if $f$ has Newton Polygon consisting of one segment $(0,0)$ to $(n,m)$ with $m,n$ coprime, then $f$ cannot be factored Let $f(X)\in 1+ X\mathbb{Z}_p[X]$ have Newton Polygon consisting of one segment joining $(0,0)$ to $(n,m)$ with $m,n$ coprime. I have to show that $f(X)$ cannot be factored as a product of two polynomials with coefficients in $\mathbb{Z}_p$. 
I know that the slope of the Newton Polygon is $m/n$ and since $m,n$ coprime, this does not lie in $\mathbb{Z}_p$. I also know there exists a theorem which says something about the number of same slopes, but I do not fully understand this theorem. Can someone help me to understand this question and hopefully to understand the theorem better? Thanks!

If you understand that every root $\rho$ of $f$ satisfies $v(\rho)=-m/n$, then you see that this will happen for both of $g$ and $h$ if $f=gh$.
Now, what can the Newton polygon of $g$ be? It will be of width $r$ for some integer with $0<r<n$, since neither $g$ nor $h$ is constant. And the right-hand vertex? Since the slope is still $m/n$, the vertex will be at $(r, \frac{rm}n)$. But the $y$-coordinate has to be an integer if the polynomial has $\Bbb Z_p$-coefficients. Thus $g\notin\Bbb Z_p[X]$.
