TITLE: The sum of same powers of all matrices modulo p
QUESTION [20 upvotes]: The following is a problem from our department algebra competition for
students:

Non-question.
   An experimental-math geek was trying to raise all matrices $17\times17$
  over the field with 17 elements to the power of 100, sum the returns,
  and observe the result, when his computer broke. Help him.

Probably, the most of us can calculate the sum without use of computer
(alternatively, Russian readers can find a solution
here).
The problem seems to become much
harder when we replace 100 by, say, 80. More generally, my question is the
following:

Question.
  What is the sum
  $$
> \sum_{A\in M_p(\mathbb F_p)}A^k,
> $$
  where $p$ is prime and $k$ is a multiple of $p-1$?

It is easy to show that the sum is a scalar matrix, but which one?
Note that the coincidence of the matrix size and the characteristic
makes trace useless.

REPLY [14 votes]: [corrections applied, per Ilya and Anton]
Consider the formal series
$$
f(x) = \sum_{k=0}^\infty (\sum_A A^k) x^k
$$
where $A$ runs through $p \times p$ matrices.  It is equal to $\sum_A (I - Ax)^{-1}$, a rational function with values in scalar matrices.  Thus, for some $d$, if the first $d$ coefficients vanish all coefficients must vanish.
To determine $d$, we need to bound the degree of the numerator and denominator of $f(x)$.  Note by Cramer's rule, each $(I - Ax)^{-1}$ is a matrix of degree $(p-1)$ polynomials divided by the degree $p$ determinant of $(I - Ax)$.
There are $p^p$ different denominators, i.e. polynomials of the form 
$$
\det(I - Ax) = c_p x^p + c_{p-1} x^{p-1} + \cdots + c_1 x + 1
$$
The $c_i$ are unconstrained, they appear at least once in case $A$ is the companion matrix with $-c_p,-c_{p-1},\ldots$ along the last column.  So we can write the sum as
$$
\sum_{c_p,\ldots,c_1} \frac{\text{something of degree $p-1$}}{c_p x^p + \cdots + c_1 x + 1} 
$$
There are $(p-1)$ linear denominators, $p (p-1)$ quadratic denominators, $p^2 (p-1)$ cubic denominators and so on, so the product of all the $c_p x^p + \cdots + 1$ has degree
$$
d = (p-1) + 2p(p-1) + 3p^2(p-1) + \cdots + p p^{p-1}(p-1)
$$
i.e. $d = p^{p+1} - \frac{p^p - 1}{p - 1}$.  So $f(x)$ is a polynomial of degree less than $d$ divided by a polynomial of degree $d$.
Ilya's answer shows that the first $p^{p+1} - p$ Taylor coefficients of $f(x)$ vanish, and when $p \geq 3$ this is larger than $d$ so all the remaining Taylor coefficients are vanish as well.  When $p = 2$ Will points out the sums are not always zero, in fact for $p=2$
$$
f(x) = \frac{x^5}{(x+1)^2 (x^2 + x + 1)} I
$$