What are numbers $n$ such that $a^2+nb^2 = c^2$ and $na^2+b^2 = d^2$?

Question

Let $n$ and $a,b,c,d,$ be in the positive integers.

I. For the system,

$$a^2-nb^2 = c^2\\a^2+nb^2=d^2$$

then $n$ is a congruent number. The sequence starts as $n=5,6,7,13,14,15,20,21,$ and so on.

II. For,

$$a^2+b^2 = c^2\\a^2+nb^2=d^2$$

then $n$ is a concordant form/number. It starts as $n=1,7, 10, 11, 17, 20, 22, 23, 24,27,$ etc.

III. Define,

$$a^2+nb^2 = c^2\\na^2+b^2=d^2$$

with an infinite number of co-prime $a,b$. (In other words, a certain elliptic curve has positive rank.)

Question:

1. Anybody knows what $n$ is called for III?
2. What is its list of $n$ below a bound? (I believe it starts $n =7, 12, 17, 19, 28, 33, 39, 40, 44, 48,$ but I am not sure if I missed some $n$ in the range.)

IV. From this post, I know $n=13$ doesn't have a solution. However, $n=12$ does. Given the nice Letac-Sinha identity,

$$(a+c)^k + (a-c)^k + (3b+d)^k + (3b-d)^k + (4a)^k = \\(3a+c)^k + (3a-c)^k + (b+d)^k + (b-d)^k + (4b)^k$$

for $k = 2,4,6,8$ where,

$$a^2+12b^2 = c^2\\12a^2+b^2=d^2$$

with non-trivial solution $a,b = 218, 11869,$ and, using an elliptic curve, an infinite more.

Answer 1

Note that I, II and III all contain the quadric $a^2+Nb^2=\Box$, which can be parameterized by $a=k^2-N$, $b=2k$, where $k$ is rational.

For I, $a^2-Nb^2=c^2$ gives the quartic $c^2=k^4-6Nk^2+N^2$, whilst for II, $a^2+b^2=c^2$ gives $c^2=k^4+(4-2N)k^2+N^2$. Both of these quartics have clear rational solutions $(k,c)=(0,\pm N)$, and so can be transformed to the well-known elliptic curves for these problems.

For III, however, $Na^2+b^2=d^2$ gives the quartic $d^2=N\,k^4+(4-2N^2)k^2+N^3$ which does NOT have an obvious rational point and so cannot be related to a family of elliptic curves. In fact, for certain values of $N$ the quartics have no rational points, such as for $N=2$ or $N=5$.

There is a rational point when $N=M^2$. In this case, the problem is equivalent to the concordant number problem for $M^4$.