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$$f(x)=\left(x-a_1\right)^2\left(x-a_2\right)^2\cdots\left(x-a_n\right)^2+1,$$ where $a_1,a_2,\dots,a_n$ are different integers.

Prove that $f(x)$ is irreducible over $\mathbb{Q}$ (rational field).

One thought:

$f(x)$ is a polynomial with integer coefficients, and assume $p$ is an root of $f(x)$, then $f(p)\geq 1$, there is no possible to make $f(p)=0$, so $f(x)$ is irreducible,

I think my thought is too easy, and is wrong...

user26857
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forlorn
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  • Sure, there is no way to make $f(p)=0$ (at least for real number $p$), but that doesn't imply that $f$ is irreducible. It could have non-trivial factors of degree greater than or equal to 2. – Prism Aug 07 '13 at 09:11
  • See this related question for links and discussion of a variant. If this is an exercise, it probably has a simpler solution. I have a vague recollection of it being done here... – Jyrki Lahtonen Aug 07 '13 at 09:18

1 Answers1

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By Gauss Lemma, it is enough to prove that $f$ is irreducible over $\mathbb{Z}$. Suppose $f(x)=p(x)q(x)$ for non-constant monic polynomials $p$ and $q$ with coefficients in $\mathbb{Z}$. For each $1\le i\le n$, evaluating at $x=a_i$, we see that there are only two cases:

Case 1. $p(a_i)=1$ and $q(a_i)=1$,

Case 2. $p(a_i)=-1$ and $q(a_i)=-1$.

Now suppose $p(a_i)=1$ and $p(a_j)=-1$ for some $i\neq j$. Then, by intermediate value theorem, we get that $p(b)=0$ for some $b$ between $a_i$ and $a_j$. However, this gives $0=p(b)q(b)=f(b)\ge 1$, contradiction.

Hence, $p$ must have take either value 1 for all $a_i$, or must take value $-1$ for all $a_i$. Suppose we are in Case 1. Now, $p(x)-1$ has at least $n$ roots, namely $a_1, a_2, ..., a_n$. Likewise for $q(x)-1$. By comparing the sum of the degrees of the terms in the equation $f(x)=p(x)q(x)$, we get that in fact $$p(x)-1=q(x)-1=(x-a_1)(x-a_2)\cdots (x-a_n)$$ Consequently, $$f(x)=(x-a_1)^2(x-a_2)^2\cdots (x-a_n)^2 + 1 = p(x)q(x)$$ $$=[(x-a_1)(x-a_2)\cdots (x-a_n)+1][(x-a_1)(x-a_2)\cdots (x-a_n)+1]$$ $$=(x-a_1)^2(x-a_2)^2\cdots (x-a_n)^2 + 2(x-a_1)(x-a_2)\cdots (x-a_n) + 1$$ We have shown that $(x-a_1)(x-a_2)\cdots (x-a_n)$ is the identically zero polynomial, contradiction. The same argument also applies if we are in Case 2.

I learnt the above solution from The USSR Olympiad Problem Book by Shklarsky, Chentzov, and Yaglom. This book is truly a gem.

Prism
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