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

[alpertron]

Originally Posted by Raman

Let N = a² + k b².

Suppose that someone uses with a cryptosystem, that uses with N as the public key,
and then a, b as the key generator elements, then I wondered if whether the designed
cryptosystem would be rather secure enough.

But, if N were to be prime - then it quite directly can be cracked with
by using the Cornacchia-Smith algorithm, or even with the GCD of x + √-k with N
inside the imaginary quadratic field Z[√-k], by itself, whereby value of x is being such that
x² = -k mod N.

But, if N is not prime, then it may be easy to crack with for k = 1, 2, 3, 4, 7, values of N into
such cases, with - unique factorization domains, etc.

For the special case k=1 and composite N, if you know how to find easily a and b such that N = a² + b², you could also factor that number N, by finding two different sums of squares and then applying Euler's factorization method.

[Raman]

Originally Posted by alpertron

For the special case k=1 and composite N, if you know how to find easily a and b such that N = a² + b², you could also factor that number N, by finding two different sums of squares and then applying Euler's factorization method.

Dear alpertron:

-> I do know that if a prime p can be written into the form a²+kb² form, then
its representation is being unique - if both of a, b are being positive,
or without that restriction - there will be four solutions for a prime for
the representation by any quadratic form.

If N is the product of m distinct primes (ALL / SOME) such that it can be written into the form with
N = a²+kb² form, then there will be 2^m different solutions to (a, b) inside the positive integers,
by itself. If in case that you would include with the negative values, then there will be
2^m+2 different solutions to (a, b).

I am not much interested in the solutions of (a, b), as since they are not being unique.
I am much interested in the numbers which are being writable into the form a²+kb² form,
by itself.

If k = 5, you need to examine the solution to p = 2a²+2ab+3b² where p is being a prime factor of N congruent to {2, 3, 7} mod 20.
The product of two primes of the form 2a²+2ab+3b² can be written into the form a²+5b², by itself.
(2a²+2ab+3b²)(2c²+2cd+3d²) = |2ac+ad+bc-2bd|²+5|ad+ac+bd|²


If k = 6, you need to examine the solution to p = 2a²+3b² where p is being a prime factor of N congruent to {2, 3, 5, 11} mod 24.
The product of two primes of the form 2a²+3b² can be written into the form a²+6b², by itself.
(2a²+3b²)(2c²+3d²) = |2ac-3bd|²+6|ad+bc|²
(2a²+3b²)(2c²+3d²) = |2ac+3bd|²+6|ad-bc|²

If k = 10, you need to examine the solution to p = 2a²+5b² where p is being a prime factor of N congruent to {2, 5, 7, 13, 23, 37} mod 40.
The product of two primes of the form 2a²+5b² can be written into the form a²+10b², by itself.
(2a²+5b²)(2c²+5d²) = |2ac-5bd|²+10|ad+bc|²
(2a²+5b²)(2c²+5d²) = |2ac+5bd|²+10|ad-bc|²

But, these polynomial forms such as those like 2a²+2ab+3b², 2a²+3b², 2a²+5b²,
a²+5b², a²+6b², a²+10b², etc.
will span with ALL the primes lying in certain set of residue classes (mod 4k).

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

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

On the other hand, consider with the case for k = 11:
The polynomial of a²+11b² form will only span with SOME of the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
(You may expand it to the residue classes mod 44, by this itself).
On the other hand, the polynomial of 3a²+2ab+4b² form seems to generate with the REMAINING primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.

As a result, the two polynomials of a²+11b² form, 3a²+2ab+4b² form TOGETHER appears to form a basis for all the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
Not it is being possible with any of these two single polynomials individually.

Is it being possible for anyone to explain why this happens? And then if possible being able to point out
off to any potentially available references, or links for the elaborating in detail, in depth towards this mechanism?

What primes are being representable in the form a²+11b² form, thereby distinguishing it from the primes of the form 3a²+2ab+4b² form,
lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11, by itself

But although, for all the primes p lying in the following residue classes {0, 1, 3, 4, 5, 9} mod 11,
it's being possible to write with 4 × p into the form a²+11b², by itself.

4 × (a²+11b²) = c²+11d² form, with, c = 2a, d = 2b, over thereby,
by itself, following with this statement of transformation, by itself

4 × (3a²+2ab+4b²) = c²+11d² form, with, c = a + 4b, d = a, over thereby,
by itself, following with this statement of transformation, by itself

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

k = 11, 19, 20, 27, 32, 35, 36, 43, 51, 52, 64, 67, 75, 84, 91, 96, 99, 100, 115, 123, 132, 147, 148, 160, 163, 180, 187, 192, 195, 228, 235,
267, 288, 315, 340, 352, 372, 403, 420, 427, 435, 448, 480, 483, 532, 555, 595, 627, 660, 672, 708, 715, 795, 928, 960, 1012, 1092, 1120, 1155,
1248, 1380, 1428, 1435, 1540, 1632, 1995, 2080, 3003, 3040, 3315, 3360, 5280, 5460, 7392

By some same way, the following rings of integers Z[(1+√k)/2],
Z[(1+√3)/2], Z[(1+√7)/2], Z[(1+√11)/2], Z[(1+√19)/2], Z[(1+√43)/2], Z[(1+√67)/2], Z[(1+√163)/2],

for k = 3, 7, 11, 19, 43, 67, 163 are being unique factorization domains, by itself.
with k being prime, of the form 3 (mod 4), by itself,
and then (k+1)/4 is being either 1 or prime,
with the polynomial of the form x²+x+(k+1)/4 form,
being with generating with all-prime values from k = 0 ... (k-7)/4.

