Factorize $f(x)=x^p−x $ into a product of irreducibles in $\mathbb ℤ_{p}[x]$

Question

So I started with
$$p=2, f(x_2)=x^2-x=x(x-1);$$ $$p=3, f(x_3)=x^3-x=x(x-1)(x+1)=x(x-1)(x-2)...\pmod 3$$

Then I need to prove it generally but taking $ p=n $ or $ n+1 $. Then I'm stucked... Please help!

Answer 1

By Fermat's little theorem, $a^p\equiv a\pmod p$ for all $a$, hence $f(a)=0$ for $a=0, 1,\ldots, p-1$, hence $x(x-1)(x-2)\cdots(x-p+1)$ divides $f$ and must in fact be equal.

Answer 2

Hint $\ $ Every element of $\,\Bbb Z_p\,$ is a root of $\rm\:f(x) = x^p\!-x,\:$ by little Fermat. Hence, invoking the Factor Theorem, we infer that $\rm\:f(x)\:$ is divisible by $\rm\:x\!-\!1,\, x\!-\!2,\, \cdots, x\!-\!p.\:$ Therefore, we deduce that $\rm\:f(x)\:$ is divisible by their lcm = product (since they are nonassociate primes in $\rm\,\Bbb Z_p[x]).$

Remark $\ $ The final inference depends crucially on the fact that $\,\Bbb Z_p\,$ is a domain. Otherwise it may fail spectacularly, e.g. over $\rm\,\Bbb Z_8\,$ the quadratic polynomial $\rm\:x^2-1\:$ has four roots $\rm\:x = \pm1,\, \pm 3,\:$ but it is not divisible by $\rm\:(x\!-\!1)(x\!+\!1)(x\!-\!3)(x\!+\!3).\:$ It's not even divisible by $\rm\ (x\!+\!1)(x\!+\!3) = x^2\!+\!4x\!+\!3.$