Self-gratification by a²+kb² forms "with ~10% FALSE STATEMENTS" (by author)

[Raman]

k = 177

Primes p of the form a²+177b²: All primes congruent to [1, 25, 49, 85, 121, 133, 145, 169, 181, 193, 205, 241, 253, 265, 277, 289, 361, 373, 433, 481, 493, 517, 529, 553, 577, 625, 661, 685, 697] mod 708.
if N is a non-negative integer that can be written as a²+177b², then p × N can be written as a²+177b²
if N is a non-negative integer that cannot be written as a²+177b², then p × N cannot be written as a²+177b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [2, 65, 77, 89, 101, 113, 149, 161, 173, 185, 209, 221, 233, 269, 305, 329, 353, 365, 377, 401, 437, 485, 509, 533, 545, 569, 581, 629, 689, 701] mod 708
Class B primes: All primes congruent to [3, 35, 59, 71, 95, 107, 119, 143, 167, 203, 239, 251, 263, 287, 299, 311, 323, 359, 371, 383, 395, 407, 479, 491, 551, 599, 611, 635, 647, 671, 695] mod 708
Class C primes: All primes congruent to [31, 43, 55, 67, 91, 103, 115, 151, 187, 211, 235, 247, 259, 283, 319, 367, 391, 415, 427, 451, 463, 511, 571, 583, 655, 667, 679, 691, 703] mod 708

N can be written as a²+177b² if and only if
- N has no prime factors congruent to [5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 47, 53, 61, 73, 79, 83, 97, 109, 125, 127, 131, 137, 139, 155, 157, 163, 175, 179, 191, 197, 199, 215, 217, 223, 227, 229, 245, 257, 271, 275, 281, 293, 301, 307, 313, 317, 325, 331, 335, 337, 341, 343, 347, 349, 355, 379, 385, 389, 397, 403, 409, 419, 421, 425, 431, 439, 443, 445, 449, 455, 457, 461, 467, 469, 473, 475, 487, 497, 499, 503, 505, 515, 521, 523, 527, 535, 539, 541, 547, 557, 559, 563, 565, 575, 587, 589, 593, 595, 601, 605, 607, 613, 617, 619, 623, 631, 637, 641, 643, 653, 659, 665, 673, 677, 683, 707] mod 708 to an odd power.
- Sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.


k = 190

Primes p of the form a²+190b²: All primes congruent to [1, 9, 39, 49, 81, 111, 119, 121, 159, 161, 169, 191, 199, 201, 239, 271, 289, 311, 321, 329, 351, 359, 391, 441, 479, 481, 511, 519, 529, 609, 631, 671, 681, 689, 719, 729] mod 760.
if N is a non-negative integer that can be written as a²+190b², then p × N can be written as a²+190b²
if N is a non-negative integer that cannot be written as a²+190b², then p × N cannot be written as a²+190b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [2, 33, 97, 103, 113, 127, 143, 167, 183, 193, 217, 223, 257, 287, 297, 303, 337, 383, 393, 407, 417, 433, 447, 487, 497, 527, 553, 583, 607, 623, 673, 687, 697, 713, 737, 743, 753] mod 760
Class B primes: All primes congruent to [5, 43, 77, 83, 93, 123, 157, 163, 187, 197, 213, 237, 253, 267, 277, 283, 347, 387, 397, 403, 427, 443, 453, 467, 517, 533, 557, 587, 613, 643, 653, 693, 707, 723, 733, 747, 757] mod 760
Class C primes: All primes congruent to [19, 21, 29, 51, 59, 69, 91, 109, 141, 179, 181, 189, 211, 219, 221, 259, 261, 269, 299, 331, 341, 371, 379, 411, 421, 451, 459, 469, 509, 531, 611, 621, 629, 659, 661, 699, 749] mod 760