Quite directly
it is being
Amazingly startling with, unbelievable with,
by itself, with, !

Following statement why it is being significant enough, as follows

Let N be a non-negative integer that can be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0).
Let p be a prime factor that divides N (raised to an odd power), i.e. pm divides N, pm+1 does not divide N, whereby m is being odd.
Then, if p can also always be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0),

then k = 1, 2, 3, 4, 7, 8, 11, 12, 19, 43, 67, 163

[Raman]

Originally Posted by henryzz

There are 3 polynomials with a discriminant of -23 for example.

I don't understand with this. Can you please elaborate with this?

What are the three polynomials being?

I find out that only two polynomial forms, actually
generate with
all the primes being congruent to {0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18} mod 23 forms.

They are being
a²+23b²
3a²+2ab+8b²

Although, but both of them having with Discriminant = -92.

[henryzz]

Originally Posted by alpertron

For the special case k=1 and composite N, if you know how to find easily a and b such that N = a² + b², you could also factor that number N, by finding two different sums of squares and then applying Euler's factorization method.

Wouldn't the same method also work for square k.
If we could try different squares we could cover most if not all numbers.

So now the challenge is representing a composite in the form a^2+k^2*b^2. Now I write it like that it looks the same as k=1 but with b=kb.
I am assuming there is no fast way to find two solutions to n=a^2+b^2.

[Raman]

Originally Posted by henryzz

I am assuming there is no fast way to find two solutions to n=a²+b² forms.

Why not? If in case that we have got with all the prime factors of n, then

we will certainly be able to write out with each, every prime factor p_j of n - that are being congruent to {1, 2} mod 4,
as a Gaussian integer, whose norm (multiple) is being equal to p_j, individually, by itself, (by using the Cornacchia-Smith algorithm), or by taking GCD of n with x±i,
whereby, x is the integer satisfying with x² = -1 (mod p_j) - GCD = a_j ± b_ji - and then, combining them together with their own product into all the possibly
available different ways as follows. If in case that the prime factors that are being congruent to 3 (mod 4), do occur in pairs, then, go ahead to
take with the product by taking them with singly, i.e., only once - actually.

product = (a₁ ± b₁i)(a₂ ± b₂i)(a₃ ± b₃i)(a₄ ± b₄i)...(a_m ± b_mi)


n = a²+b² form, with,

a = real part(product)
b = imaginary part(product)





The Gaussian integers are being a unique factorization domain - such that many classical properties do work out with into the representation of n as the sum of two squares, by itself,

-> All the prime numbers that are being congruent to {1, 2} mod 4, are being the sum of two squares.
-> No prime numbers that are being congruent to {3} mod 4, are being the sum of two squares.
-> Numbers that can be written as the sum of two squares will have no prime factors congruent to {3} mod 4, being raised to an odd power.
-> Every prime representation as sum of two squares is being unique.
-> The product of a prime of the form {1, 2} mod 4, along with a number that is the sum of two squares - will be the sum of two squares.
-> The product of a prime of the form {1, 2} mod 4, along with a number that is not the sum of two squares - will not be the sum of two squares.
-> The product of m distinct primes congruent to 1 (mod 4) - can be written as the sum of two squares - into 2^m different ways.

, ..., etc.,

[Raman]

Dear alpertron:

-> I do know that if a prime p can be written into the form a²+kb² form, then
its representation is being unique - if both of a, b are being positive,
or without that restriction - there will be four solutions for a prime for
the representation by any quadratic form.

If N is the product of m distinct primes (ALL / SOME) such that it can be written into the form with
N = a²+kb² form, then there will be 2^m different solutions to (a, b) inside the positive integers,
by itself. If in case that you would include with the negative values, then there will be
2^m+2 different solutions to (a, b).

I am not much interested in the solutions of (a, b), as since they are not being unique.
I am much interested in the numbers which are being writable into the form a²+kb² form,
by itself.

If k = 5, you need to examine the solution to p = 2a²+2ab+3b² where p is being a prime factor of N congruent to {2, 3, 7} mod 20.
The product of two primes of the form 2a²+2ab+3b² can be written into the form a²+5b², by itself.
(2a²+2ab+3b²)(2c²+2cd+3d²) = |2ac+ad+bc-2bd|²+5|ad+bc+bd|²
(2a²+2ab+3b²)(2c²+2cd+3d²) = |2ac+ad+bc+3bd|²+5|ad-bc|²

If k = 6, you need to examine the solution to p = 2a²+3b² where p is being a prime factor of N congruent to {2, 3, 5, 11} mod 24.
The product of two primes of the form 2a²+3b² can be written into the form a²+6b², by itself.
(2a²+3b²)(2c²+3d²) = |2ac-3bd|²+6|ad+bc|²
(2a²+3b²)(2c²+3d²) = |2ac+3bd|²+6|ad-bc|²

If k = 10, you need to examine the solution to p = 2a²+5b² where p is being a prime factor of N congruent to {2, 5, 7, 13, 23, 37} mod 40.
The product of two primes of the form 2a²+5b² can be written into the form a²+10b², by itself.
(2a²+5b²)(2c²+5d²) = |2ac-5bd|²+10|ad+bc|²
(2a²+5b²)(2c²+5d²) = |2ac+5bd|²+10|ad-bc|²

But, these polynomial forms such as those like 2a²+2ab+3b², 2a²+3b², 2a²+5b²,
a²+5b², a²+6b², a²+10b², etc.
will span with ALL the primes lying in certain set of residue classes (mod 4k).

