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Sa se rezolve in \( M_3(\mathbb{Z}) \) ecuatia \( X^{2008}=\left(\begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) \).

Fie \(
A = \left( {\begin{array}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}} \right) \Rightarrow AX = XA{\rm sau }\left( {\begin{array}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}} \right)\left( {\begin{array}
x_{11} & x_{12} & x_{13} \\
{x_{21} } & {x_{22} } & {x_{23} } \\
{x_{31} } & {x_{32} } & {x_{33} } \\
\end{array}} \right) = \left( {\begin{array}
x_{11} & x_{12} & x_{13} \\
{x_{21} } & {x_{22} } & {x_{23} } \\
{x_{31} } & {x_{32} } & {x_{33} } \\
\end{array}} \right)\left( {\begin{array}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}} \right)
\) si obtinem ca
\( {\rm X = }\left( {\begin{array}
a & b & c \\
c & a & b \\
b & c & a \\
\end{array}} \right)
\)
\( \left| {\begin{array}
a & b & c \\
c & a & b \\
b & c & a \\
\end{array}} \right| = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 1\ {\mbox sau } - 1 \).
\( \det X\in \{\pm 1\} \)
Daca detX=1 si a+b+c=1 atunci \( {\rm (a + b + c)}^{\rm 2} = 1
\) si gasim \( a^2 + b^2 + c^2 = 1 \) cu a=c=0 si b=1 si e solutie a ecuatiei.
Pentru detX=-1 se cauta analog.