how to find all the solutions to $I+A+\cdots+A^n=0.$ Let $GL_3(\mathbb{Z}[i])$ be the group of invertible $3\times 3$ matrices whose coefficients are Gaussian integers.I want to find all the pair $(A\in GL_3(\mathbb{Z}[i]),n\in\mathbb{Z})$ satisfying
$$I+A+\cdots+A^n=0.$$
It is easy to dope out some obvious solutions.Say $(-I,1)$.But how to find all the solutions to this equation?
 A: The only hope of solving this is the small size $3\times3$ imposed on the matrices. You need a (short) list of invariant factors, polynomials in $\def\Z{\Bbb Z}\def\i{\mathbf i}\def\Zi{\Z[\i]}\Zi$ each dividing the next and the last (the minimal polynomial) dividing the annihilating polynomial $P=1+X+X^2+\cdots+X^n=\frac{X^{n+1}-1}{X-1}$ of$~A$, and whose degrees add up to$~3$.
Now $P$ factors over $\def\Q{\Bbb Q}\Q$ as a product of cyclotomic polynomials $P=\prod_{{k\mid n+1\atop k\neq1}}\Phi_k$. Those cyclotomic polynomials are the irreducible factors in $\Z[X]$ of $P$, but they might factor over the Gaussian integers $\Zi$ (as is clear in the case of $\Phi_4=X^2+1$); however a decomposition into two equal-degree factors is the only possibility, since any nontrivial divisor in $\Zi[X]$ of $\phi_k$ when multiplied by its complex conjugate will give a divisor in $\Z[X]$ of $\Phi_k$. 
The minimal polynomial$~\mu$ (over $\Q[\i]$) of $A$ must divide $P$, and lie in $\Zi[X]$ (since by Gauss's lemma a monic divisor in $\Q[\i][X]$ of your monic $P\in\Z[X]$ must have its coefficients in $\Zi$), and of course it can have degree$~3$ at most. This leads to the question which irreducible factors over$~\Zi$ of cyclotomic polynomials have degree at most$~3$. First the cyclotomic polynomials that themselves have degree $\phi(k)\leq3$ (in fact $\phi(k)\leq2$, since beyond $1$ no odd values of $\phi(k)$ occur) are limited to $\Phi_2=X+1$, $\Phi_3=X^2+X+1$, $\Phi_4=X^2+1$ and $\Phi_6=X^2-X+1$, of which $\Phi_4=(X-\i)(X+\i)$ splits over $\Zi$. Then there are are the cases with $\phi(k)\in\{4,6\}$ where $\Phi_k$ might, if it factors over $\Zi$, have irreducible factors of degree $2$ or $3$; these are $k\in\{8,12,5,10\}$ for $\phi(k)=4$ and $k\in\{9,18,7,14\}$ for $\phi(k)=6$. However since $\Phi_k$ splits over the cyclotomic field $K=\Q[X]/(\Phi_k)$ ($K$ is normal over$~\Q$), having a decomposition $\Phi_k=F\overline F$ with $F\in\Zi[X]$ (but of course $F\notin\Z[X]$) requires that $K$ contains$~\i$, and this only happens when $k$ is a multiple of$~4$. This means we only need to consider $k\in\{8,12\}$; in those cases one indeed has
$$\begin{align}\Phi_8=X^4+1&=(X^2+\i)(X^2-\i),
\qquad\text{and}\\
\Phi_{12}=X^4-X^2+1&=(X^2+\i X-1)(X^2-\i X-1).\end{align}
$$
Therefore $\mu$ cannot be irreducible of degree$~3$, nor can it be irreducible of degree$~2$ since then the remaining invariant factor of degree$~1$ would have to divide it, but cannot. So either $\mu$ has only (distinct) factors of degree$~1$, so that $A$ is diagonalisable over $\Zi$ with eigenvalues among $\{-1,+\i,-\i\}$, or else $\mu$ is the product of a quadratic factor chosen among $X^2+X+1$, $X^2-X+1$, $X^2+\i$, $X^2-\i$, $X^2+\i X-1$, $X^2-\i X-1$ and a linear factor chosen among $X+1$, $X-\i$, $X+\i$. In the former (diagonalisable) case there can be just one eigenvalue (a scalar matrix), two eigenvalues among which one is double, or three distinct eigenvalues; in the latter case (presence of a quadratic irreducible factor) $\mu$ equals the characteristic polynomial (this is also the case for three distinct eigenvalues), so there is a cyclic vector for$~A$. Each polynomial factor used comes with the obligation for $n+1$ to be a multiple of$~k$ when the factor was obtained from$~\Phi_k$ (for instance using $X^2-\i X-1$ requires $n+1$ to be divisible by$~12$).
This gives, up to similarity over$~\Q[\i]$, all possibilities for $n$ and $A$. In the non-scalar cases one can take $A$ to be a companion matrix of$~\mu$, or if you prefer a block diagonal matrix with the companion matrices of the factors of$~\mu$ as diagonal blocks. For instance for $\mu=(X+\i)(X^2-\i)=X^3+\i X^2-\i X+1$ the following two matrices are similar over$~\Q[\i]$ and satisfy your equation $P[A]=0=P[B]$ whenever $n+1$ is a multiple of$~8$:
$$
  A=\begin{pmatrix}0&0&-1\\1&0&\i\\0&1&-\i\end{pmatrix}
\qquad\text{and}\qquad
  B=\begin{pmatrix}-\i&0&0\\0&0&-\i\\0&1&0\end{pmatrix}
$$
I'll leave the (natural) question of similarity over$~\Zi$ for another occasion, as it is quite a bit harder. I believe that the above $A,B$ are not similar over$~\Zi$.
A: You can use the formula : $1 + x + x^2 + ... + x^n$ = $\frac{x^{n+1}-1}{x-1}$ on matrices ($1 = I$)
This would work on all matrices where A-I is invertible i.e that don't have 1 as an eigen value, but it is easy to show that those that have are not solution.
This leads you to $A^{n+1} = I$
