Let the columns of $A$ and $B$ be $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ respectively. By definition, the rank of $A$ and $B$ are the dimensions of the linear spans $\langle a_1, \ldots, a_n\rangle$ and $\langle b_1, \ldots, b_n\rangle$. Now the rank of $A + B$ is the dimension of the linear span of the columns of $A + B$, i.e. the dimension of the linear span $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$. Since $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$ the result follows.
Edit: Let me elaborate on the last statement. Any vector $v$ in $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$ can be written as some linear combination $v = \lambda_1 (a_1 + b_1) + \ldots + \lambda_n (a_n + b_n)$ for some scalars $\lambda_i$. But then we can also write $v = \lambda_1 (a_1) + \ldots + \lambda_n (a_n) + \lambda_1 (b_1) + \ldots + \lambda_n (b_n)$. This implies that also $v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$. We can do this for any vector $v$, so
$$\forall v \in \langle a_1 + b_1, \ldots, a_n + b_n\rangle: v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$$
This is equivalent to saying $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$.
Since $A^3=A$, the eigenvalues of $A$ are from $\{0,+1,-1\}$.
In addition, the $A$ is diagonalizable, as the minimal polynomial of $A$ divides $t^3-t=(t-1)(t+1)t$, hence it factors into linear factors.
Let me denote there algebraic multiplicities by $n_0,n_+,n_-$, respectively.
Then the rank of $A$ is equal to
$$
rank(A)=n_++n_-,
$$
while the trace is (the sum of the eigenvalues)
$$
tr(A) = n_+-n_-.
$$
So
$$
rank(A) + tr(A) = n_++n_-+n_+-n_- = 2n_+,
$$
which is even.
Best Answer
It is not true that the square of a matrix has the same rank as the matrix itself. For instance, the $2\times 2$ matrix $A= \left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$ has rank $1$ and satisfies $A^2=0$.