How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$? How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?
Note: $I$ denotes the identity matrix of size $3 \times3$ and $A^{T}$ represents the transpose of the matrix $A$.
I took transpose on both sides in given equation to get $A=(A^T)^2-I$ and then I put value of $A^T$ in this equation using $A^{T}=A^2-I$ to get $A^3-2A-I=0$ which ultimately gave $(A^T-A)(A+I)=0$ . How to deal the problem from here? And is there any better approach to tackle this problem? The given answer is $0$.

Since $A^3-2A-I=0$ , any eigenvalue of $A$ satisfies
$$\lambda^3-2\lambda-1=0 \Rightarrow \lambda=-1, \frac{1 \pm \sqrt{5}}{2}$$
Let $\lambda_{1,2,3}$ be the eigenvalues of $A$. Then 
$$tr(A^T)=tr(A^2-I) \Rightarrow
\lambda_1+\lambda_2+\lambda_3 = \lambda_1^2+\lambda_2^2+\lambda_3^2-3 \\
\Rightarrow \sum_{j=1}^3(\lambda_j^2 -\lambda_j -1) =0 \qquad (*)$$ 
Note that $\lambda= \frac{1 \pm \sqrt{5}}{2}$ satisfy $\lambda^2 -\lambda-1=0$, while $\lambda=-1$ satisfies $\lambda^2 -\lambda-1>0$
It follows from $(*)$ that $-1$ is NOT an eigenvalue of $A$.
Therefore, the minimal polynomial of $A$ is 
$$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})^\alpha (x- \frac{1 -\sqrt{5}}{2})^\beta$$
Moreover, since $A^3-2A-I=0$ we have $\mu_A(x) | X^3-2X-1$ and hence $\alpha, \beta \leq 1$.
Therefore, we have three choices left:
$$\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2})\\
\mu_A(x)= (x- \frac{1 -\sqrt{5}}{2})\\
\mu_A(x)=(x- \frac{1 +\sqrt{5}}{2}) (x- \frac{1 -\sqrt{5}}{2})$$
The first two lead to the solutions
$$A=\lambda I \qquad \lambda= \frac{1 \pm +\sqrt{5}}{2}$$
While the last gives
$$A^2-A-I =0$$
and hence 
$$A=A^T$$
