Showing that $f(x) \mid f(x^{m + i}) - f(x^{m})$ given that $f(x) \mid x^m - 1$ Hypothesis: 


*

*$f = a_0 + a_1 x + \ldots + a_n x^n \in \mathbb{Z}[x]$

*$g = x^m - 1 \in \mathbb{Z}[x]$

*$f \mid g$

*$f^{(m)} = f(x^m) = a_0 + a_1 x^m + \ldots + a_n (x^m)^n  \in \mathbb{Z}[x]$

*$f^{(m+i)} = f(x^{m + i}) = a_0 + a_1 x^{m+i} + \ldots + a_n (x^{m+i})^n \in \mathbb{Z}[x]$


Goal: Show that $f \mid f^{(m+i)} - f^{(m)}$ given that $f \mid g$.
Attempt:


*

*$f \mid f^{(m+i)} - f^{(m)}$ iff $f \mid f^{(m)}$ and $f \mid f^{(m+i)}$.

*If we can show $g \mid f^{(m)}$ and $g \mid f^{(m+i)}$, then our desired result will follow.

*Then our goal is to now show that $g \mid f^{(m)}, f^{(m+i)}$.


But why should (3) be true?
 A: This is not true as it stands: for a counterexample consider $m=2,\ f(x):=x+1$ and $i=1$. Then $f(x^{m+i})=f(x^3)=x^3+1$ and $f(x^m)=f(x^2)=x^2+1$, so $f(x^3)-f(x^2)=x^3-x^2=x^2(x-1)$ which is not dividable by $x+1$.
You probably want to conclude instead that $f(x)\ |\ \,f(x^{m+i})-f(x^i)$.
Your attempt 1. is wrong!!! Just consider divisablity among numbers: $3\,|\,4-1$ though $3$ neither divides $4$ nor $1$.
Instead, perhaps simplest way is to consider the quotient ring $R:=\Bbb Z[x]\,/\,(f(x))$. The elements of this quotient ring are basically those of $\Bbb Z[x]$, only that $f(x)=0$ (and all its consequences -- using ring structure) holds in $R$. In particular, as $x^m-1=h(x)\cdot f(x)$ for some polynomial $h\in\Bbb Z[x]$, we have that 
$$x^m=1\quad\text{ in }R\,.$$
Then, in $R$ we have
$$f(x^{m+i}) =\ 
a_0+a_1x^mx^i+a_2(x^m)^2x^{2i}+\dots a_n(x^m)^nx^{ni} = \\
= a_0+a_1x^i+a_2x^{2i}+\dots a_nx^{ni} \ =\ f(x^i)
$$
So, we conclude $f(x^{m+i})-f(x^i)=0$ in $R$ but that proves exactly the statement.

 (Alternatively, we can basically use that $x^m-1\ |\ (x^m)^k-1$.)

