Characterizing an integer in terms of associated binary quadratic forms

Question

Let’s say I have a positive integer $n$, and I know the following:

1. $4n+1$ is of the form $a^2+b^2$
2. $16n+3$ is of the form $a^2+2b^2$.
3. $16n+1$ is of both forms $a^2+b^2$ and $a^2+2b^2$.

QUESTION: Is there a simple/elegant way to characterize all such integers $n$? I’m imagining that quadratic reciprocity will give some constraints… Are there other elementary means one can apply?

EDIT: A brute search on $n ≤ 200$ yields $$n \in \{1, 3, 6, 15, 18, 21, 28, 36, 39, 42, 48, 55, 58, 60, 61, 78, 81, 88, 105, 106,\\ 111, 120, 126, 130, 133, 139, 150, 151, 175, 189, 190, 193, 196, \dots\}.$$ There are some obvious patterns/members in there (e.g., most [but not all!] triangular numbers), but I’d like a complete characterization, if possible.

Answer 1

Suppose a is odd and:

$a^2\equiv 1 \bmod 16$

and b is even, then:

$A=4n+1=a^2+b^2\equiv 1\bmod 16$

$B=16n+3=c^2+2d^2$

c must be odd, suppose:

$c^2\equiv 1\bmod 16$

let d be odd and:

$d^2\equiv 1\bmod 16$

therefore:

$c^2+2d^2\equiv 3\bmod 16$

$C=16n+1$

It can be seen that a family of n can be of the form $n=16m+1$ and we have:

$A=4n+1=64m+5$

$B=16n+3=256m+19$

$C=16n+1=256m+17$

For example for $m=5$ we have:

$A=325=15^2+10^2$

$B=1299=7^2+2\times 25^2$

$C=1297=1^2+36^2=35^2+2\times 6^2$

These values of A, B and C are primitive solutions, clearly any multiple such as $k^2$ will give another solution.

Update:

For $m=2^6+1=65$ we obtain:

$A=4165=42^2+49^2$

$B=16659=39^2+2\times 87^2$

$C=16657=67^2+2\times 78^2=129^2+4^2$

I checked more powers up to $m=2^{14}+1$ and could not find more solutions. May be there are some more results with higher powers.