Is there someone who can give at a least an idea for solving this problem?
Determine the matrices $ X_{1} , X_{2} , ..., X_{9} \in \mathrm{M}_{2}(\mathbb{Z})$ such that:
$$(X_{1})^{4} + \cdots +(X_{9})^{4}=(X_{1})^2 + \cdots + (X_{9})^2 +18\cdot I_{2}$$
and $\det X_{k}=1$ for $ k=1,\ldots,9$.
Thank you very much for any help or any comments.
Let $X_1=\cdots X_9= \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} $. We then have $X_i^2=-I$ and $X_i^4=I$ for all $i$ so your equation holds.
More generally, we can pick $X_1,\dots,X_9$ to be any nine solutions to the equation $X^4-X^2-2I=0$, and there are infinitely many of these. We could also distribute the 18 in different ways between these equations... and so on. And this only takes care of the solutions in which the different matrices «do not interact».