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.
 A: I have a list of all $N$ up to $9999$, which, I think, are concordant numbers and corresponding values of $a,b$. I also have negative $N$ solutions down to $-999$.
These were all computed based on assuming the Birch and Swinnerton-Dyer conjecture for the elliptic curve
\begin{equation*}
y^2=x(x+1)(x+N)
\end{equation*}
to predict the rank. Then various computational methods were applied to curves with predicted rank greater than $0$, to find a rational point.
I can send you the results file if you wish. They are much too large for here.
A: For the system of equations:  
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.$$
Solution 1:
$$a=p-s,\quad b=2t,\quad c=p+s$$
$$w=\mp2q+p+s\pm2$$
$$q=(p\pm1)(s\pm1)$$  
$$ps=t^2$$  
Solution 2:
$$a=t^2-1,\quad b=2t,\quad c=t^2+1$$
$$w=3t^2-1$$
$$q=2t^2-1$$
A: 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.
A: If we consider two Diophantine equations:
$$4\,{\left( k\,p\,t+n\,{p}^{2}+{k}^{2}\,m\right) }^{2}+\frac{{k}^{2}\,{\left( k\,t+2\,n\,p\right) }^{2}\,\left( {t}^{2}-4\,m\,n\right) }{{n}^{2}}=\frac{{\left( {k}^{2}\,{t}^{2}+2\,k\,n\,p\,t+2\,{n}^{2}\,{p}^{2}-2\,{k}^{2}\,m\,n\right) }^{2}}{{n}^{2}}$$
and
$${\left( h\,{k}^{2}\,t-n\,{p}^{2}\,s+{k}^{2}\,m\,s+2\,h\,k\,n\,p\right) }^{2}\,\left( {t}^{2}-4\,m\,n\right) +4\,n\,{\left( k\,p\,t+n\,{p}^{2}+{k}^{2}\,m\right) }^{2}\cdot\,\left( h\,s\,t+m\,{s}^{2}+{h}^{2}\,n\right) ={\left( h\,{k}^{2}\,{t}^{2}+n\,{p}^{2}\,s\,t+{k}^{2}\,m\,s\,t+2\,h\,k\,n\,p\,t+4\,k\,m\,n\,p\,s+2\,h\,{n}^{2}\,{p}^{2}-2\,h\,{k}^{2}\,m\,n\right) }^{2}$$
You can write down some solution:
$${k}^{2}\,{\left( n-m\right) }^{2}\,{\left( 2\,n\,p+k\,n+k\,m\right) }^{2}+4\,a\,{n}^{2}\,{\left( p+k\right) }^{2}\,{\left( n\,p+k\,m\right) }^{2}$$
$a$ - some concordant form:
$$\frac{\left( {n}^{2}\,{\left( p+k\right) }^{2}\,s\pm k\,\left( 2\,n\,p+k\,n+k\,m\right) \right) \,\left( {\left( n\,p+k\,m\right) }^{2}\,s\pm k\,\left( 2\,n\,p+k\,n+k\,m\right) \right) }{{k}^{2}\,{\left( 2\,n\,p+k\,n+k\,m\right) }^{2}}$$
