Factor A Degree 100 Polynomial (III)

Relevant wiki: Eisenstein's Irreducibility Criterion

This one falls to Eisenstein's Irreducibility criterion. Note that $101$ is prime. If $1 \le j \le 100$ then $j^{100} \,\equiv\, 1 \pmod{101}$ , and hence $x^{100} - 1 \; \equiv \; \prod_{j=1}^{100}(x-j) \pmod{101}$ Thus if we write $f(x) \; = \; \prod_{j=1}^{100}(x-j) + 1 + 101! \; = \; \sum_{j=0}^{100} a_j x^j$ then the prime $101$ divides all of $a_0,a_1,\ldots,a_{99}$ , while $101$ does not divide $a_{100} = 1$ , and $101^2$ does not divide $a_0 = 100! + 101! + 1$ . Thus $f(x)$ is irreducible over $\mathbb{Z}[x]$ .