Let $$A = \begin{pmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$$ and $$B = A - I$$. If $$\omega = \dfrac{\sqrt{3}i - 1}{2}$$, then the number of elements in the set $$\{n \in \{1, 2, \ldots, 100\} : A^n + (\omega B)^n = A + B\}$$ is equal to ______.

Given $$A = \begin{pmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$$, $$B = A - I$$, and $$\omega = \dfrac{\sqrt{3}i - 1}{2}$$. Find the number of $$n \in \{1, 2, \ldots, 100\}$$ such that $$A^n + (\omega B)^n = A + B$$.

Subtracting the identity matrix from $$A$$ gives:

$$B = \begin{pmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix}$$

Since every row of $$B$$ is $$(1, -1, -1)$$, the matrix $$B$$ has rank 1.

Multiplying $$B$$ by itself yields:

$$B^2 = \begin{pmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & 1 & 1 \end{pmatrix} = -B$$

Thus $$B^2 = -B$$, which implies $$B^3 = B$$. More generally, $$B^n = B$$ if $$n$$ is odd and $$B^n = -B$$ if $$n$$ is even (for $$n \ge 1$$).

Using the relation $$A = I + B$$ together with $$B^2 = -B$$, we compute:

$$A^2 = (I + B)^2 = I + 2B + B^2 = I + 2B - B = I + B = A$$

Therefore $$A^2 = A$$, and by induction $$A^n = A$$ for all $$n \ge 1$$.

Considering $$(\omega B)^n$$, we note that:

$$(\omega B)^n = \omega^n B^n$$

When $$n$$ is odd, $$B^n = B$$ so $$(\omega B)^n = \omega^n B$$.

When $$n$$ is even, $$B^n = -B$$ so $$(\omega B)^n = -\omega^n B$$.

Substituting into the equation $$A^n + (\omega B)^n = A + B$$ and using $$A^n = A$$ leads to:

$$A + (\omega B)^n = A + B$$ $$(\omega B)^n = B$$

In the case of odd $$n$$, this requires $$\omega^n B = B \implies \omega^n = 1$$.

In the case of even $$n$$, it requires $$-\omega^n B = B \implies \omega^n = -1$$.

Since $$\omega = \dfrac{-1 + \sqrt{3}i}{2} = e^{i\cdot 2\pi/3}$$ is a primitive cube root of unity, we have $$\omega^3 = 1$$.

For odd $$n$$, the condition $$\omega^n = 1$$ implies $$n \equiv 0 \pmod{3}$$, and combined with oddness gives $$n \equiv 3 \pmod{6}$$.

Within the set $$\{1, \ldots, 100\}$$, these values are $$n = 3, 9, 15, \ldots, 99$$, whose number is $$\dfrac{99 - 3}{6} + 1 = 17$$.

For even $$n$$, there is no solution because $$\omega^n$$ can only be $$1, \omega, \omega^2$$, none of which equals $$-1$$.

Hence there are 17 such values of $$n$$.