Vrftsjtryk

I've done some (maybe incorrect) algebra, which has led me to a strange equality:
$$frac{1}{x^p-x}$$
$$frac{1}{(1+(x+p-1))^p+(p-1)x}$$
$$frac{1}{1+p(x+p-1)+cdots+p(x+p-1)^{p-1}+(x+p-1)^p+(p-1)x}$$
$p(x+p-1)+cdots+p(x+p-1)$ are $0$ in $mathbb{Z}_p$.
$$frac{1}{(x-1)^p +1 -x}$$
$$frac{1}{(x-1)^p -(x-1)}$$

Which gave me an idea to use induction on $x$ (I don't know if this idea is legal), but I'm struggling to find the base.

I also see that $frac{1}{x^p-x}=frac{1}{x(x^{p-1}-1)}$, but that doesn't seem to lead anywhere.

The answer in the book is $-sum_{a=0}^{p-1}frac{1}{x-a}$.

Hints or full answers, I will be very grateful for any help.

Thank you.

You cannot use induction on $x$, since in $mathbb Z_p[x]$, $x$ does not actually represent a number, let along an integer; it is merely a placeholder, or an algebraic object. You can only induct over things like integers or natural numbers, so this is out of the question.

Here is a more beneficial approach. Because $x^p-x$ equals $0$ for all $xinmathbb Z_p$, you may factor it as $x(x-1)...(x-p+1),$ which suggests (if you are familiar with partial fractions) that the decomposition is of the form
$$frac{a_0}{x}+frac{a_1}{x-1}+...+frac{a_{p-1}}{x-p+1}$$
But by symmetry (or, by substituting $xto x+1$), you can see that all of these coefficients must be equal. So instead, you have a decomposition in the form
$$a_0bigg(frac{1}{x}+frac{1}{x-1}+...+frac{1}{x-p+1}bigg)$$
In which case all you have to do is show that this constant $a_0$ is equal to $-1$. Does this help?

I shall find the partial fraction decomposition of $$f(x):=frac{1}{x^q-x}$$
over $mathbb{F}_q$, where $q=p^r$ for some prime natural number $p$ and for some positive integer $r$. In particular, when $r=1$, $mathbb{F}_q=mathbb{F}_p=mathbb{Z}_p$, which is what the OP asks for.

Note that $x^q-xinmathbb{F}_q[x]$ factors into linear factors $prodlimits_{tinmathbb{F}_q},(x-t)$, with each linear factor occurring with multiplicity $1$. Therefore, the partial fraction decomposition of $f(x)$ is a sum of simple fractions
$$f(x)=sum_{tinmathbb{F}_q},frac{s_t}{x-t}$$
for some $s_tinmathbb{F}_q$ for each $tinmathbb{F}_q$.

Now, $g_tau(x):=dfrac{1}{(x-tau),f(x)}$ is a polynomial in $mathbb{F}_q[x]$ for each $tauinmathbb{F}_q$. It can be easily seen that $g_tau(tau)$ is the (first) derivative of $x^q-x$ evaluated at $x:=tau$, which is $-1$. From
$$frac{1}{g_tau(x)}=s_tau+sum_{tinmathbb{F}_qsetminus{tau}},frac{s_t(x-tau)}{x-t}text{ for all }tauinmathbb{F}_q,,$$
we evaluate this expression at $x:=tau$ to get
$$s_tau=-1text{ for every }tauinmathbb{F}_q,.$$
This shows that
$$frac{1}{x^q-x}=f(x)=-sum_{tinmathbb{F}_q},frac{1}{x-t},.$$