N can be written as a²+190b² if and only if
- N has no prime factors congruent to [3, 7, 11, 13, 17, 23, 27, 31, 37, 41, 47, 53, 61, 63, 67, 71, 73, 79, 87, 89, 99, 101, 107, 117, 129, 131, 137, 139, 147, 149, 151, 153, 173, 177, 203, 207, 227, 229, 231, 233, 241, 243, 249, 251, 263, 273, 279, 281, 291, 293, 301, 307, 309, 313, 317, 319, 327, 333, 339, 343, 349, 353, 357, 363, 367, 369, 373, 377, 381, 389, 401, 409, 413, 419, 423, 429, 431, 439, 449, 457, 461, 463, 471, 473, 477, 483, 489, 491, 493, 499, 501, 503, 507, 521, 523, 537, 539, 541, 543, 547, 549, 559, 561, 563, 567, 569, 571, 573, 577, 579, 581, 591, 593, 597, 599, 601, 603, 617, 619, 633, 637, 639, 641, 647, 649, 651, 657, 663, 667, 669, 677, 679, 683, 691, 701, 709, 711, 717, 721, 727, 731, 739, 751, 759] mod 760 to an odd power.
- Sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.


k = 253

Primes p of the form a²+253b²: All primes congruent to [1, 9, 25, 49, 81, 93, 133, 141, 169, 177, 185, 213, 225, 257, 265, 269, 289, 301, 317, 353, 357, 361, 377, 397, 441, 445, 449, 485, 489, 509, 533, 537, 553, 565, 577, 581, 625, 653, 669, 685, 729, 749, 785, 817, 829, 837, 841, 905, 929, 933, 949, 961, 969, 993, 1005] mod 1012.
if N is a non-negative integer that can be written as a²+253b², then p × N can be written as a²+253b²
if N is a non-negative integer that cannot be written as a²+253b², then p × N cannot be written as a²+253b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [2, 35, 39, 87, 95, 123, 127, 131, 139, 151, 167, 211, 215, 219, 239, 255, 259, 271, 303, 307, 315, 347, 351, 371, 395, 403, 415, 439, 491, 519, 535, 547, 579, 591, 607, 611, 623, 679, 699, 703, 739, 767, 783, 791, 811, 831, 855, 875, 887, 899, 915, 923, 959, 967, 975, 1007] mod 1012
Class B primes: All primes congruent to [11, 15, 23, 67, 91, 103, 111, 135, 155, 159, 191, 199, 203, 235, 247, 251, 267, 287, 291, 295, 339, 355, 367, 375, 379, 383, 411, 419, 467, 471, 511, 543, 551, 559, 595, 603, 619, 631, 643, 663, 687, 707, 727, 735, 751, 779, 815, 819, 839, 895, 907, 911, 927, 939, 971, 983, 999] mod 1012
Class C primes: All primes congruent to [17, 21, 57, 61, 65, 109, 129, 145, 149, 153, 189, 205, 217, 237, 241, 249, 281, 293, 321, 329, 337, 365, 373, 413, 425, 457, 481, 497, 505, 513, 525, 549, 557, 569, 585, 589, 613, 677, 681, 689, 701, 733, 769, 789, 833, 849, 865, 893, 937, 941, 953, 965, 981, 985, 1009] mod 1012

N can be written as a²+253b² if and only if
- N has no prime factors congruent to [3, 5, 7, 13, 19, 27, 29, 31, 37, 41, 43, 45, 47, 51, 53, 59, 63, 71, 73, 75, 79, 83, 85, 89, 97, 101, 105, 107, 113, 117, 119, 125, 137, 147, 157, 163, 171, 173, 175, 179, 181, 183, 193, 195, 197, 201, 221, 223, 227, 229, 233, 243, 245, 261, 263, 273, 277, 279, 283, 285, 305, 309, 311, 313, 323, 325, 327, 331, 333, 335, 343, 349, 359, 369, 381, 387, 389, 393, 399, 401, 405, 409, 417, 421, 423, 427, 431, 433, 435, 443, 447, 453, 455, 459, 461, 463, 465, 469, 475, 477, 479, 487, 493, 499, 501, 503, 507, 515, 521, 523, 527, 531, 541, 545, 555, 563, 567, 571, 573, 587, 593, 597, 599, 601, 609, 615, 617, 629, 633, 635, 637, 639, 641, 645, 647, 651, 655, 657, 659, 661, 665, 673, 675, 683, 691, 695, 697, 705, 709, 711, 717, 719, 721, 723, 725, 731, 741, 743, 745, 747, 753, 755, 757, 761, 763, 765, 771, 773, 775, 777, 787, 793, 795, 797, 799, 801, 807, 809, 813, 821, 823, 827, 835, 843, 845, 853, 857, 859, 861, 863, 867, 871, 873, 877, 879, 881, 883, 885, 889, 901, 903, 909, 917, 919, 921, 925, 931, 945, 947, 951, 955, 963, 973, 977, 987, 991, 995, 997, 1003, 1011] mod 1012 to an odd power.
- Sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.

