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I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks

Show that the polynomial $h(x) = (x-1)(x-2)\cdots(x-n) + 1$ is irreducible over $\mathbb Z$ for all $n\ge1$ and $ n\ne4$.

Suppose $h(x) = f(x) g(x)$, then we must have $f(i)g(i) = 1$ for all $i = 1,2,...n$. So both $f(i)$ and $g(i)$ are $1$ or $-1$. In either case, $m(x) = f(x) - g(x)$ has degree smaller than $n$ and have $n$ different roots ($1,2,...,n$). So we must have $m(x) = 0$. Then $h(x) = f(x)^{2}$. So $n$ must be even. Let $n = 2k$. Because $f(x)$ has degree $k$, there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.

This is where I got stuck. Hope some one can help me solve this. Thanks.

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    $\begingroup$ Great work. Now use the fact that $a-b | f(a) - f(b)$ for integer polynomials. $\endgroup$ – Calvin Lin Sep 24 '13 at 17:27
  • $\begingroup$ Oh, I forgot that property. Thanks so much. Got it clear right now... $\endgroup$ – le duc quang Sep 24 '13 at 17:31
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    $\begingroup$ In that case, please write up an answer, so that others can learn from it. $\endgroup$ – Calvin Lin Sep 24 '13 at 17:32
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    $\begingroup$ @DietrichBurde I think you misunderstand. The intention is for this question to also have a complete answer attached to it, so that if someone stumbles upon this they do not have to click elsewhere to read it. Further, note that the link you provided deals with $f(x) = \ldots -1$, opposed to this which has $+1$. $\endgroup$ – Calvin Lin Sep 24 '13 at 20:23
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    $\begingroup$ @Jim: Yes, you are right. Anyway, we have many duplicates also inside of MSE. See also here now. $\endgroup$ – Dietrich Burde Nov 19 '14 at 14:52
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AS @Calvin Lin suggest, I write down the complete answer here for someone who need.

From my deduction above, we must have $h(x) = f(x)^{2}$. And there are $k$ values from $\{1,2,...,2k\}$ at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$.

Let $I$ is subset of $\{1,2,...,2k\}$ consists of elements at which $f(x)$ is $1$ and $J$ is subset of $\{1,2,...,2k\}$ consists elements at which $f(x)$ is $-1$. It's easy to see that if $n = 2$, then $h(x)$ is irreducible, and $n \neq 4$, so we just consider the case $n \ge 6$, which means that $k \ge 3$.

Suppose $I$ consists of $1$. $J$ has at least $3$ distinct elements greater than $1$, so there exists an elment $u$ in $J$ such that $u - 1 \ge 3$. But we must have $u - 1 | f(u) - f(1) = -2$, contradiction.

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    $\begingroup$ Why must there be k values from ${1,2...,2k}$ at which at which $f(x)$ is $1$ and $k$ values at which $f(x)$ is $-1$. $\endgroup$ – user162089 Feb 28 '16 at 13:22
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For $n\geq 7$ the irreducibility follows from following criterion of Pólya (see Polynomials by Prasolov).

(Pólya). Let $f$ be a polynomial of degree $n$ with integer coefficients and define $m=\lfloor\frac{n+1}{2} \rfloor$. Suppose that, for $n$ different integers $a_1,\dots,a_n$ we have $|f(a_i)|<2^{-m}m!$ and the numbers $a_1,\dots,a_n$ are not roots of $f$. Then $f$ is irreducible.

Choose $a_i=i$ and for $n \geq 7$ the claim follows.

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