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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.

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    $\begingroup$ Hi and welcome to MSE! Your question doesn't show any effort of your own. What have you tried? Did you consider examples or special cases? Have a look at How to ask a good question and edit your question to make improvements. As of now, the question is likely to get closed. $\endgroup$
    – Christoph
    Dec 26, 2020 at 11:25
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    $\begingroup$ Maths Lover’s question is very interesting and shows efforts to answer it, indeed the user says he is unable to solve it and it means he tried but he could not. Hence, there is not any reason to close his question. $\endgroup$
    – Angelo
    Dec 26, 2020 at 11:27
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    $\begingroup$ @Angelo Question authors should share their own thoughts on the problem and if they made failed attempts, include those in the question as well. This post doesn't show any effort whatsoever. Writing "I can't solve this" is not showing effort. $\endgroup$
    – Christoph
    Dec 26, 2020 at 11:32
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    $\begingroup$ @Christoph I agree with you. In fact, those are the rules here at MathSE. $\endgroup$
    – user798113
    Dec 26, 2020 at 11:33
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    $\begingroup$ Maths Lover is a new user and it is obvious that even now he does not manage to solve questions that are difficult for him and also he does not manage to express his efforts because he does not know the rules quite well so far. That is why we should help him instead of closing his questions and ignoring his efforts. $\endgroup$
    – Angelo
    Dec 26, 2020 at 11:41

2 Answers 2

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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 $P+Q$.

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  • $\begingroup$ I am unable to prove that each ai is a root of P+Q. Please help. $\endgroup$ Dec 26, 2020 at 11:35
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    $\begingroup$ What are the possible values of $P(a_1)$ and $Q(a_1)$? $\endgroup$
    – Aphelli
    Dec 26, 2020 at 11:38
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    $\begingroup$ That is almost certainly not correct. Again, what can you say about $P(a_1)$ and $Q(a_1)$? $\endgroup$
    – Aphelli
    Dec 26, 2020 at 12:06
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    $\begingroup$ Maths Lover, can you tell me if my answer is clear and understandable? $\endgroup$
    – Angelo
    Dec 26, 2020 at 12:21
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    $\begingroup$ $P(a_1)+Q(a_1)\;$ is not a polynomial, it is an integer number, so you do not need to consider any root $\;m\;$. You just have to notice that if the product of two integers $\;P(a_1)\;$ and $\;Q(a_1)\;$ is equal to$\;-1\;$ there are only two possibilities that are $$P(a_1)=1\quad\land\quad Q(a_1)=-1$$ or $$P(a_1)=-1\quad\land\quad Q(a_1)=1$$ $\endgroup$
    – Angelo
    Dec 26, 2020 at 12:53
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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:

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

  2. $\;P(a_i)=-1\;$ 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.

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    $\begingroup$ This is nice but I think there should be some sharpening there: since $\;\deg(P+Q)<n\;$ and $\;(P+Q)(a_i)=0\;\;\forall\,i=1,...,n\implies (P+Q)(x)\equiv 0\;$ , and this possibility hasn't yet been covered above...This is a small nitpick, yet the OP may want to address it. $\endgroup$
    – DonAntonio
    Dec 26, 2020 at 13:18
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    $\begingroup$ I have discarded the possibility that $(P+Q)(x)\equiv0$ in the beginning of my answer when I wrote that $P(x)$ and $Q(x)$ are monic polynomials so it is impossible that $P(x)\equiv -Q(x)\;.$ $\endgroup$
    – Angelo
    Dec 26, 2020 at 13:38

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