Divisibility in $\mathbb F_p[x]$ For any prime $p$ and nonzero $a\in \mathbb{F}_p$ prove that for all positive integers $n$, $x^{p^n}-x+na$ is divisible by $x^p-x+a$ in $\mathbb{F}_p[x]$. Can anyone give me some hints to proceed?
 A: Well, note that $$x^{p^n} - x + n a = \sum_{i=1}^n(x^{p^i}-x^{p^{i-1}} + a)$$
I claim that every term of the sum is divisible by your polynomial. In fact, the first term of the sum is divisible by the second term, and so on. Better than that, the first term is the $p$-th power of the second term, and so on. Can you prove this?
A: Hint $\ {\rm mod}\ \ x^p\!-x+a:\ \color{#c00}{x^p\! \equiv x\!-\!a},\ $ therefore, by $\rm\color{#0a0}{induction,}$ and Frobenius $\rm\color{blue}{(Frob)}$
$\ \ \ \begin{eqnarray}&&\ x^{p^2}\! = (\color{#c00}{x^p})^p\equiv (\color{#c00}{x\!-\!a})^p \overset{\color{blue}{\rm Frob}}\equiv \color{#c00}{x^p}\! - a \equiv \color{#c00}{x\!-\!a}-a \equiv x\!-\!2a \\
&\Rightarrow&\ \cdots \\
 &\Rightarrow&\ x^{p^n}\! = (\color{#0a0}{x^{p^{n-1}}})^p \equiv (\color{#0a0}{x\!-\!(n\!-\!1)a})^p \overset{\color{blue}{\rm Frob}}\equiv \color{#c00}{x^p}\! - (n\!-\!1) a \equiv (\color{#c00}{x\!-a})-\!(n\!-\!1)a) \equiv x\!-\!na  \end{eqnarray} $
Thus $\ x^{p^n}\!-x+na\equiv 0\quad $ QED
