Let $R$ be an arbitrary ring with maximal ideal $I$, such that the quotient ring $\frac{R}{I}$ is finite. For each integer $i$, we define the quotient ring number as below:
$$N\_i(R,I)=\sum\_{a\in \frac{R}{I}}{a^i}.$$


It is easy to see that if $\frac{R}{I}=\mathbb{F}\_q$ be a finite field with $q$ elements, then we have two case for $N\_i(R,I)$. Actually, if $i$ is not a multiple of $q-1$, this summation is zero and if $i$ is a multiple of $q-1$, this summation is $-1$. So, for commutative ring with maximal ideal $I$, we have:
$$N\_i(R,I)\in \{0,-1\}.$$


Now, my questions is:


Is there a non-commutative ring $R$ with maximal ideal $I$ such that we have $N\_i(R,I)\in \{0,-1\}$?


Also, for arbitrary non-commutative ring $R$ with maximal ideal $I$, what can be the maximum number of elements of the set $\{N\_i(R,I); i\in \mathbb{N}\}$?



Quotient Ring number
The definition of $N\_i(R,I)$ depends only on $R/I$, which is a ring isomorphic to $M\_n(q)$ as HeinrichD has already observed.
Thus, the question is: what is the sum
$$
S = \sum\_{A\in M\_n(q)} A^i \quad?
$$
The answer that has been proposed is: either $0$ or $-I$.


This answer is correct. It is proved in


Brawley, J. V.; Carlitz, L.; Levine, J.
Power sums of matrices over a finite field.
Duke Math. J. 41 (1974), 9-24.


that, if $n>1$, the sum $S$ is $0$ unless 
$n=q=2$, $i>1$, and $i\equiv -1, 0, 1\pmod{6}$, in which 
case $S=-I=I$ is the $2\times 2$ identity matrix over the $2$-element field.