[Raman]

k = 24

Primes p of the form a²+24b²: All primes congruent to [1] mod 24.
if N is a non-negative integer that can be written as a²+24b², then p × N can be written as a²+24b²
if N is a non-negative integer that cannot be written as a²+24b², then p × N cannot be written as a²+24b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [3, 11] mod 24
Class B primes: All primes congruent to [5] mod 24
Class C primes: All primes congruent to [7] mod 24

N can be written as a²+24b² if and only if
- N is not congruent to 2 (mod 4)
- N has no prime factors congruent to [13, 17, 19, 23] mod 24 to an odd power.
- If N is odd, then the sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.
- If the highest power of 2 dividing N is an even number that is two or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of the same parity.
- If the highest power of 2 dividing N is an odd number that is three or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of different parity.


k = 40

Primes p of the form a²+40b²: All primes congruent to [1, 9] mod 40.
if N is a non-negative integer that can be written as a²+40b², then p × N can be written as a²+40b²
if N is a non-negative integer that cannot be written as a²+40b², then p × N cannot be written as a²+40b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [5, 13, 37] mod 40
Class B primes: All primes congruent to [7, 23] mod 40
Class C primes: All primes congruent to [11, 19] mod 40

N can be written as a²+40b² if and only if
- N is not congruent to 2 (mod 4)
- N has no prime factors congruent to [3, 17, 21, 27, 29, 31, 33, 39] mod 40 to an odd power.
- If N is odd, then the sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.
- If the highest power of 2 dividing N is an even number that is two or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of the same parity.
- If the highest power of 2 dividing N is an odd number that is three or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of different parity.


k = 88

Primes p of the form a²+88b²: All primes congruent to [1, 9, 25, 49, 81] mod 88.
if N is a non-negative integer that can be written as a²+88b², then p × N can be written as a²+88b²
if N is a non-negative integer that cannot be written as a²+88b², then p × N cannot be written as a²+88b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [11, 19, 35, 43, 51, 83] mod 88
Class B primes: All primes congruent to [13, 21, 29, 61, 85] mod 88
Class C primes: All primes congruent to [15, 23, 31, 47, 71] mod 88

N can be written as a²+88b² if and only if
- N is not congruent to 2 (mod 4)
- N has no prime factors congruent to [3, 5, 7, 17, 27, 37, 39, 41, 45, 53, 57, 59, 63, 65, 67, 69, 73, 75, 79, 87] mod 88 to an odd power.
- If N is odd, then the sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.
- If the highest power of 2 dividing N is an even number that is two or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of the same parity.
- If the highest power of 2 dividing N is an odd number that is three or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of different parity.


k = 232

Primes p of the form a²+232b²: All primes congruent to [1, 9, 25, 33, 49, 57, 65, 81, 121, 129, 161, 169, 209, 225] mod 232.
if N is a non-negative integer that can be written as a²+232b², then p × N can be written as a²+232b²
if N is a non-negative integer that cannot be written as a²+232b², then p × N cannot be written as a²+232b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [21, 29, 37, 61, 69, 77, 85, 101, 133, 157, 189, 205, 213, 221, 229] mod 232
Class B primes: All primes congruent to [15, 31, 39, 47, 55, 79, 95, 119, 127, 135, 143, 159, 191, 215] mod 232
Class C primes: All primes congruent to [35, 51, 59, 67, 83, 91, 107, 115, 123, 139, 179, 187, 219, 227] mod 232

