How to show that there is no $3\times3$ real matrix $A$ such that $A^2+I=0$? Question: show that there is no $3\times3$ real matrix $A$ such that $A^2+I=0$?
Is it because: 
$$\det(A^2)=\det(-I)\\
\implies \det(A)\det(A)=-1\\
\implies \det(A)=-i$$
How to continue?
 A: Every real $3 \times 3$ matrix, being of odd size, has a real eigenvalue since the characteristic polynomial is of odd degree.  If $A^2 + I = 0$, and $A\mathbf v = \lambda \mathbf v$ with $\mathbf v \ne 0$, then $0 = (A^2 + I) \mathbf v = (\lambda^2 + 1) \mathbf v \Rightarrow \lambda^2 + 1 = 0$.  Applying this notion to a real eigenvalue of $A$ leads to an immediate contradiction, since no real $\lambda$ satisfies $\lambda^2 + 1 = 0$.  Hence, no real $3 \times 3$, or indeed $n \times n$ for $n$ odd, matrix $A$ satisfies $A^2 + I = 0$.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: OP, I like your proof way better than ^^.  Intuitively it seems like you want to show that it implies existence of solution to $X^2 + 1 = 0$ in $\Bbb{R}$.
You had:
$$
\det(A^2)=\det(-I)\\
\implies \det(A)\det(A)=-1\\
$$
What you have now is, assuming that there is such a $3\times 3$ then there exists a real number $X = \det(A)$ such that $X^2 = -1$.  But you know that's not true from studying algebra & polynomials, so contradiction!
