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Question is to prove :

  • Irreducibility of $(x-1)(x-2)\cdots (x-n)- 1$ over $\mathbb{Z}$ for all $n\geq 1$
  • Irreducibility of $(x-1)(x-2)\cdots (x-n)+ 1$ over $\mathbb{Z}$ for all $n\geq 1$ and $n\neq 4$

Hint for first bullet is

If the polynomial factors consider the value of the factors at $x=1,2,\dots,n$

For second bullet :

Suppose $p(x)=(x-1)(x-2)\cdots (x-n)-1$ is reducible we have:

$p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}<n$

Hint is suggesting me to use that $p(i)=-1$ for all $1\leq i\leq n$

i.e., $q(i)r(i)=-1$ for all $1\leq i\leq n$

i.e., $q(i)=-1; r(i)=1$ or $q(i)=1;r(i)=-1$ for all $1\leq i\leq n$

For second bullet :

Suppose $p(x)=(x-1)(x-2)\cdots (x-n)+1$ is reducible we have:

$p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}<n$

Hint is suggesting me to use that $p(i)=1$ for all $1\leq i\leq n$

i.e., $q(i)r(i)=1$ for all $1\leq i\leq n$

i.e., $q(i)=r(i)=1$ or $q(i)=r(i)=-1$ for all $1\leq i\leq n$

I am getting some vague ideas but could not bind them to prove this.

I would be thankful if some one can help me to clear this.

Thank you.

P.S : Please give "just hints". Do not write whole answer at once. This is a "request". Thank you :)

Edit : I have changed the title from

Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$

to

Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree

for two reasons :

  • Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$ for any $n\geq 1$ implies Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree

  • The title looks atractive

1 Answers1

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Extended hints as requested

Assume a factorization $p(x)=q(x)r(x)$.

1: $p(x)=(x-1)(x-2)\cdots (x-n)-1$

As you observed, in this first case you get $q(i)=-r(i)=\pm1$ for all $i=1,2,\ldots,n$, so $q(x)+r(x)$ has at least $n$ zeros. This is a problem, because the leading coefficients of $q(x)$ and $r(x)$ have the same sign.

2: $p(x)=(x-1)(x-2)\cdots (x-n)+1$

In the second case $q(i)=r(i)=\pm1$ for all $i=1,2,\ldots,n$, so we get that $q(x)-r(x)$ has at least $n$ zeros. As observed in the comments this leaves only the possibility $q(x)=r(x)$. Indeed, when $n=4$ we have $$ p(x)=(x^2-5x+5)^2. $$ Anyway, the remaining case is that $n=2k$, $q(x)=r(x)$, $k=\deg q(x)$. The factorization $p(x)=q(x)^2$ shows that $p(x)\ge0$ for all real numbers $x$. Estimate $p(2k-\dfrac12)$ when $k>2$.

Jyrki Lahtonen
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  • Thank you for your interest... I am also thinking on that way... I am not getting any positive results... No hope :( –  Jan 02 '14 at 14:10
  • @Praphulla, Another idea came while dining with the family :-) – Jyrki Lahtonen Jan 02 '14 at 14:39
  • I did not get you... :) –  Jan 02 '14 at 14:40
  • Can you prove that $p(x)$ has a negative value at some point? In the second case, assuming $n=2k>4$. The case $n=2$ can be checked separately. – Jyrki Lahtonen Jan 02 '14 at 14:41
  • $p(1)=-1 <0$ so $p(x)$ have negative value at $1,2,\dots,n$ –  Jan 02 '14 at 14:43
  • This is only needed for the second polynomial, when $p(i)=1$ for all $i=1,2,\ldots,n$. – Jyrki Lahtonen Jan 02 '14 at 14:45
  • I do not understand your question.. Am i missing something? do you want me to prove that $p(x)=(x-1)(x-2)\dots (x-n)+1$ have negative value at some point? –  Jan 02 '14 at 14:48
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    Yes, precisely. – ccorn Jan 02 '14 at 14:49
  • @JyrkiLahtonen $p(-1)<0$ :O so. how does this lead me to conlcusion :O –  Jan 02 '14 at 14:57
  • $p(-1)<0$ only for odd $n$, and these are already ruled out by $n=2k$. Besides, the way to the conclusion is already in the answer. – ccorn Jan 02 '14 at 14:58
  • @ccorn Yes Yes you are correct $p(-1)<0$ only for odd $n$... –  Jan 02 '14 at 15:00
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    Jyrki has given $x=n-\frac{1}{2}$ to try: Look at $\cdots\frac{7}{2},\frac{5}{2},\frac{3}{2},\frac{1}{2},\frac{-1}{2}+1$. – ccorn Jan 02 '14 at 15:03
  • @Praphulla: If you have the time, plot $p(x)$ for $n=4,6,8$. You will see, why I suggested $n-\dfrac12$ :-) – Jyrki Lahtonen Jan 02 '14 at 15:09
  • I am extremely sorry, I was so confused... What i have understood is $p(x)=q(x)r(x)$ and as we have $q(x)=r(x)$ we would have $p(x)=(q(x))^2$ which means $p(x)$ is always positive but you have written $p(x)^2\geq 0$ There i have got confused by noyt knowing why did you write that... Now as we are getting $p(x)\geq 0$ for all $x$ we would end up with contradiction if we get some point where $p(x)$ takes negative value... and that is what i have to do –  Jan 02 '14 at 15:09
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    please edit $p(x)^2$ to $p(x)$ (If that is so) That has confused me a lot :D –  Jan 02 '14 at 15:11
  • OOps. Corrected. Sorry about leading you astray to that extent. – Jyrki Lahtonen Jan 02 '14 at 15:13
  • Now every thing is perfect except for two points... how did you guess that $n-\frac{1}{2}$ gives negative value for $p$ :O and.. In answer for first bullet you said "This is a problem, because the leading coefficients of q(x) and r(x) have the same sign." I do not understand this :O –  Jan 02 '14 at 15:16
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    As shown above, $x=n-\frac{1}{2}$ yields $p(x)<0$ because the product $(x-1)\cdots(x-n)$ has exactly one negative factor, and there are enough factors to make its absolute value greater than $1$, therefore the final $+1$ cannot change the sign any more. – ccorn Jan 02 '14 at 15:18
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    I plotted $p(x)$, when $n=6$ and $n=8$. You do the same (Wolfram Alpha is your friend), and you see this deep valley there. My first thought actually was to prove that $p(x)$ has $n$ zeros, but I had forgotten about that while trying to make that "modulo 4" argument work. – Jyrki Lahtonen Jan 02 '14 at 15:18
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    As for the leading coefficients, well if $q(x)=-r(x)$, the leading coefficient of $q(x)r(x)$ must be negative. But $p(x)$ has positive leading coefficient. – ccorn Jan 02 '14 at 15:20
  • @JyrkiLahtonen : oh yes... I have seen it just now.... Thank you so much for your wonderful support (as always) :) –  Jan 02 '14 at 15:20
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    @ccorn : I am so thankful for your wonderful support :) Please join in the club of people in MSE helping me from day 1... :D Thank you so much once again! –  Jan 02 '14 at 15:25