# Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree [duplicate]

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

## marked as duplicate by Sil, Xander Henderson, Jose Arnaldo Bebita-Dris, user334732, Claude LeiboviciMay 23 '18 at 9:19

• $p(i)=-1$ for the first bullet – ccorn Jan 2 '14 at 12:55
• why? I dont understand your point.... It would be $1$ :O Am i missing something? – user87543 Jan 2 '14 at 13:01
• Hint: How many zeros does $q(x)-r(x)$ have? – Jyrki Lahtonen Jan 2 '14 at 13:05
• That's what I thought. It does leave the possibility $q(x)=r(x)$, so there's more to it. – Jyrki Lahtonen Jan 2 '14 at 13:08
• @Prism : This is from Dummit Foote Abstract algebra.. – user87543 Jan 2 '14 at 13:31

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 $k>2$.

• Yes, precisely. – ccorn Jan 2 '14 at 14:49
• 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 2 '14 at 15:03
• please edit $p(x)^2$ to $p(x)$ (If that is so) That has confused me a lot :D – user87543 Jan 2 '14 at 15:11
• 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 2 '14 at 15:18
• 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 2 '14 at 15:20