N can be written as a²+232b² if and only if
- N is not congruent to 2 (mod 4)
- N has no prime factors congruent to [3, 5, 7, 11, 13, 17, 19, 23, 27, 41, 43, 45, 53, 63, 71, 73, 75, 89, 93, 97, 99, 103, 105, 109, 111, 113, 117, 125, 131, 137, 141, 147, 149, 151, 153, 155, 163, 165, 167, 171, 173, 175, 177, 181, 183, 185, 193, 195, 197, 199, 201, 207, 211, 217, 223, 231] mod 232 to an odd power.
- If N is odd, then the sum of exponents of Class A prime factors of N, sum of exponents of Class B prime factors of N, sum of exponents of Class C prime factors of N are all even, or are all odd.
- If the highest power of 2 dividing N is an even number that is two or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of the same parity.
- If the highest power of 2 dividing N is an odd number that is three or more, then the sum of exponents of Class A prime factors of N and the sum of exponents of Class B prime factors of N are of different parity.

[henryzz]

How long does it take to check if a number can be represented in the form a^2+k*b^2 for a certain k? Is this different to finding the values a and b? Is this feasible for large numbers(factorizable numbers so upto maybe 200 digits).
Is there any particular reason you haven't touched on negative values of k?
Have you found any external information on this sort of expression?

[Raman]

k = 16

Primes p of the form a²+16b² : All primes congruent to {1} mod 8.
if N is a non-negative integer that can be written as a²+16b², then p × N can be written as a²+16b²
if N is a non-negative integer that cannot be written as a²+16b², then p × N cannot be written as a²+16b²

N can be written as a²+16b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16).
- N has no prime factors congruent to {3, 7} mod 8 to an odd power.
- If N is odd, then the sum of exponents of {5} mod 8 prime factors of N is even.

k = 28

Primes p of the form a²+28b² : All primes congruent to {1, 9, 25} mod 28.
if N is a non-negative integer that can be written as a²+28b², then p × N can be written as a²+28b²
if N is a non-negative integer that cannot be written as a²+28b², then p × N cannot be written as a²+28b²

N can be written as a²+28b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16).
- N has no prime factors congruent to {3, 5, 13, 17, 19, 27} mod 28 to an odd power.
- If N is odd, then the sum of exponents of {7, 11, 15, 23} mod 28 prime factors of N is even.

[Raman]

Originally Posted by henryzz

How long does it take to check if a number can be represented in the form a^2+k*b^2 for a certain k? Is this different to finding the values a and b? Is this feasible for large numbers (factorizable numbers so upto maybe 200 digits).
Is there any particular reason you haven't touched on negative values of k?
Have you found any external information on this sort of expression?

I think that it is possible to generate expressions like this for the sixty-five values of k, for which all the representable primes are being all the primes that lie in a certain set of residue classes (mod 4k). Such prime representations are being unique!

k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 21, 22, 24, 25, 28, 30, 33, 37, 40, 42, 45, 48, 57, 58, 60, 70, 72, 78, 85, 88, 93, 102, 105, 112, 120, 130, 133, 165, 168, 177, 190, 210, 232, 240, 253, 273, 280, 312, 330, 345, 357, 385, 408, 462, 520, 760, 840, 1320, 1365, 1848

Once we have got with this expression, we can check out instantly by applying this expression, if in case we have got with the prime factorization of a big number, by itself.