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

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

On the other hand, consider with the case for k = 11:
The polynomial of a²+11b² form will only span with SOME of the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
(You may expand it to the residue classes mod 44, by this itself).
On the other hand, the polynomial of 3a²+2ab+4b² form seems to generate with the REMAINING primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.

As a result, the two polynomials of a²+11b² form, 3a²+2ab+4b² form TOGETHER appears to form a basis for all the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
Not it is being possible with any of these two single polynomials individually.

Is it being possible for anyone to explain why this happens? And then if possible being able to point out
off to any potentially available references, or links for the elaborating in detail, in depth towards this mechanism?

What primes are being representable in the form a²+11b² form, thereby distinguishing it from the primes of the form 3a²+2ab+4b² form,
lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11, by itself

But although, for all the primes p lying in the following residue classes {0, 1, 3, 4, 5, 9} mod 11,
it's being possible to write with 4 × p into the form a²+11b², by itself.

4 × (a²+11b²) = c²+11d² form, with, c = 2a, d = 2b, over thereby,
by itself, following with this statement of transformation, by itself

4 × (3a²+2ab+4b²) = c²+11d² form, with, c = a + 4b, d = a, over thereby,
by itself, following with this statement of transformation, by itself

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

k = 11, 19, 20, 27, 32, 35, 36, 43, 51, 52, 64, 67, 75, 84, 91, 96, 99, 100, 115, 123, 132, 147, 148, 160, 163, 180, 187, 192, 195, 228, 235,
267, 288, 315, 340, 352, 372, 403, 420, 427, 435, 448, 480, 483, 532, 555, 595, 627, 660, 672, 708, 715, 795, 928, 960, 1012, 1092, 1120, 1155,
1248, 1380, 1428, 1435, 1540, 1632, 1995, 2080, 3003, 3040, 3315, 3360, 5280, 5460, 7392

By some same way, the following rings of integers Z[(1+√k)/2],
Z[(1+√3)/2], Z[(1+√7)/2], Z[(1+√11)/2], Z[(1+√19)/2], Z[(1+√43)/2], Z[(1+√67)/2], Z[(1+√163)/2],

for k = 3, 7, 11, 19, 43, 67, 163 are being unique factorization domains, by itself.
with k being prime, of the form 3 (mod 4), by itself,
and then (k+1)/4 is being either 1 or prime,
with the polynomial of the form x²+x+(k+1)/4 form,
being with generating with all-prime values from k = 0 ... (k-7)/4.

Quite directly
it is being
Amazingly startling with, unbelievable with,
by itself, with, !

Following statement why it is being significant enough, as follows

Let N be a non-negative integer that can be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0).
Let p be a prime factor that divides N (raised to an odd power), i.e. pm divides N, pm+1 does not divide N, whereby m is being odd.
Then, if p can also always be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0),

then k = 1, 2, 3, 4, 7, 8, 11, 12, 19, 43, 67, 163

[Raman]

For a given k,
it is being a good question to ask how many, (what all) polynomial forms are being needed in order to represent with
all the prime numbers that are being congruent to x² (mod 4k), x²+k (mod 4k).

The answer is being given below as follows: All of the following polynomials will have got with the discriminant value of -4k,
by itself, over thereby,

