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
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 $k>2$.