Matrix version of Pythagoras theorem Can I find a solution for $C_{n\times n}$, explicitly, for the given $A_{n\times n}$ and $B_{n\times n}$ such that
$AA^{T} + BB^{T} = CC^{T}$? Here $A^{T}$ denotes the transpose of $A$ and all the matrices ($A$, $B$, and $C$) are real. 
I think there is a solution for $C$ as there are $n^2$ unknowns (the $n^2$ components $C$) and $n^2$ equations (comparing the components of LHS and RHS). 
Note: If it can not be solved (explicitly), then one can assume mild conditions (like positive definiteness, symmetry) on $A$ and $B$.
 A: I think you are saying: if we know A and B can we find a C such that the equation holds.
A Hermitian Positive Definite matrix M has a Cholesky decomposition $M=CC^{\ast}$.  So, if $AA^T+BB^T$ is positive definite symmetric (implying real), then a C exists that satisfies the equation.
A sufficient condition is that A and B are both positive definite.
A: This can always be done.
The following is, by and large, written under the assumption that the matrices are taken over the complex field, assuming that 
$A' = A^\dagger, \tag 1$
that is, $A'$ denotes the adjoint matrix of $A$, which of course is the transpose of the complex conjugate:
$A^\dagger = (\bar A)^T; \tag 2$
all the key steps of the argument, however, are valid when restricted solely to the reals, provided we limit the unitary matrix $U$ of (10), which diagonalizes $AA^\dagger + BB^\dagger$, to be in fact orthogonal; that such an orthogonal $U$ exists in the real case is a well-known result; then we of course also have $\bar A = A$, so (2) becomes $A^\dagger = A^T$.
By using the notation (1) we may write the given equation as
$AA^\dagger + BB^\dagger = CC^\dagger; \tag 3$
we are given $A$ and $B$ and we seek $C$ satisfying (3).  
Now for any complex matrix such as $A$ we observe that $AA^\dagger$ is self-adjoint:
$(AA^\dagger)^\dagger = (A^\dagger)^\dagger A^\dagger = AA^\dagger, \tag 4$
since
$(A^\dagger)^\dagger = A; \tag 5$
furthermore, $AA^\dagger$ is positive semidefinite, viz,
$x^\dagger (AA^\dagger x) = (x^\dagger A)(A^\dagger x) = (A^\dagger x)^\dagger (A^\dagger x) \ge 0. \tag 6$ 
Now the matrix $AA^\dagger + BB^\dagger$ also enjoys these two properties, self-adjointness and positive semi-definiteness, as may easily be see in a fashion similar to (4)-(6):
$(AA^\dagger + BB^\dagger)^\dagger = (AA^\dagger)^\dagger + (BB^\dagger)^\dagger = AA^\dagger + BB^\dagger, \tag 7$
where we have deployed (4), and also
$x^\dagger ((AA^\dagger + BB^\dagger)x) = x^\dagger(AA^\dagger x + BB^\dagger x) = x^\dagger(AA^\dagger x) + x^\dagger (BB^\dagger x) \ge 0, \tag 8$
since each term on the right of this equation is non-negative by way of (6).  
At this point we pause to note once again that if $A$ and $B$ are real matrices, everything we have done so far still binds; in the real case, we simply have $\bar A = A$ and hence
$A^\dagger = (\bar A)^T = A^T \tag 9$
etc.  
The matrix $AA^\dagger + BB^\dagger$, being self-adjoint, may diagonalized by a unitary matrix $U$, as is well-known; that is, there is a complex matrix $U$ such that
$U^\dagger U = UU^\dagger = I, \tag{10}$
and
$U(AA^\dagger + BB^\dagger)U^\dagger = D, \tag{11}$
where $D$ is a diagonal matrix; furthermore, $D$ is itself is self-adjoint:
$D^\dagger = (U^\dagger)^\dagger(AA^\dagger + BB^\dagger)^\dagger U^\dagger = U(AA^\dagger + BB^\dagger)U^\dagger = D  \tag{12}$
(where we have made use of (7)), which implies that all the entries of $D$ are real; obviously, the off-diagonal elements, being $0$, are real; furthermore, the diagonal elements of $D$ satisfy
$D_{ii} = (D^\dagger)_{ii} = ((\bar D)^T)_{ii} = \bar D_{ii}, \tag{13}$
hence they are also all real, and since it follows from (8) and (11) that 
$x^\dagger D x = x^\dagger U^\dagger (AA^\dagger + BB^\dagger)Ux = (Ux)^\dagger (AA^\dagger + BB^\dagger)Ux \ge 0; \tag{14}$
we thus find that $D$ is positive semi-definite, and from here it is but an easy step to see that the diagonal entries of $D$ obey
$D_{ii} \ge 0. \tag{15}$
By virtue of (15) we may define the diagonal matrix $\sqrt D$ such that
$(\sqrt D)_{ij} = [\delta_{ij} (\sqrt D)_{ii}]; \tag{16}$
it is clear that
$(\sqrt D)^\dagger = \sqrt D, \tag{17}$
and furthermore that
$(\sqrt D)^2 = D; \tag{18}$
thus, from (11),
$(U^\dagger \sqrt D U)(U^\dagger \sqrt D U) = U^\dagger \sqrt D UU^\dagger \sqrt D U = U\dagger \sqrt D  I \sqrt D U = U^\dagger (\sqrt D)^2 U = U^\dagger D U$
$= U^\dagger (U(AA^\dagger + BB^\dagger)U^\dagger)U = (U^\dagger U)(AA^\dagger + BB^\dagger)(U^\dagger U)$
$= I(AA^\dagger + BB^\dagger)I = AA^\dagger + BB^\dagger; \tag{19}$
if we now set
$C = U^\dagger \sqrt D U, \tag{20}$
we see that $C$ is self-adjoint:
$C^\dagger = (U^\dagger \sqrt D U)^\dagger = U^\dagger (\sqrt D)^\dagger (U^\dagger)^\dagger = U^\dagger \sqrt D U = C, \tag{21}$
and (19) yields
$C^\dagger C = C^2 = ( U^\dagger \sqrt D U)^2 = AA^\dagger + BB^\dagger, \tag{22}$
as desired.  We note that the procedure followed above provides a de facto recipe for explicitly computing $C$.
A: We assume that the matrices are real.
About a (false) remark by the OP; in fact, we consider a system of only $\approx n^2/2$ equations in $n^2$ unknowns, because $S=AA^T+BB^T$ is symmetric. Thus, there are an infinity of solutions (except when $A=B=0$).
In general, $\ker(A^T)\cap \ker(B^T)=\{0\}$ and $S$ is symmetric $>0$; note that there are two particular solutions that are interesting: 
The symmetric $>0$ matrix $C=S^{1/2}$ and the triangular matrix $C$ obtained by the standard Choleski method.
Exceptionally, $S$ is not invertible; in this case, we can again calculate the above two matrices $C$, the second one, using the extended Choleski method. About that, the OP gave a good reference (dated 1990) by Higham. Higham is a very good mathematician who also writes very well; he has written a book on this subject that I recommend everyone read:
Handbook of writing for the mathematical sciences, Siam.
Our friend @Robert Lewis also writes perfectly; I do not know if he read this book...
Remark. Higham, as many other researchers, uses the expression "positive semidefinite". In my opinion, it's a fault of English; we should say "non-negative semidefinite"...