Program Output

k = 1: a² + b²
k = 2: a² + 2b²
k = 3: a² + 3b²
k = 4: a² + 4b²
k = 5: a² + 5b²
k = 6: a² + 6b²
k = 7: a² + 7b²
k = 8: a² + 8b²
k = 9: a² + 9b²
k = 10: a² + 10b²
k = 11: a² + 11b², 3a² + 2ab + 4b²
k = 12: a² + 12b²
k = 13: a² + 13b²
k = 14: a² + 14b², 2a² + 7b²
k = 15: a² + 15b²
k = 16: a² + 16b²
k = 17: a² + 17b², 2a² + 2ab + 9b²
k = 18: a² + 18b²
k = 19: a² + 19b², 4a² + 2ab + 5b²
k = 20: a² + 20b², 4a² + 5b²
k = 21: a² + 21b²
k = 22: a² + 22b²
k = 23: a² + 23b², 3a² + 2ab + 8b²
k = 24: a² + 24b²
k = 25: a² + 25b²
k = 26: a² + 26b², 3a² + 2ab + 9b²
k = 27: a² + 27b², 4a² + 2ab + 7b²
k = 28: a² + 28b²
k = 29: a² + 29b², 5a² + 2ab + 6b²
k = 30: a² + 30b²
k = 31: a² + 31b², 5a² + 4ab + 7b²
k = 32: a² + 32b², 4a² + 4ab + 9b²
k = 33: a² + 33b²
k = 34: a² + 34b², 2a² + 17b²
k = 35: a² + 35b², 4a² + 6ab + 11b²
k = 36: a² + 36b², 4a² + 9b²
k = 37: a² + 37b²
k = 38: a² + 38b², 6a² + 4ab + 7b²
k = 39: a² + 39b², 3a² + 13b²
k = 40: a² + 40b²
k = 41: a² + 41b², 2a² + 2ab + 21b², 5a² + 4ab + 9b²
k = 42: a² + 42b²
k = 43: a² + 43b², 4a² + 2ab + 11b²
k = 44: a² + 44b², 5a² + 2ab + 9b²
k = 45: a² + 45b²
k = 46: a² + 46b², 2a² + 23b²
k = 47: a² + 47b², 3a² + 2ab + 16b², 7a² + 6ab + 8b²
k = 48: a² + 48b²
k = 49: a² + 49b², 2a² + 2ab + 25b²
k = 50: a² + 50b², 6a² + 8ab + 11b²
k = 51: a² + 51b², 4a² + 2ab + 13b²
k = 52: a² + 52b², 4a² + 13b²
k = 53: a² + 53b², 6a² + 10ab + 13b²
k = 54: a² + 54b², 7a² + 6ab + 9b²
k = 55: a² + 55b², 5a² + 11b²
k = 56: a² + 56b², 8a² + 8ab + 9b²
k = 57: a² + 57b²
k = 58: a² + 58b²
k = 59: a² + 59b², 3a² + 2ab + 20b², 5a² + 2ab + 12b², 7a² + 4ab + 9b², 4a² + 6ab + 17b²
k = 60: a² + 60b²
k = 61: a² + 61b², 5a² + 4ab + 13b²
k = 62: a² + 62b², 2a² + 31b², 7a² + 2ab + 9b²
k = 63: a² + 63b², 7a² + 9b²
k = 64: a² + 64b², 4a² + 4ab + 17b²
k = 65: a² + 65b², 9a² + 10ab + 10b²
k = 66: a² + 66b², 3a² + 22b²
k = 67: a² + 67b², 4a² + 2ab + 17b²
k = 68: a² + 68b², 8a² + 12ab + 13b², 4a² + 17b²
k = 69: a² + 69b², 6a² + 6ab + 13b²
k = 70: a² + 70b²
k = 71: a² + 71b², 3a² + 2ab + 24b², 5a² + 4ab + 15b², 8a² + 2ab + 9b²
k = 72: a² + 72b²
k = 73: a² + 73b², 2a² + 2ab + 37b²
k = 74: a² + 74b², 3a² + 2ab + 25b², 9a² + 10ab + 11b²
k = 75: a² + 75b², 4a² + 2ab + 19b²
k = 76: a² + 76b², 5a² + 4ab + 16b²
k = 77: a² + 77b², 9a² + 14ab + 14b²
k = 78: a² + 78b²
k = 79: a² + 79b², 5a² + 2ab + 16b², 8a² + 6ab + 11b²
k = 80: a² + 80b², 9a² + 16ab + 16b²
k = 81: a² + 81b², 9a² + 12ab + 13b²
k = 82: a² + 82b², 2a² + 41b²
k = 83: a² + 83b², 3a² + 2ab + 28b², 7a² + 2ab + 12b², 9a² + 8ab + 11b², 4a² + 6ab + 23b²
k = 84: a² + 84b², 4a² + 21b²
k = 85: a² + 85b²
k = 86: a² + 86b², 6a² + 8ab + 17b², 9a² + 4ab + 10b²
k = 87: a² + 87b², 7a² + 4ab + 13b²
k = 88: a² + 88b²
k = 89: a² + 89b², 2a² + 2ab + 45b², 5a² + 2ab + 18b², 9a² + 16ab + 17b²
k = 90: a² + 90b², 9a² + 10b²
k = 91: a² + 91b², 4a² + 2ab + 23b²
k = 92: a² + 92b², 9a² + 10ab + 13b²
k = 93: a² + 93b²
k = 94: a² + 94b², 2a² + 47b², 7a² + 4ab + 14b²
k = 95: a² + 95b², 5a² + 19b², 9a² + 4ab + 11b²
k = 96: a² + 96b², 4a² + 4ab + 25b²
k = 97: a² + 97b², 2a² + 2ab + 49b²
k = 98: a² + 98b², 2a² + 49b², 9a² + 2ab + 11b²
k = 99: a² + 99b², 4a² + 2ab + 25b²
k = 100: a² + 100b², 4a² + 25b²

[Raman]

Some common and / then classical observations / questions...

