Matrix equation $A^2+A=I$ when $\det(A) = 1$ I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.

The fact $A^2 + A = I$, means that  $X^2 + X -1$ is an annihilating polynomial of $A$. 
The minimal polynomial of $A$ thus divides that polynomial. Thus all the eigenvalues of $A$ are among the roots of $X^2 + X - 1$, which are
$$\frac{-1 \pm  \sqrt{5}}{2}.$$
Since the determinant is the product of all eigenvalues (with multiplicity) it follows that both have the same multiplicity and this multiplicity is even. 
Since the sum of all multiplicities is the dimension, it follows that the dimension must be divisible by $4$. 
This condition suffices. A matrix has this property if and only it is similar to a diagonal matrix with half the diagonal entries $\frac{-1 +  \sqrt{5}}{2}$ and the other $\frac{-1 - \sqrt{5}}{2}$. 
The matrix is diagonalisable as the minimal polynomial has only distinct roots.
