According to Brahmagupta-Fibonacci Identity, for $p=q\cdot r$ we can prove if any two of the integers $p,q,r$ are of the form $a^2+n\cdot b^2,$ the third must of the same form
This is probably a generalization of this problem or this
Now, I want to determine the $n$ such that if $M=r\cdot s=a^2+n\cdot b^2$ where $(r,s)=1, r$ and $s$ must be of the same form for any integer pair $a,b$
Using Program, it seems that $n=1,2,3,7$ satisfy this
I think the proof for $2$ will be required here
We can safely assume $n$ to be square-free, as its square part(if any) can easily be merged with $b$
Now, if $(a,b)=d$ and $\frac aA=\frac bB=d\implies (A,B)=1$ and $ a^2+n\cdot b^2=d^2(A^2+n\cdot B^2)$
As $n$ is square-free, $(A^2,n)=(A,n)=D$(say) and $\frac A{A_1}=\frac nN=D\implies (A_1,N)=1$
Subsequently, $A^2+n\cdot B^2$ becomes $D^2\cdot A_1^2+N\cdot D\cdot B^2=D(D\cdot A^2_1+N\cdot B^2)$ which is not of the form $a^2+n\cdot b^2$ unless $D$ or $N=1$
So, we can focus on $a^2+n\cdot b^2,$ where $(a,n\cdot b)=1$
Some observations can be made:
$(1):$ If $p^c$ divides $M=a^2+n\cdot b^2,$ where integer $c\ge1$ and $p$ is prime,
$a^2\equiv-n\cdot b^2\pmod {p^c}\iff (a\cdot b^{-1})^2\equiv-n\pmod {p^c}$ $\implies -n$ must be a Quadratic residue of $p^c$
This is a necessary condition for $p^c$ to be of the form $a^2+n\cdot b^2$
So, if $2^c$ (where $c\ge3,$) divides $M,n\equiv-1\pmod 8$ as $x^2\equiv e\pmod {2^c}$ is solvable with exactly $4$ solutions $\iff e\equiv1\pmod 8$
$(2):$ Generalizing the solution of this problem,
Let's consider $2^x=a^2+n\cdot b^2$
As $n$ is odd and $(a,b)=1, a\cdot b$ must be odd
One value of $x$ is $y,$ i..e, $2^y=a_1^2+nb_1^2$
and if the smallest value of $x$ is $x_\text{min},$ i.e., $2^{x_\text{min}}=a_2^2+n\cdot b_2^2$
$2^{x_\text{min}+y}=(a_1^2+n\cdot b_1^2)(a_2^2+n\cdot b_2^2)=(a_1a_2\pm n\cdot b_1b_2)^2+n(a_1b_2\mp a_2b_1)^2$
Observe that $a_1a_2\pm n\cdot b_1b_2,a_1b_2\mp a_2b_1$ are even and the highest powers of $2$ that divides each will be same $=2^k$(say).
So, $2^{x_\text{min}+y-2k}=\left(\frac{a_1a_2\pm n\cdot b_1b_2}{2^k}\right)^2+n\left(\frac{a_1b_2\mp a_2b_1}{2^k}\right)^2$
Following this line, we can be prove
$4^k,k\ge 2$ can be represented as $a^2+15b^2$
$2^{3k+2},k\ge 1$ can be represented as $a^2+31b^2$