1. I had been deeply analyzing with the truth of the following statements.

Suppose S(k) be the set of all numbers that can be written in the form a²+kb².
N, p, q are any arbitrary numbers satisfying with the below stated conditions,
they are being freshly defined, to be independent from the other statements, for
a new statement. Then,
Which of the following statements are true, which of them are false?

i. If a prime p belongs to S(k) and N belongs to S(k), then p × N belongs to S(k). TRUE. (a²+kb²)(c²+kd²) = |ac+kbd|² + k|ad-bc|² = |ac-kbd|² + k|ad+bc|²
ii. If a prime p belongs to S(k) and N does not belong to S(k), then p × N does not belong to S(k). TRUE. Consequence of III
iii. If a prime p belongs to S(k) and p × N belongs to S(k), then N belongs to S(k). TRUE. Consequence of Unique Factorization of Ideals?
iv. If a prime p belongs to S(k) and p × N does not belong to S(k), then N does not belong to S(k). TRUE. Consequence of I
v. For a prime p, if N belongs to S(k) and p × N belongs to S(k), then p belongs to S(k). FALSE. Classical counterexample: 28 belongs to S(24), 196 belongs to S(24), but 7 does not belong to S(24)
vi. For a prime p, if N belongs to S(k) and p × N does not belong to S(k), then p does not belong to S(k). TRUE. Consequence of I
vii. For a prime p, if N does not belong to S(k) and p × N belongs to S(k), then p does not belong to S(k). TRUE. Consequence of III
viii. For a prime p, if N does not belong to S(k) and p × N does not belong to S(k), then p belongs to S(k). FALSE. Counterexample: 7 does not belong to S(1), 21 does not belong to S(1), but 3 does not belong to S(1)



2. Primes of form a²+b²: All primes congruent to 1, 2 (mod 4)
Primes of form a²+2b²: All primes congruent to 1, 2, 3 (mod 8)
Primes of form a²+3b²: All primes congruent to 1, 7 (mod 12)
Primes of form a²+4b²: All primes congruent to 1, 5, 9, 13 (mod 16)
Primes of form a²+5b²: All primes congruent to 1, 5, 9 (mod 20)
Primes of form a²+6b²: All primes congruent to 1, 7 (mod 24)
Primes of form a²+7b²: All primes congruent to 1, 7, 9, 11, 15, 23, 25 (mod 28)
Primes of form a²+8b²: All primes congruent to 1, 9, 17, 25 (mod 32)
Primes of form a²+9b²: All primes congruent to 1, 13, 25 (mod 36)
Primes of form a²+10b²: All primes congruent to 1, 9, 11, 19 (mod 40)

Question.
Why primes of form a²+kb² = Primes congruent to certain set of some residue classes (mod 4k)?
Although "all primes" does not work out with always, it works out precisely for the sixty-five values of k alone - within such cases.
Why - mod 4k?

Answer.
My own guess is that if a prime p can be written into the form a²+kb² then certainly
that -k needs to be a quadratic residue (mod p).
If k is congruent to 3 (mod 4), then by the law of quadratic reciprocity, Legendre Symbol
(-k|p) = (p|k) always,
so the quadratic character of p is the same of quadratic character of p + k...
into this case, it can certainly be simplified into certain set of some residue classes (mod k)

On the other hand,
if k is congruent to 1 (mod 4), then by the law of quadratic reciprocity, Legendre Symbol
(-k|p) = -(p|k), if p = 1 mod 4
(-k|p) = (p|k), if p = 3 mod 4
thus the quadratic character of p will be the same of quadratic character of p + (4 × k)...

In general, by using what factor do you think that can (mod 4k) be simplified by, for the some haphazard k value?
For this example, consider with
n = 1 mod 4, or n = 2 mod 4, n is being square-free - The certain set of some residue classes (mod 4k)...
n = 3 mod 4, or n = 0 mod 4, The certain set of some residue classes (mod k)...
n = 0 mod 16, The certain set of some residue classes (mod k/2)...
n = 0 mod 9, The certain set of some residue classes (mod 4k/3)...
n = 0 mod 25, The certain set of some residue classes (mod 4k/5)...
n = 0 mod 49, The certain set of some residue classes (mod 4k/7)...
n = 0 mod 36, The certain set of some residue classes (mod k/3)...

Why so?
Thus



3. Question.
Is any prime number being uniquely generated by using a²+kb² form? With the following condition that a, b are non-negative?

If the condition that a, b are non-negative is not being imposed, then certainly that any prime number can be generated by using
FOUR different ways, as since a²+kb² = a²+k(-b)² = (-a)²+kb² = (-a)²+k(-b)².

Similarly, that can any prime number can be generated by using some arbitrary quadratic form (with negative discriminant) into exactly FOUR different ways?
Does there exist with some random quadratic form (with negative discriminant) for which a prime number can be generated by using more or
less than FOUR different ways?

For this example, consider with f(a,b) = 3a²+10ab+11b², then
with f(a,b) = f(-a,-b)
19 = f(4,-1) = f(-4,1) = f(5,-2) = f(-5,2)



4. Question.
Can the same prime number be generated by TWO different quadratic forms of the same discriminant,
that are inequivalent?
For example, the set of primes being generated by using a²+11b², 3a²+2ab+4b² are disjoint.
For example, the set of primes being generated by using a²+14b², 2a²+7b² are disjoint.
For example, the set of primes being generated by using a²+17b², 2a²+2ab+9b² are disjoint.
For example, the set of primes being generated by using a²+19b², 4a²+2ab+5b² are disjoint.

Does this following condition does rather hold out with - always?



