Vtyjkyuk

Let $A$ be a given $m times n$ matrix with $m geq n$ and Rank$(A)=n$. Does there always exist an $m times n$ matrix $X$ (with Rank$(X)=n$) that solves,
$$
(X+A)(X+A)'=left[beginarrayccI_n & 0_n times (m-n) \ 0_(m-n) times n & 0_(m-n) times (m-n)\ endarrayright]+AA'?
$$

Simplifying, I get
$$
XA'+AX'+ XX'=left[beginarrayccI_n & 0_n times (m-n) \ 0_(m-n) times n & 0_(m-n) times (m-n)\ endarrayright]
$$

but not sure where to go from here.

I was trying to investigate the case where $m=n$ but didn't make progress.

I suspect the answer is yes (but this is pure speculation from the case when all objects are scalars). This reminds me of Sylvester equations https://en.wikipedia.org/wiki/Sylvester_equation and pseudo inverses https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse

Let $Y=X+A$; then $YY^T=diag(I_n,0_m-n)+AA^T=B$.

Since $rank(YY^T)leq n$, if we want that a solution exists, then necessarily $rank(B)leq n$.

EDIT 1. Since $AA^T$ is symmetric $geq 0$, $rank(B)geq n$ and we assume that $rank(B)=n$. Since $rank(AA^T)=n$, the spd symmetric matrix $AA^T$ is in the form $diag(U_n,0_m-n)$, where $U$ is a pd symmetric matrix. Finally we may assume that $B=diag(b_1,cdots,b_n,0_m-n)$ where $b_i>0$. We can choose $Y=beginpmatrixdiag(pmsqrtb_i)\0_m-n,nendpmatrix$.

EDIT 2. Answer to user. Necessarily, for any $y$, $[0_n,y^T]B[0_n,y]=0$; then $AA^T$ is in the form $beginpmatrixU_n&V\V^T&0_m-nendpmatrix$. Thus if $A=[P,Q]^T$,then $QQ^T=0$, that implies that $Q=0$. Finally $V$ is also a zero matrix and we are done.

Conclusion. If $A$ is not in the form $[P,0]^T$, then $0$ solution; otherwise the solutions are $Y=[F,0]^T$ with $FF^T=I_n+PP^T=Wdiag(b_i)W^T$ where $Win O(n)$. Thus there are at least $2^n$ solutions: $X=beginpmatrixWdiag(pmsqrtb_i)W^T\0_m-n,nendpmatrix-A$.