Prove that exists matrices $B,C \in \mathcal{M}_{3}(\mathbb{R})$ such that: $A=B^2+C^2$ Let $A \in \mathcal{M}_3(\mathbb{R})$. Prove that exists matrices $B,C \in \mathcal{M}_{3}(\mathbb{R})$ such that:
$$A=B^2+C^2$$
 A: This problem has been discussed in several papers, e.g., see here. For the field of real numbers, $-I_n$ is not the sum of two squares in $M_n(\mathbb{R})$ for $n$ odd. So the answer is positive if $n$ is even, and negative if $n$ is odd. The question also has been asked on MO here.
A more interesting case is perhaps what happens for integral matrices in $M_n(\mathbb{Z})$. Leonid Vaserstein proved that for $n\ge 2$ every integral matrix is the sum of three squares.
A: @ Road Human , your question is very poorly drafted. The example due to Alex is not a counter-example.
If $K $ is a field, then any $n\times n$ matrix (where $n\geq 2$) can be expressed as a sum of $3$ squares. cf. 
Matrices as sums of squares. Linear and Multilinear Algbra 5 (1977), 33-44.
Moreover,the only matrices that cannot be expressed as sum of $2$ squares are the $cI_n$ where $n$ is odd and $c\in K$ cannot be expressed as sum of $2$ squares. cf.
Matrices as sums of squares ; a conjecture of Griffin and Krusemeyer. Linear and Multilinear Algbra 17 (1985), 289-294.
EDIT 1. I fixed 2 mistakes reported by Marc, user1551 and Dietrich.
EDIT 2. Answer to Road Human ($n=4$). We assume that a decomposition is known for any matrix with dimension $2,3$. We assume that $A$ is in its real Jordan form. Only $3$ cases cannot be reduced to block matrices of dimension $2,3$ or a block $\geq 0$ of dimension $1$:
Case 1: $A=aI_4+N$ where $N^4=0$. Let $B=diag(U,U)$ where $U=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and let $u$ s.t. $a+u^2>0$. Then $A-(uB)^2=(a+u^2)I_4+N=(a+u^2)(I_4+M)$ where $M^4=0$. Finally $A=(uB)^2+C^2$ where $C=\sqrt{a+u^2}(I+1/2M-1/8M^2+\cdots)$ (my Taylor is rich).
Case 2. $A=\begin{pmatrix}U&I_2\\0&U\end{pmatrix}$ where $U=\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\rho R(\theta)$ where $\rho >0$ and $R$ is a rotation. In fact $A$ is a square because it has no $\leq 0$ eigenvalues. More precisely $A=B^2$ where $B=\begin{pmatrix}\sqrt{\rho}R(\theta/2)&\dfrac{R(-\theta/2)}{2 \sqrt{\rho}}\\0&\sqrt{\rho}R(\theta/2)\end{pmatrix}$.
Case 3. $A=diag(aI_3+N,b)$ where $N^3=0$ and $b<0$.
