What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$

Part II. Strangely, for the similar concordant forms/numbers $n$ such that,
$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$
is not even in the OEIS, 
$$n = 1,7,10,11,17,20,22,23,24,27,30,31,34,\dots$$
The list of $104$ prime $n<10^3$ is known (by Kevin Brown, David Einstein (hm?), and Allan MacLeod) though several troublesome primes were excluded assuming the Birch/Swinnerton-Dyer conjecture.

Question: Anybody knows how to generate the list of all concordant forms $n<1000$? (Elkies describes a method here.)

P.S. Incidentally, the special case $n=52$ appears in equal sums of like powers. Let,
$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$
Then for $k=1,2,4,6,8,10,$ 
$$(8b)^k + (5a-4b)^k + (-a-2d)^k +  (a-2d)^k + (-5a-4b)^k  + (-12b+4c)^k + (12b+4c)^k =\\ 
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$ 
found by J. Wroblewski and yours truly. An initial primitive solution is $a,b,c,d = 3,4,5,29$ and an infinite more.

While individ in his answer has shown there is an infinite number of concordant forms $n$, it can also be shown there is an infinite number of the special case $n = \pm N^2$. Given the system,
$$a^2+b^2 = c^2\tag1$$
$$a^2+nb^2 = d^2\tag2$$
First, let $a,\,b,\,c = p^2-q^2,\,2pq,\,p^2+q^2$ to satisfy $(1)$. The second becomes,

$$(p^2-q^2)^2+n(2pq)^2 =d^2\tag3$$

Two parametric solutions to $(3)$ are,
$$n = -(2t^2-2)^2,\quad p =t(2t^2-1),\quad q=1$$
found by Will Jagy and,
$$n = (4t^2+3)^2,\quad p =t(4t^2+1),\quad q=1$$
by this OP.
