Factor A Degree 100 Polynomial (II)

If we can write this polynomial as a product $A(x)B(x)$ of two nonconstant polynomials in $\mathbb{Z}[x]$ , then $A(j)B(j) = 1$ for all $1 \le j \le 100$ , so that $A(j) = B(j) = \pm 1$ for all $1 \le j \le 100$ . Thus $A(x) - B(x)$ is a polynomial of degree at most $99$ which has at least $100$ distinct zeros. Thus $A(x) - B(x)$ must be identically zero, and hence $A(x) \equiv B(x)$ , so that $f(x) \; \equiv \; \prod_{j=1}^{100}(x-j) + 1 \; \equiv \; A(x)^2$ But this would imply that $f(99.5) \,=\, A(99.5)^2 \ge 0$ , while in fact $f(99.5) \approx -2.6426 \times 10^{154}$ . No such factorization is possible.