Proof or disprove: There exists a real $n \times n$ matrix $A$ that satisfies the equation when $n$ is even. Prove or disprove: There exists a real $n \times n$ matrix $A$ with
$$
A^2+2\cdot A+5\cdot I_n = 0
$$
if and only if n is even.
I could not find a counterexample for an odd $n$. Therefore, I suspect that the statement is true, but I have not yet found a solution.
If $n$ is odd, then $A$ has a a real eigenvalue $\lambda$. But then $\lambda^2+2\lambda+5=0$ and no such real number exists.
If $n=2$, consider$$A=\begin{bmatrix}-1&amp;-2\\2&amp;-1\end{bmatrix}.$$It will work because its eigenvalues are $-1\pm2i$, which are the roots of $\lambda^2+2\lambda+5$.
Can you generalize this for $n$ even and greater than $2$?
A possible generalization of Santos's answer:
Note that $A^T$ also satisfies the equation. Hence $A^2+2A+5I_n = {A^T}^2+2A^T+5I_n$ =&gt; $A^2-{A^T}^2=-2(A-A^T)$ =&gt; $(A+A^T)(A-A^T)+AA^T-A^TA = -2(A-A^T)$ =&gt; $(A+A^T+2I_n)(A-A^T)=-AA^T+A^TA$. Now take transpose on both sides: we now have:
$-(A+A^T+2I_n)(A-A^T)=-AA^T+A^TA$. Hence $(A+A^T+2I_n)(A-A^T)=0$. Since $A\neq A^T$ as this would mean $A$ is symmetric and real eigen values. A guess (not the answer) is to assume $A$ is skew symmetric and hence we have: $A= -A^T-2I_n$ and hence we have $a_{ii}=-a_{ii}-2$. Hence $a_{ii}=-1$. This is a possibility.
Also we have from original equation: $A(A+2I_n)/5 = -I_n$. Hence we have $-5A^{-1}=A+2I_n$. If we substitute this in $A= -A^T-2I_n$ then we get $-5A^{-1}=-A^T$=&gt; $5I_n = A^TA$. hence one possibility is to choose $A$ as a skew symmetric orthogonal matrix with diagonal entries $-1$ with norm if each column $5$. This matches the idea of solution given by Santos.