Let $$f(x)=a_nx^n + a_{n-1}x^{n-1}+...+a_0$$ be a polynomial with integer coefficients. If there exist a prime number $p$ such that $$a_{n-1} \equiv a_{n-1} \equiv...\equiv a_0 \equiv 0 \pmod p$$ but $a_n \not\equiv 0 \pmod p $ and $a_0 \not\equiv 0 \pmod {p^2}$, then $f(x)$ is irreducible over the field of rational numbers.
I understand the theorem and the proof, its elegant and simple. However, I am having troubles to understand the motivation of the theorem, I simply can't see where it comes from or how one gets the idea to formulate the theorem. Can anyone help me with these details?
Here’s an informal discussion based on $p$-adic theory. Consider the Newton polygon of the polynomial $f$ in question. It has a vertex at $(0,1)$ for the constant term, and at $(n,0)$ for the highest term. So there’s a single segment, of slope $-1/n$. In any factorization $f=gh$, the factor $g$, say of degree $d$ with $0<d<n$, will have some of the roots of $f$, so all of valuation $1/n$, but what will the vertices of $g$ be? They’d be at $(d,0)$ and $(0,d/n)$, but $d/n$ isn’t an integer, impossible.