5. Question.
The Cornacchia-Smith algorithm can be used to test out if a prime number p can be written
into the form a²+kb² almost instantly. Or by using the modular square root algorithm, you can
get the solutions (a,b) such that a²+kb² = p, by taking the GCD of x+√-k with N into Z[√-k], whereby x² = -k (mod p).

Does a similar test exist for getting with the solutions (a,b) of p = Xa²+Yab+Zb²,
for the following arbitrary random quadratic form?



6.
The following two quadratic forms are being equivalent as since one can be got / obtained by using a linear transformation from the another!
u(a,b) = 8a²+9b²; v(a,b) = 17a²+50ab+41b²;
v(a,b) = u(a+2b,a+b)

The following three quadratic forms are being equivalent as since one can be got / obtained by using a linear transformation from the another!
f(a,b) = 3a²+2ab+3b²; g(a,b) = 3a²+4ab+4b²; h(a,b) = 3a²+10ab+11b²;
g(a,b) = f(a+b,b); h(a,b) = g(a+b,b)
Similarly, that g(a,b) = h(a-b,b); f(a,b) = g(a-b,b)

Thus, two or more equivalent quadratic polynomial forms will always generate with the same set of values, always, in... Why so?

Question.
Two or more equivalent polynomials will always have sharing the same discriminant.
That thing is being a necessary condition, not being a sufficient condition at all...
Is there being some "sufficient condition" test, in order to experiment out, if the two or more polynomials are being equivalent? Things and stuff, then, simultaneously, in order to check them out within almost instantly...



7. If a number N, in general, can be written into the form N = a²+kb² = c²+kd² into two different ways, simultaneously,
then certainly that GCD((a+c)²+k(b+d)², N) will generate with a non-trivial factor for the number N, yielding in...



8. All primes being congruent to 1 (mod 24) can certainly be generated by using the following forms - a² + 24b², a² + 72b²
All primes being congruent to 3, 11 (mod 24) can certainly be generated by using the following forms - 3a² + 8b²
All primes being congruent to 5 (mod 24) can certainly be generated by using the following forms - 5a² + 2ab + 5b²
All primes being congruent to 7 (mod 24) can certainly be generated by using the following forms - 7a² + 10ab + 7b²
All primes being congruent to 11 (mod 24) can certainly be generated by using the following forms - 11a² + 14ab + 11b²
All primes being congruent to 17 (mod 24) can certainly be generated by using the following forms - 8a² + 9b²
All primes being congruent to 19 (mod 24) can certainly be generated by using the following forms - 4a² + 4ab + 19b²



9. Regarding the prime numbers being generated by using the polynomial a²+kb² for the negative k values,
for the such cases,

A prime number p can be written into the form a²-Db² form, if and only if,
D is being a quadratic residue mod p. Certainly, that there should be infinitely many number of solutions (a,b)
into towards p = a²-Db² for the D ≥ 2, as since, there will be infinitely many units into the real number field Z[√D], as such,

for the multiplication by using integer solutions into the Pell's equation a²-Db² = 1.

Modular arithmetic modulo 4D, will certainly, reveal that almost instantly that there can be no integer solutions for the ordered pair (a,b)
into the equation a²-Db² = p, if p is being prime number; D is being quadratic non-residue mod p.

[Raman]

10.
There does not exist non-zero (positive) integers N that can be simultaneously written into two following forms, as follows, as a, b, c, d are being integers, always

i. N = a² + 5b² = 2c² + 2cd + 3d²
ii. N = a² + 6b² = 2c² + 3d²
iii. N = a² + 10b² = 2c² + 5d²
iv. N = a² + 13b² = 2c² + 2cd + 7d²
v. N = a² + 22b² = 2c² + 11d²
vi. N = a² + 26b² = 2c² + 13d²
vii. N = a² + 38b² = 2c² + 19d²
viii. N = a² + 58b² = 2c² + 29d²
ix. N = a² + 15b² = 3c² + 5d²
x. N = a² + 21b² = 3c² + 7d²
xi. N = a² + 24b² = 3c² + 8d²
xii. N = a² + 30b² = 2c² + 15d²
xiii. N = a² + 30b² = 3c² + 10d²
xiv. N = a² + 30b² = 5c² + 6d²
xv. N = a² + 33b² = 3c² + 11d²
xvi. N = a² + 40b² = 5c² + 8d²
xvii. N = a² + 42b² = 2c² + 21d²
xviii. N = a² + 42b² = 3c² + 14d²
xix. N = a² + 42b² = 6c² + 7d²
xx. N = a² + 51b² = 3c² + 17d²
xxi. N = a² + 57b² = 3c² + 19d²
xxii. N = a² + 60b² = 3c² + 20d²
xxiii. N = a² + 60b² = 5c² + 12d²
xxiv. N = a² + 70b² = 2c² + 35d²
xxv. N = a² + 70b² = 5c² + 14d²
xxvi. N = a² + 70b² = 7c² + 10d²
xxvii. N = a² + 78b² = 2c² + 39d²
xxviii. N = a² + 78b² = 3c² + 26d²
xxix. N = a² + 78b² = 6c² + 13d²
xxx. N = a² + 85b² = 5c² + 17d²
xxxi. N = a² + 88b² = 8c² + 11d²
xxxii. N = a² + 93b² = 3c² + 31d²
xxxiii. N = a² + 66b² = 6c² + 11d²
xxxiv. N = a² + 77b² = 7c² + 11d²

