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.

Let the eigenvalues of $A$ be $\{e_i\}$ and let $P$ be a similarity transformation that diagonalizes $A$.  Then $P$ also diagonalizes $A^2$ and the equation becomes 
$$
\forall i: e_i^2 + e_i = 1
$$
Since there are two real roots to that equation, there are $2^n$ candidates for $A$, before imposing that the determinant is one.  
When we impose that the diagonal is one, we see that the eigenvalues  have to come in pairs 
$$
e_i = \frac{-1-\sqrt{5}}{2}. e_{i'} = \frac{-1+\sqrt{5}}{2}. 
$$
and that there must be an even number of such pairs.  So no such $A$ exists, unless $N=4k$ for$k\in\Bbb{Z}^+$.  For a given such $N$ there are 
$$
\binom{N}{N/2}
$$
possible diagonal determinant one $N\times N$ matrices satisfying the equation.
However, if $A$ satisfies the equation, then so does any similar matrix 
$$
p^{-1}AP
$$
so as long as $N=4k$ there are a continuum of solutions for $A$.
