What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$?

Question

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)

$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$

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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.

Answer 1

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

Answer 2

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.

Answer 3

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:

$$a=t^2-1,\quad b=2t,\quad c=t^2+1$$

$$w=3t^2-1$$

$$q=2t^2-1$$

Answer 4

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.

Answer 5

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:

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