I am reading the paper "On an irreducibility criterion of M. Ram Murty" by Kurt Girstmair.
I have a question regarding an assumption of the proof, which I have stated at the end of the post.
Theorem: Let $f(x)=a_mx^m+\cdots+a_{m-1}x^{m-1}+\cdots+a_1x+a_0$ be a primitive polynomial of degree $m \ge 1$ in $\mathbb{Z}[X]$ and set $H$ as $$ H:=\operatorname{max}_{0 \ \le i\le \ m-1} \left|\frac{a_i}{a_m}\right| $$ If for some natural number $d$ , $n$ and a prime $p$ we have $$ f(n)=\pm d.p \ \ \ \ \ ; \ \ \ \ \ H+d+1 \le n \ \ \ \text{and} \ \ \ p \ \ \text{does not divide} \ \ d $$
then $f(x)$ is irreducible in $\mathbb{Z}[X]$.
Proof:
1. $f(x)=g(x)h(x)$, where $g(x)$ and $h(x)$ are of positive degree. $$ f(n)=g(n)h(n)=\pm d\cdot p \ \ \ \text{and} \ \ \ p \nmid d $$ One of the factors $g(n)$ or $h(n)$ cannot divisible by $p$, and therefore must divide $d$. WLOG we assume $g(n) \mid d$
2. Now we know that $g(x)$ is of the form $$g(x)=c \prod_i (x-\alpha_i)$$ where $c$ : leading coefficient of $g \ \ $ & $\ \ \alpha_i$ runs over a subset of zeros of $f$.
3. For each root $\alpha$ of $f(x)$ we have $$ \mathbf{\mid n- \alpha \mid} \ge n-|\alpha| \ \ > \ \ H+d+1-(H+1)=\mathbf{d} $$ Therefore we contradict the fact $\ \ g(n) \mid d \ \ $ as $$ \mathbf{|g(n)|} =|c| \prod_i |n-\alpha_i| \ \ \mathbf{> \ d} $$
Hence we arrive at a contradiction.
My Question:
Where are we using the fact that $f$ is a primitive polynomial?