There does exist non-zero (positive) integers N that can be simultaneously written into two following forms, as follows, as a, b, c, d are being integers, always

xxxv. N = a² + 18b² = 2a² + 9b²
xxxvi. N = a² + 14b² = 2c² + 7d²
xxxvii. N = a² + 34b² = 2c² + 17d²
xxxviii. N = a² + 46b² = 2c² + 23d²
xxxix. N = a² + 12b² = 3c² + 4d²
xl. N = a² + 20b² = 4c² + 5d²
xli. N = a² + 72b² = 8c² + 9d²
xlii. N = a² + 28b² = 4c² + 7d²
xliii. N = a² + 45b² = 5c² + 9d²
xliv. N = a² + 48b² = 3c² + 16d²
xlv. N = a² + 60b² = 4c² + 15d²
xlvi. N = a² + 11b² = 3c² + 2cd + 4d²
xlvii. N = a² + 17b² = 2c² + 2cd + 9d²
xlviii. N = a² + 19b² = 4c² + 2cd + 5d²
xlix. N = a² + 23b² = 3c² + 2cd + 8d²
l. N = a² + 26b² = 3c² + 2cd + 9d²
li. N = a² + 29b² = 5c² + 2cd + 6d²
lii. N = a² + 31b² = 5c² + 4cd + 7d²
liii. N = a² + 63b² = 7c² + 9d²
liv. N = a² + 44b² = 4c² + 11d²
lv. N = a² + 99b² = 9c² + 11d²
lvi. N = a² + 52b² = 4c² + 13d²
lvii. N = a² + 68b² = 4c² + 17d²
lviii. N = a² + 76b² = 4c² + 19d²
lix. N = a² + 84b² = 4c² + 21d²
lx. N = a² + 92b² = 4c² + 23d²

Why so?
Thus

[Raman]

Regarding the equation N = a² + 6b² = 2c² + 3d²,
a² + 6b² = 2c² + 3d²,

a² - 2c² = 3 (d² - 2b²).

3 divides a²-2c².
Considering modulo 3 arithmetic,
3 should divide a, 3 should divide c.

As since, if a²-2c² = 0 (mod 3),
then let a = m × c (mod 3), then certainly we will have that,
c²(m²-2) = 0 (mod 3)
THERE EXIST NO SOLUTION TO THE EQUATION m²-2 = 0 mod 3, as such,
as since, 2 is a quadratic non-residue mod 3.
So that, c MUST BE CONGRUENT TO 0 (mod 3).

Thus, a MUST ALSO BE CONGRUENT TO 0 (mod 3).

Let a = 3x, c = 3y,
Then, 9x² + 6b² = 18y² + 3d².
3x² + 2b² = 6y² + d².
So, 3 divides b, 3 divides d.

Let b = 3z; d = 3w.

3x² + 2b² = 6y² + d².
3x² + 18z² = 6y² + 9w².

x² + 6z² = 2y² + 3w².

(a/3)² + 6(b/3)² = 2(c/3)² + 3(d/3)² = N/9.

So, if N is writable simultaneously as a² + 6b², 2a² + 3b²
both the forms, then certainly N/9 is being writable as well.

We will get that GCD(a, b, c, d) ≥ 3. If in case that we claim
a primitive minimal solution for the following equation, a² + 6b² = 2c² + 3d²,
then, and this argument will certainly lead to a contradiction.

So, by using the method of infinite descent, the only possible
solutions into the following equation, a² + 6b² = 2c² + 3d², actually are
being a = 0, b = 0, c = 0, d = 0.



Similarly, that concerning for the following equation
a² + 10b² = 2c² + 5d²

This equation has got with no non-trivial solutions
As since 2 is being a quadratic non-residue mod 5,
a² - 2c² = 0 (mod 5).

This certainly leads into following
5 divides a, 5 divides c.

Similar argument will apply → no non-zero solutions.



On the other hand, for the following equation
a² + 14b² = 2c² + 7d²

As since 2 is a quadratic residue mod 7,
a² - 2c² = 0 (mod 7),
has certainly got with some solutions without necessarily that
7 dividing a, 7 dividing c.

So, the same argument for the above mentioned two equations over thereby
does not work out over hereby.

a² + 14b² = 2c² + 7d²
Applying that a = 1, b = 1, c = 2, d = 1, suits out!



In general, I claim that for the following equation,
a² + 2kb² = 2c² + kd²
if in case that 2, k are being co-prime to each other
in, then within,
the equation has got with non-trivial, i.e. non-zero solutions
if and only if 2 is being a quadratic residue mod k.



In general, I guess that for the following equation,
a² + pkb² = pc² + kd²
if in case that p, k are being co-prime to each other
in, then within,
the equation has got with non-trivial, i.e. non-zero solutions
if and only if p is being a quadratic residue mod k.

[Raman]

Dear alpertron:

-> I do know that if a prime p can be written into the form a²+kb² form, then
its representation is being unique - if both of a, b are being positive,
or without that restriction - there will be four solutions for a prime for
the representation by any quadratic form.

