1. Given matrix

We have

$$A=\begin{pmatrix}2&-1\\1&0\end{pmatrix}.$$

We need integers $m$ such that

$$A^{m^2}+A^m=3I-A^{-6}.$$

We must find the number of integers in

$$S=\{m\in \mathbb Z: A^{m^2}+A^m=3I-A^{-6}\}.$$

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2. Find a simple relation satisfied by $A$

Compute $A^2$:

$$A^2=\begin{pmatrix}2&-1\\1&0\end{pmatrix} \begin{pmatrix}2&-1\\1&0\end{pmatrix} =\begin{pmatrix}3&-2\\2&-1\end{pmatrix}.$$

Then

$$A^2-2A+I =\begin{pmatrix}3&-2\\2&-1\end{pmatrix} -2\begin{pmatrix}2&-1\\1&0\end{pmatrix} +\begin{pmatrix}1&0\\0&1\end{pmatrix} =0.$$

So

$$A^2-2A+I=0 \quad\Rightarrow\quad (A-I)^2=0.$$

Let

$$N=A-I.$$

Then $N^2=0$, and

$$A=I+N.$$

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3. Formula for powers of $A$

Since $N^2=0$,

$$(I+N)^k=I+kN$$

for every integer $k$ (this also works for negative integers because $A$ is invertible).

Hence

$$A^k=I+k(A-I).$$

Now

$$A-I=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}.$$

Therefore

$$A^k=I+k(A-I) =\begin{pmatrix}1+k&-k\\k&1-k\end{pmatrix}.$$

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4. Compute both sides of the equation

Using the formula,

$$A^{m^2}=I+m^2(A-I),$$ $$A^m=I+m(A-I).$$

So

$$A^{m^2}+A^m=2I+(m^2+m)(A-I).$$

Also,

$$A^{-6}=I-6(A-I)$$

because $k=-6$. Thus

$$3I-A^{-6}=3I-[I-6(A-I)]=2I+6(A-I).$$

Therefore the given equation becomes

$$2I+(m^2+m)(A-I)=2I+6(A-I).$$

So

$$(m^2+m)(A-I)=6(A-I).$$

Since $A-I\neq 0$, we get

$$m^2+m=6.$$

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5. Solve for $m$

$$m^2+m-6=0$$ $$(m+3)(m-2)=0.$$

Hence

$$m=-3 \quad \text{or} \quad m=2.$$

So

$$S=\{-3,2\}.$$

Therefore,

$$n(S)=2.$$

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6. Comparison with stored answer

Stored correct answer = $2$.

Our derived answer is also $2$, so they agree.