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 $n&lt;10^4$ 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 $n&lt;10^3$ 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 $n&lt;1000$? (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.
As requested, here is some limited data:

N = 7, 10, 11, 17, 20, 22, 23, 24, 27, 30, 31, 34, 41, 42, 45, 
47, 49, 50, 52, 53, 57, 58, 59, 60, 61, 68, 71, 72, 74, 76, 
77, 79, 82, 83, 85, 86, 90, 92, 93, 94, 97, 99, 100, 101, 
102, 104, 105, 107, 110, 111, 112, 113, 114, 115, 119, 120, 
121, 122, 124, 126, 127, 130, 133, 134, 137, 138, 140, 142, 
144, 146, 148, 149, 150, 151, 152, 153, 154, 155, 157, 161, 
162, 164, 165, 167, 170, 171, 174, 175, 176, 178, 179, 181, 
182, 183, 185, 187, 188, 189, 192, 193, 195, 196, 198, 199, 
200, 202, 206, 209, 210, 211, 212, 214, 217, 218, 219, 224, 
227, 230, 232, 233, 234, 236, 237, 239, 240, 241, 244, 246, 
247, 249, 250, 251, 252, 253, 254, 255, 257, 258, 259, 260, 
261, 263, 265, 267, 268, 270, 276, 280, 281, 282, 286, 287, 
289, 290, 291, 292, 293, 294, 297, 298, 302, 304, 307, 309, 
313, 315, 316, 318, 319, 321, 323, 325, 326, 330, 331, 333, 
337, 338, 339, 340, 341, 343, 344, 347, 348, 349, 350, 351, 
353, 354, 355, 357, 358, 359, 360, 361, 362, 364, 367, 370, 
371, 372, 374, 375, 377, 380, 382, 383, 384, 385, 387, 388, 
389, 391, 393, 394, 399, 401, 402, 403, 406, 407, 408, 409, 
410, 411, 412, 413, 417, 420, 421, 424, 427, 428, 429, 430, 
433, 434, 436, 438, 440, 442, 444, 446, 449, 450, 451, 452, 
454, 455, 459, 461, 462, 463, 467, 471, 473, 475, 479, 481, 
482, 483, 484, 485, 486, 487, 488, 489, 491, 492, 494, 496, 
497, 498, 500

The three largest in size up to N=1000 are

$N = 978 = 6\times163$
a = 97724305086705990999491587041233313315191
b = 298737834302995183844618508475212828517680 

N = 983
a = 25612319152259738402372448240896341241531
b = 2927481175425024504484732240429126750140 

N = 863
a = 21697973611729663760123617224693905231
b = 140467357958644482600871394399613917520 

These are miniscule compared to the solution for

N = 8699     

a = 4835764154698933396807843882812763442889072738330844
7166783688776201461179031885978526561254300685692050
7298019993009790405598898439658017242050368089283786
155905093356149      

b = 7049312124254670086896841031232730133055532969718772
8828074452757842046709670329707566837793998606078626
2405134112587097831696284256965284093632390944423311
668372687220
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:

$$\frac{\left( {n}^{2}\,{\left( p+k\right) }^{2}\,s\pm k\,\left( 2\,n\,p+k\,n+k\,m\right) \right) \,\left( {\left( n\,p+k\,m\right) }^{2}\,s\pm k\,\left( 2\,n\,p+k\,n+k\,m\right) \right) }{{k}^{2}\,{\left( 2\,n\,p+k\,n+k\,m\right) }^{2}}$$