38

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.

Batominovski
  • 49,629
le duc quang
  • 4,815
  • 8
    Great work. Now use the fact that $a-b | f(a) - f(b)$ for integer polynomials. – Calvin Lin Sep 24 '13 at 17:27
  • Oh, I forgot that property. Thanks so much. Got it clear right now... – le duc quang Sep 24 '13 at 17:31
  • 1
    In that case, please write up an answer, so that others can learn from it. – Calvin Lin Sep 24 '13 at 17:32
  • @CalvinLin: the solution is written up at many places. Helpful discussion is here: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2007;task=show_msg;msg=3779.0001.0001 – Dietrich Burde Sep 24 '13 at 17:56
  • 1
    @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 $+1$. – Calvin Lin Sep 24 '13 at 20:23
  • No, it does apply as well to $+1$. The text discusses two questions , the first with $-1$, the second with $+1$. And why not to click "elsewhere", where you can learn more ? – Dietrich Burde Sep 24 '13 at 20:49
  • There's no problem with clicking elsewhere, your link will remain in the comments for anyone who wants to see it. There's also no problem with having an answer here. – Jim Sep 24 '13 at 23:09
  • 1
    @Jim: Yes, you are right. Anyway, we have many duplicates also inside of MSE. See also here now. – Dietrich Burde Nov 19 '14 at 14:52
  • http://math.stackexchange.com/questions/579971/proof-that-a-polynomial-is-irreducible-for-all-n-ne-4 – user26857 May 31 '15 at 15:59
  • Alternative strategy: show f(n-1/2)<0. – Clay Thomas Nov 10 '16 at 15:55

2 Answers2

22

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.

Reinstate Monica
  • 5,209
le duc quang
  • 4,815
  • 2
    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 $-1$. – user162089 Feb 28 '16 at 13:22
  • @user162089 suppose, for instance, there are at least $k+1$ distinct values at which $f(x)$ is $1$. Then in $\mathbb Q[x]$, $f(x)-1$ is divisible by (at least) $k+1$ distinct linear factors since $\mathbb Q$ is a field; but this is impossible because that would imply that $f(x)-1$, and hence $f(x)$, would have at least degree $k+1$. – Squirrel-Power Nov 24 '21 at 05:04
4

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.

Sil
  • 16,612