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The equations I want to solve looks like this: $$ X^n = 6 I $$ where $I$ is the identity matrix, $n$ is $1,2,3,..."$ and the square-matrix $$ X=\begin{pmatrix} x_{11} & x_{12} & \cdots \\ x_{21} & x_{22} & \cdots \\ \vdots & \vdots & \ddots \\ \end{pmatrix} $$ with the $x_{ij} \in \Bbb Z$. If the entries of the matrix $X$ would be allowed to be reals, the problem would be easy to solve. The solution would be $$ X=\begin{pmatrix} \sqrt[n]{6} & 0 & \cdots \\ 0 & \sqrt[n]{6} & \cdots \\ \vdots & \vdots & \ddots \\ \end{pmatrix} $$ However for $x_{ij}$ being integer the solution seems to be much harder. With guessing and some computer help I was able to find solutions for $n=2$ $$ X=\begin{pmatrix} 0 & 3 \\ 2 & 0 \\ \end{pmatrix} $$ and for $n=3$ $$ X=\begin{pmatrix} 2 & 3 & 1 \\ -1 & -1 & -2 \\ -2 & -1 & -1 \end{pmatrix} $$ I noticed that it didn't seem to be possible to find a two-dimensional matrix $X$ to solve the equation $X^3 = 6 I$. Could it be a general rule, that if you are searching for the $n$th root of a square matrix with integer component, the solution matrix needs to have $n$ dimensions?

And I would like to know if there is a better method than just guessing to solve these kind of nonlinear matrix equations over integer numbers?

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Notice that if the order of $X$ is $k$ then $\det(X)^n=6^k$. Since $\det(X)\in\mathbb{Z}$ then $\det(X)$ must be $\pm 6^m$. So $\det(X)^n=(\pm 1)^n6^{mn}=6^k$. So $mn=k$. Thus if we assume that there is a solution we must have $k=mn$, for some $m$.

Let $C(p)_{n\times n}$ be the companion matrix of the polymonial $p(x)=-6+t^n$.

Since the characteristic polynomial of $C(p)_{n\times n}$ is $p(x)$, by Cayley-Hamilton we have $0=-6Id+C(p)^n$. Thus, $C(p)^n=6Id$. Notice that the entries of $C(p)$ are intergers $(1$ and $6)$.

Now consider a matrix $X=\begin{pmatrix}C(p)&0&\ldots&0\\ 0&C(p)&\ldots& 0\\ \vdots &\vdots&\ddots&\vdots\\ 0&0&\ldots&C(p)\end{pmatrix}_{mn\times mn}$.

Notice that $X^n=\begin{pmatrix}C(p)^n&0&\ldots&0\\ 0&C(p)^n&\ldots& 0\\ \vdots &\vdots&\ddots&\vdots\\ 0&0&\ldots&C(p)^n\end{pmatrix}_{mn\times mn}=6Id_{k\times k}$.

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