Similarly I think it is being possible to do so for the other values of k given below in post #6 for the form (a²+kb²)/4 form.

k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 19, 20, 21, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 37, 40,
42, 43, 45, 48, 51, 52, 57, 58, 60, 64, 67, 70, 72, 75, 78, 84, 85, 88, 91, 93, 96, 99, 100, 102, 105, 112, 115, 120, 123,
130, 132, 133, 147, 148, 160, 163, 165, 168, 177, 180, 187, 190, 192, 195, 210, 228, 232, 235, 240, 253, 267, 273, 280, 288,
312, 315, 330, 340, 345, 352, 357, 372, 385, 403, 408, 420, 427, 435, 448, 462, 480, 483, 520, 532, 555, 595, 627, 660, 672,
708, 715, 760, 795, 840, 928, 960, 1012, 1092, 1120, 1155, 1248, 1320, 1365, 1380, 1428, 1435, 1540, 1632, 1848, 1995, 2080,
3003, 3040, 3315, 3360, 5280, 5460, 7392

For a prime, finding out the values of a & b is being instantaneous by using the Cornacchia-Smith algorithm, but for a general number, if it is the product of representable primes, it is obviously being easy by using the following identity
(a²+kb²)(c²+kd²) = |ac+kbd|² + k|ad-bc|²
(a²+kb²)(c²+kd²) = |ac-kbd|² + k|ad+bc|²

but for other numbers (most of them thereby forming with the majority), I will need to study more. For this example consider with 21 = 3×7 being representable as a²+5b² form, but neither 3 nor 7 can be written in that form, by itself.

For other values of k, some but not all the primes from within a certain set of residue classes (mod 4k) are being representable.

For example, consider with 11,
The primes 53 & 97 belong to the same residue class (mod 44),
but 53 is being representable as a²+11b² form, but 97 is not.
This cannot happen (mod 4k), for the sixty-five values of k, being mentioned above.

(Although 53 & 97 both can be written uniquely in the form (a²+11b²)/4.)

I am being interested from different point of view - ideal classes, class number problem.

For the negative values of k, it is not being possible to get unique representation of primes. For the equation x²-ky² = 1, there are being infinitely many solutions to this - the solution of the Pell's equation - given by using the continued fraction convergents of √k, by itself.

Thereby, there are being infinitely many units in inside the quadratic field Z[√k] for real fields, for positive values of k.

So, if a prime is being representable, then it is being possible to multiply by using one of these units, in order to get with another representation, although I would think that it is being a good problem to find out what primes are being representable / writable by using the form a² - kb² = p, by itself !
to study what relationship it has got with to do with the residue classes, (mod 4k)

[henryzz]

I was thinking if we can represent a composite number in the form a^2+k*b^2 for k=2 but not k=4 then we know that either N is 2 mod 4(trivially tested) or N has a factor 3 mod 8 to an odd power.
If N is not 2 mod 4, then we know there is a factor 3 mod 8 to an odd power. Rather than searching through all possible factors which have the forms {1,3,5,7} mod 8 we can search through just a quarter of possible factors and know we can find a result.

Similar things can be done with other combinations of k.

For reference:

N can be written as a²+2b² if and only if
- N has no prime factors congruent to {5, 7} mod 8 to an odd power.

N can be written as a²+4b² if and only if
- N is not congruent to 2 (mod 4).
- N has no prime factors congruent to {3} mod 4 to an odd power.

[Raman]

Originally Posted by Raman

What is the lowest value of k, for which there exist residue classes (mod 4k), that
contains with two different types of primes: SOME / NONE?
Answer: k = 96

Is it possible that there exist a value of k, for which there exist residue classes (mod 4k), that
contains with two different types of primes: ALL / NONE?

Or even containing with all the three different types of primes: ALL / SOME / NONE?
Answer: k = 107

again that unfortunately that these statements are being false
enough, being caused by the arbitrary artificial limit being used in
inside my own code, by itself

Edit - I don't think that it is being possible to have with a
SOME / NONE residue class Or ALL / NONE residue class Or ALL / SOME / NONE residue class for any k value (mod 4k) value, by itself.

This is being a minor correction only, thereby, let it be as it is being OK not to be corrected
it is not being highlighted, they are being only given in
inside the spoiler text alone, only actually, thereby!
by itself

[chalsall]

Originally Posted by Raman

again that unfortunately that these statements are being false
enough, being caused by the arbitrary artificial limit being used in
inside my own code, by itself

If I may share...

