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

We are given $$A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$$ and $$S = \left\{m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6}\right\}$$.

The characteristic equation of A: $$\det(A - \lambda I) = (2-\lambda)(0-\lambda) - (-1)(1) = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0$$.

By the Cayley-Hamilton theorem: $$(A - I)^2 = 0$$, so $$A^2 - 2A + I = 0$$, meaning $$A^2 = 2A - I$$.

Since $$(A - I)^2 = 0$$, let $$N = A - I$$, so $$N^2 = 0$$. Then $$A = I + N$$.

$$A^n = (I + N)^n = I + nN = I + n(A - I) = (1 - n)I + nA$$

Let us verify: $$A^2 = -I + 2A = 2A - I$$. This matches $$A^2 = 2A - I$$. Good.

So $$A^n = (1-n)I + nA$$ for all integers $$n$$.

$$A^{-6} = (1-(-6))I + (-6)A = 7I - 6A$$

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

$$[(1 - m^2)I + m^2 A] + [(1 - m)I + mA] = 3I - (7I - 6A)$$ $$[(1 - m^2) + (1 - m)]I + [m^2 + m]A = -4I + 6A$$ $$[2 - m^2 - m]I + [m^2 + m]A = -4I + 6A$$

Since $$I$$ and $$A$$ are linearly independent (as $$A \ne kI$$):

Coefficient of $$A$$: $$m^2 + m = 6$$, so $$m^2 + m - 6 = 0$$, giving $$(m+3)(m-2) = 0$$, so $$m = -3$$ or $$m = 2$$.

Coefficient of $$I$$: $$2 - m^2 - m = -4$$, so $$m^2 + m = 6$$. This gives the same equation.

Both conditions are identical, so $$m \in \{-3, 2\}$$ and $$n(S) = 2$$.

The answer is $$\boxed{2}$$.