If N is the product of m distinct primes(ALL / SOME) such that it can be written into the form with
N = a²+kb² form, then there will be 2^m different solutions to (a, b) inside the positive integers,
by itself. If in case that you would include with the negative values, then there will be
2^m+2 different solutions to (a, b).

I am not much interested in the solutions of (a, b), as since they are not being unique.
I am much interested in the numbers which are being writable into the form a²+kb² form,
by itself.

If k = 5, you need to examine the solution to p = 2a²+2ab+3b² where p is being a prime factor of N congruent to {2, 3, 7} mod 20.
The product of two primes of the form 2a²+2ab+3b² can be written into the form a²+5b², by itself.
(2a²+2ab+3b²)(2c²+2cd+3d²) = |2ac+ad+bc-2bd|²+5|ad+bc+bd|²
(2a²+2ab+3b²)(2c²+2cd+3d²) = |2ac+ad+bc+3bd|²+5|ad-bc|²

If k = 6, you need to examine the solution to p = 2a²+3b² where p is being a prime factor of N congruent to {2, 3, 5, 11} mod 24.
The product of two primes of the form 2a²+3b² can be written into the form a²+6b², by itself.
(2a²+3b²)(2c²+3d²) = |2ac-3bd|²+6|ad+bc|²
(2a²+3b²)(2c²+3d²) = |2ac+3bd|²+6|ad-bc|²

If k = 10, you need to examine the solution to p = 2a²+5b² where p is being a prime factor of N congruent to {2, 5, 7, 13, 23, 37} mod 40.
The product of two primes of the form 2a²+5b² can be written into the form a²+10b², by itself.
(2a²+5b²)(2c²+5d²) = |2ac-5bd|²+10|ad+bc|²
(2a²+5b²)(2c²+5d²) = |2ac+5bd|²+10|ad-bc|²

But, these polynomial forms such as those like 2a²+2ab+3b², 2a²+3b², 2a²+5b²,
a²+5b², a²+6b², a²+10b², etc.
will span with ALL the primes lying in certain set of residue classes (mod 4k).

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

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

On the other hand, consider with the case for k = 11:
The polynomial of a²+11b² form will only span with SOME of the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
(You may expand it to the residue classes mod 44, by this itself).
On the other hand, the polynomial of 3a²+2ab+4b² form seems to generate with the REMAINING primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.

As a result, the two polynomials of a²+11b² form, 3a²+2ab+4b² form TOGETHER appears to form a basis for all the primes lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11.
Not it is being possible with any of these two single polynomials individually.

Is it being possible for anyone to explain why this happens? And then if possible being able to point out
off to any potentially available references, or links for the elaborating in detail, in depth towards this mechanism?

What primes are being representable in the form a²+11b² form, thereby distinguishing it from the primes of the form 3a²+2ab+4b² form,
lying in the residue classes {0, 1, 3, 4, 5, 9} mod 11, by itself

But although, for all the primes p lying in the following residue classes {0, 1, 3, 4, 5, 9} mod 11,
it's being possible to write with 4 × p into the form a²+11b², by itself.

4 × (a²+11b²) = c²+11d² form, with, c = 2a, d = 2b, over thereby,
by itself, following with this statement of transformation, by itself

4 × (3a²+2ab+4b²) = c²+11d² form, with, c = a + 4b, d = a, over thereby,
by itself, following with this statement of transformation, by itself

Similarly, so it happens, thus for the following values of k,
over thereby, by itself,

k = 11, 19, 20, 27, 32, 35, 36, 43, 51, 52, 64, 67, 75, 84, 91, 96, 99, 100, 115, 123, 132, 147, 148, 160, 163, 180, 187, 192, 195, 228, 235,
267, 288, 315, 340, 352, 372, 403, 420, 427, 435, 448, 480, 483, 532, 555, 595, 627, 660, 672, 708, 715, 795, 928, 960, 1012, 1092, 1120, 1155,
1248, 1380, 1428, 1435, 1540, 1632, 1995, 2080, 3003, 3040, 3315, 3360, 5280, 5460, 7392

By some same way, the following rings of integers Z[(1+√-k)/2],
Z[(1+√-3)/2], Z[(1+√-7)/2], Z[(1+√-11)/2], Z[(1+√-19)/2], Z[(1+√-43)/2], Z[(1+√-67)/2], Z[(1+√-163)/2],

for k = 3, 7, 11, 19, 43, 67, 163 are being unique factorization domains, by itself.
with k being prime, of the form 3 (mod 4), by itself,
and then (k+1)/4 is being either 1 or prime,
with the polynomial of the form x²+x+(k+1)/4 form,
being with generating with all-prime values from k = 0 ... (k-7)/4.

Quite directly
it is being
Amazingly startling with, unbelievable with,
by itself, with, !

Following statement why it is being significant enough, as follows

Let N be a non-negative integer that can be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0).
Let p be a prime factor that divides N (raised to an odd power), i.e. p^m divides N, p^m+1 does not divide N, whereby m is being odd.
Then, if p can also always be written as (a/2)²+k(b/2)² (with a ≥ 0, b ≥ 0),

then k = 1, 2, 3, 4, 7, 8, 11, 12, 19, 43, 67, 163