I was sixteen years old. I just had my drivers' licence. I'd been given a car to drive. I didn't really know how to drive, but I had been playing many video games.

One morning I lost control of the car going down an icy hill. My natural instinct was to reach up with my left hand and try to hit the "Esc" button.

Much to my (and my passenger's) surprise, there wasn't an "Escape" key built into the car. Thankfully I knew enough to prevent us from dying.

I never played video games again.

Perhaps it's time for you to press the "reset button", and as we say here in Bim "come and wheel again?"

[R.D. Silverman]

Originally Posted by chalsall

If I may share...

I was sixteen years old. I just had my drivers' licence. I'd been given a car to drive. I didn't really know how to drive, but I had been playing many video games.

One morning I lost control of the car going down an icy hill. My natural instinct was to reach up with my left hand and try to hit the "Esc" button.

Much to my (and my passenger's) surprise, there wasn't an "Escape" key built into the car. Thankfully I knew enough to prevent us from dying.

I never played video games again.

Perhaps it's time for you to press the "reset button", and as we say here in Bim "come and wheel again?"

There is a saying: a little knowledge is dangerous.

The content of this entire thread is already well known. Read Vol 2 of
H. Cohen's book on Computational Algebraic Number Theory along with
David Cox's book: Primes of the form x^2 + ny^2.

It was well known two centuries ago by Gauss.

[Raman]

Originally Posted by Raman

For a prime, finding out the values of a & b is being instantaneous by using the Cornacchia-Smith algorithm, but for a general number, if it is the product of representable primes, it is obviously being easy by using the following identity
(a²+kb²)(c²+kd²) = |ac+kbd|² + k|ad-bc|²
(a²+kb²)(c²+kd²) = |ac-kbd|² + k|ad+bc|²

but for other numbers (most of them thereby forming with the majority), I will need to study more. For this example consider with 21 = 3×7 being representable as a²+5b² form, but neither 3 nor 7 can be written in that form, by itself.

If I understand correctly, then all the primes congruent to {1, 5, 9} mod 20 can be uniquely written in the form a²+5b².
All the primes congruent to {2, 3, 7} mod 20 can be uniquely written in the form 2a²+2ab+3b².

Product of two primes of the form 2a²+2ab+3b² can be written in the form a²+5b².
Product of a prime of the form 2a²+2ab+3b² and a prime of the form a²+5b² can be written in the form 2a²+2ab+3b².

So, if you look at the general condition, then
you can be able to see this following structure, with

N can be written in the form 2a²+2ab+3b² if and only if
- N has no prime factors congruent to {11, 13, 17, 19} mod 20 to an odd power.
- Sum of exponents of {2, 3, 7} mod 20 prime factors of N is odd.

Compare this form with its own companion form - namely the a²+5b² form!

So, in order to write with a big factored number N - into the a²+5b² form - i.e. finding out with the following a, b values,
by using combining of products - by making use of the following identity,

(a²+kb²)(c²+kd²) = |ac+kbd|² + k|ad-bc|²
(a²+kb²)(c²+kd²) = |ac-kbd|² + k|ad+bc|²

then certainly, you would need to get the a²+5b² representation (a,b) values for the {1, 5, 9} mod 20 prime factors of N,
and then certainly, next again, with the 2a²+2ab+3b² form (a,b) values for the {2, 3, 7} mod 20 prime factors of N,

to be combined, thereby, being, to be put altogether, with!

[henryzz]

Originally Posted by R.D. Silverman

There is a saying: a little knowledge is dangerous.

The content of this entire thread is already well known. Read Vol 2 of
H. Cohen's book on Computational Algebraic Number Theory along with
David Cox's book: Primes of the form x^2 + ny^2.

It was well known two centuries ago by Gauss.

Part of what I think Raman is doing is going on his own journey of discovery though this area of maths. This can be quite valuable for helping some people learn. It's not just about the knowledge. Learning how to learn is important.
Thanks for the references though. They should be useful.