Factor A Degree 100 Polynomial

If this factorization were possible, we could find $A(x),B(x) \in \mathbb{Z}[x]$ , both with degrees less than $100$ , such that $A(x)B(x) \; \equiv\; \prod_{j=1}^{100}(x-j) - 1$ and hence $A(j)B(j) \; = \; -1 \hspace{2cm} 1 \le j \le 100$ Thus we can find $\epsilon_j \in \{-1,1\}$ for $1 \le j \le 100$ such that $A(j) \; = \; \epsilon_j \hspace{1cm} B(j) \; = \; -\epsilon_j \hspace{2cm} 1 \le j \le 100$ so that $A(j) + B(j) = 0$ for $1 \le j \le 100$ . Thus $A(x) + B(x)$ is a polynomial of degree at most $99$ with at least $100$ distinct zeros, and hence is identically zero. Thus $B(x) \equiv - A(x)$ , and so $-A(x)^2 \; \equiv\; \prod_{j=1}^{100}(x-j) - 1$ Putting $x=101$ , this would imply that $100! - 1 = -A(101)^2$ was negative, which is nonsense. Thus no such factorization exists.

$(x-1)(x-2)(x-3) \cdots (x-100)-1=p(x)q(x)$ where p(x) and q(x) are 2 non-constant polynomials with integer coefficients.

$p(1)q(1)=-1$ So, either $p(1)=1,q(1)=-1 \, \text{or}\, p(1)=-1,q(1)=1$

$p(2)q(2)=-1$ .So,either $p(2)=1,q(2)=-1 \, \text{or}\, p(2)=-1,q(2)=1$

Now it continues...

$p(100)q(100)=-1$ .So,either $p(100)=1,q(100)=-1 \, \text{or}\, p(100)=-1,q(100)=1$ .

Each p(i) or q(i) will be $\pm1$ where $1 \leq i \leq 100$ ..

We know for a polynomial $p(x)$ we know $(i-j) |p(i)-p(j)$ where $i,j$ are integers.Now we can find $i,j$ with $i-j >2,p(i)=1,p(j)=-1$ ,This will become $i-j>2$ does not divide $p(i)-p(j)=2$ .it will not satisfy above condition because p(x) is an polynomial with integer coefficients.So, there is no $p(x)$ exists.This logic is also valid for $q(x)$ .

Again if every $p(i)=1$ ,then we can write the $p(x)=(x-1)(x-2)....(x-100)+1 \\ =(x-1)(x-2)....(x-100)-1+2=p(x)q(x)+1$

That's not possible .So, there must be at least one $p(i)=-1$ .hence we can find $i,j$ for above logic.Same logic for q(x) also.

Again if every $p(i)=-1$ ,then we can write the $p(x)=(x-1)(x-2)....(x-100) -1=p(x)q(x)$

That's not possible because $q(x)$ will become constant polynomial. .So, there must be at least one $p(i)=1$ .hence we can find $i,j$ for above logic.Same logic for q(x) also.

That's why factorization is not possible.