Let S=mZ: A^m^2+A^m=3I-A^-6 where A=pmatrix2&-11&0pmatrix.

Question

Let $S=\{m\in\mathbb{Z}: A^{m^2}+A^m=3I-A^{-6}\}$ where $A=\begin{pmatrix}2&-1\1&0\end{pmatrix}$. Then $n(S)$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Direct computation gives the pattern $A^k=\begin{pmatrix}k+1&-k\k&-k+1\end{pmatrix}$ for all integers $k$ (checked for $k=1,2,3,4$ and extends to negative $k$), so in particular $A^6=\begin{pmatrix}7&-6\6&-5\end{pmatrix}$.

Writing $A^m=\begin{pmatrix}m+1&-m\m&-m+1\end{pmatrix}$ and $A^{m^2}=\begin{pmatrix}m^2+1&-m^2\m^2&-m^2+1\end{pmatrix}$, the equation $A^{m^2}+A^m=3I-A^{-6}$ becomes, entrywise (top-left entry):
$$m^2+1+m+1 = 3-7+ \cdots$$
which after simplification gives $m^2+m-6=0\Rightarrow m=-3,\,2$.

So $S=\{-3,2\}$, and $n(S)=2$.

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