This question is, in a sense, homework. I'm taking a problem-solving seminar which uses questions like these, the first question on the 2010 Virginia Tech Regional Math Competition, as fodder. The syllabus tells me that correct solutions do not factor into grading, so asking this on MSE is kosher.
The exam asks about a particular case:
Let $d$ be a positive integer and let $A$ be a $d \times d$ matrix with integer entries. Suppose $I + A + A^2 + \cdots + A^{100} = O$ (where I denotes the identity $d \times d$ matrix, so $I$ has 1’s on the main diagonal, and $O$ denotes the zero matrix, which has all entries $0$). Determine the positive integers $n \le 100$ for which $A^n + A^{n+1} + \cdots + A^{100}$ has determinant $\pm 1$.
I'm really quite stumped on this one. Of course (multiplying by $I - A$) we will have $A^n = I$. Let me note explicitly that such $A$ do exist; we can put
$$A = \left[\begin{array}{cc} 1&-1\\ 3&-2 \end{array} \right]$$
and this $A$ satisfies the hypotheses for $n = 3$.
I have worked at this a bit but am really up a tree here. Does anybody have any nice hints to give me a push???
This is a nice problem, involving quite a few tricks. The answer turns out to be for all $1 \le n \le 100$, and one should be able to replace $100$ in the problem statement with $p - 1$ for any prime $p$.
Set $\phi_n(x) := x^{n-1} + \ldots + 1 = \dfrac{x^n-1}{x-1}$. Then $\phi_{101}(A) = 0$ and $\phi_{101}(x)$ is irreducible (since $101$ is prime), so $\phi_{101}(x)$ is the minimal polynomial of $A$. Let $\psi_A$ be the characteristic polynomial of $A$. Note that $\psi_A$ has the same roots as $\phi_{101}$, and also $\phi_{101} \mid \psi_A$ in $\mathbb{Z}[x]$. Write $\psi_A = \phi_{101}^m \cdot g$ for some $m \ge 1$, $g \in \mathbb{Z}[x]$ with $\phi_{101} \nmid g$. We show that $g = 1$:
If $\zeta$ is a root of $g$ (hence of $\psi_A$), then $\zeta$ is a $101^\text{st}$ root of unity, so $\phi_{101}$ (which is the minimal polynomial of $\zeta$ over $\mathbb{Q}$) divides $g$ in $\mathbb{Q}[x]$, hence also in $\mathbb{Z}[x]$ by Gauss' lemma, contradicting choice of $m$. Thus $g$ is constant, and since $\psi_A, \phi_{101}$ are monic, $g = 1$. Thus $\psi_A = \phi_{101}^m$, so the eigenvalues of $A$ are precisely the $101^\text{st}$ roots of unity, each of which appear with common multiplicity $m \ge 1$.
Now for any polynomial $f$, the eigenvalues of $f(A)$ are precisely $f(\lambda)$, where $\lambda$ is an eigenvalue of $A$. Let $S$ be the multiset of eigenvalues of $A$. Then for any $n$, the eigenvalues of $\phi_n(A)$ are precisely $\phi_n(s)$ for $s \in S$, so
$$\det(\phi_n(A)) = \prod_{s \in S} \phi_n(s) = \left(\prod_{k=1}^{100} \phi_n(e^{2\pi i k/101}) \right)^m = \left( \prod_{k=1}^{100} \frac{e^{2\pi i nk/101} - 1}{e^{2\pi i k/101} - 1} \right)^m$$
But for any $1 \le n \le 100$, $s \mapsto s^n$ is a permutation of $S$ (identifying the distinct elements of $S$ with the nonidentity elements of the cyclic group $\mathbb{Z}/101\mathbb{Z}$), so for such $n$ the above quotient is always $1$. Finally, note that $\phi_n(A) = I + A + \ldots + A^{n-1} = -(A^n + \ldots + A^{100})$, so $\det(\phi_n(A)) = \pm \det (A^n + \ldots + A^{100})$.