\,.$$ Also, we have $$q(x)=\prod_{i\in X}\,\left(x-a_i\right)+1\,.$$ You should see a contradiction at this point.
Batominovski
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Why $q(x)$ is of this form ? Also then what will be the contradiction ? – user371231 Sep 07 '18 at 18:10
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This is an alternative answer to your question.
Consider the polynomial $f^2$ instead. So $f^2|_{\{a_i\}}=1$. If $f$ is reducible over $\mathbb{Z}[x]$, then it must be reducible to $f^2=g^2h^2$ for $g,h \in \mathbb{Z}[x]$. So $f^2=g'h'$. Suppose $\deg(g')\leq n$. $g'|_{\{a_i\}}=1$ as $g$ is positive and $g'(a_i)$'s are all invertibles in $\mathbb{Z}$. So $(g'-1)|_{\{a_i\}}=0$ implies $a_i$ are all its roots. Since $\deg(g)\leq n$, $g'=\prod_i(x-a_i)+1$ by Bézout's theorem. Similarly for $h'$. Now $h'g'\neq f^2$. Thus $f$ is irreducible over $\mathbb{Z}[x]$.
