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

[Raman]

Originally Posted by Raman

I had been waiting for this, to be posted, with.
It was being much of work effort to generate for these cases - for these k values, as since
they all behave quite differently with - from each other cases.

In particular, it was quite difficult specifically to generate for the k = 72 case, as since
2² divides with 72; 3² divides with 72.

I think that I will certainly stop generating these forms with these above mentioned cases.
I have covered with all k values cases - for which the discriminant in the quadratic field
Z[√-k] has got with class numbers 1, 2, 4.

The other k values will have
Class Number 8: k = 105, 120, 165, 168, 210, 240, 273, 280, 312, 330, 345, 357, 385, 408, 462, 520, 760
Class Number 16: k = 840, 1320, 1365, 1848

For the cases with k values having with class number of 8 (or even 16),
my own program / coding style can classify these different equivalence classes for the various residue classes (mod 4k),
by studying that two primes p and q, (neither of them being writable in the form a²+kb², by itself
but some non-negative integer N being writable in the form a²+kb², does contain with prime factor of p raised to an odd power
and then some same / another (different) non-negative integer M, again being writable in the form a²+kb², does contain with prime factor of q raised to an odd power)
such that p² does not divide with k value, q² does not divide with k value

p and q - they belong to the same equivalence class if and only if p × q is being writable in the form a²+kb².

But,

Although these seven / fifteen different equivalence classes can be easily generated by my own
JAVA implementation - for the k values cases having with class numbers of eight / sixteen respectively
for the "SOME" primes being mentioned as in above cases,

But, the interaction of the combined product from among these different equivalence classes are being
harder enough to be studied; it is being needed to demonstrate with the interaction of the seven / fifteen
different equivalence classes into 127 or 32767 different ways, by itself,

Better that it may be automated, but should be verified by testing for the every N value, at last.

But, whenever that 4 divides with the value of k, then the behaviour of the prime factor of 2 needs to be studied, in depth, with, in detail.
But, whenever that 9 divides with the value of k, then the behaviour of the prime factor of 3 needs to be studied, in depth, with, in detail.
But, whenever that 36 divides with the value of k, then the behaviour of the prime factors of 2 and then 3 needs to be studied, in depth, with, in detail.

some
again certainly
but that
by that some same way
by itself
some

k = 45

Primes p of the form a²+45b²: All primes congruent to [1, 49] mod 60.
if N is a non-negative integer that can be written as a²+45b², then p × N can be written as a²+45b²
if N is a non-negative integer that cannot be written as a²+45b², then p × N cannot be written as a²+45b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [2, 23, 47] mod 60
Class B primes: All primes congruent to [5, 29, 41] mod 60
Class C primes: All primes congruent to [7, 43] mod 60

N can be written as a²+45b² if and only if
- N is not congruent to {3, 6} (mod 9).
- N has no prime factors congruent to [11, 13, 17, 19, 31, 37, 53, 59] mod 60 to an odd power.
- If N is not a multiple of 3, 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 3 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 C prime factors of N are of the same parity.
- If the highest power of 3 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 C prime factors of N are of different parity.


k = 48

Primes p of the form a²+48b²: All primes congruent to [1] mod 24.
if N is a non-negative integer that can be written as a²+48b², then p × N can be written as a²+48b²
if N is a non-negative integer that cannot be written as a²+48b², then p × N cannot be written as a²+48b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [3, 19] mod 24
Class B primes: All primes congruent to [7] mod 24
Class C primes: All primes congruent to [13] mod 24

N can be written as a²+48b² if and only if
- N has no prime factors congruent to [2, 5, 11, 17, 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 N is congruent to 4 (mod 8), 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.


k = 60

Primes p of the form a²+60b²: All primes congruent to [1, 49] mod 60.
if N is a non-negative integer that can be written as a²+60b², then p × N can be written as a²+60b²
if N is a non-negative integer that cannot be written as a²+60b², then p × N cannot be written as a²+60b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [3, 23, 47] mod 60
Class B primes: All primes congruent to [5, 17, 53] mod 60
Class C primes: All primes congruent to [19, 31] mod 60

N can be written as a²+60b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16).
- N has no prime factors congruent to [7, 11, 13, 29, 37, 41, 43, 59] mod 60 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 five 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 = 72

Primes p of the form a²+72b²: All primes congruent to [1] mod 24.
if N is a non-negative integer that can be written as a²+72b², then p × N can be written as a²+72b²
if N is a non-negative integer that cannot be written as a²+72b², then p × N cannot be written as a²+72b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [11] mod 24
Class B primes: All primes congruent to [17] mod 24
Class C primes: All primes congruent to [19] mod 24

N can be written as a²+72b² if and only if
- N is not congruent to 2 (mod 4).
- N is not congruent to {3, 6} (mod 9).
- N has no prime factors congruent to [5, 7, 13, 23] mod 24 to an odd power.
- If N is odd and not a multiple of 3, 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 N is not a multiple of 3, and 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 N is not a multiple of 3, and 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.
- If N is odd, and the highest power of 3 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 C prime factors of N are of the same parity.
- If N is odd, and the highest power of 3 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 C prime factors of N are of different parity.


k = 112

Primes p of the form a²+112b²: All primes congruent to [1, 9, 25] mod 56.
if N is a non-negative integer that can be written as a²+112b², then p × N can be written as a²+112b²
if N is a non-negative integer that cannot be written as a²+112b², then p × N cannot be written as a²+112b²

Let us define Class A, Class B, Class C primes as follows:
Class A primes: All primes congruent to [7, 15, 23, 39] mod 56
Class B primes: All primes congruent to [11, 43, 51] mod 56
Class C primes: All primes congruent to [29, 37, 53] mod 56

N can be written as a²+112b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16) or 32 (mod 64).
- N has no prime factors congruent to [3, 5, 13, 17, 19, 27, 31, 33, 41, 45, 47, 55] mod 56 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 N is congruent to 4 (mod 8), 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.

[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. But for k = 11, 14, 17, 19, etc. it may not be that easier enough, especially
if N is not being all composed of product of prime factors that can be written into the form
a²+kb² form - such prime factors being raised to an odd power, by itself!

Finding out with the representability of the a²+kb² form is being possible enough for the above mentioned
sixty-five values of k, with such cases.

By same way, newly found out Mersenne Prime 57885161 = 2444² + 7205² = 5469² + 2 × 3740².

It is being good enough to observe that the companion polynomial for a²+kb² form of a certain prime
residue class (mod 4k), can be given by using
f(a,b) = Xa² + Yab + Zb², whereby
X = first prime of the residue class (mod 4k),
Z = second prime of the residue class (mod 4k),
Y² - 4XZ = discriminant (-4k),
thereby, Y = √(4XZ - 4k).

We will certainly have that with
f(1,0) = first prime
f(0,1) = second prime

For example, consider this example with,
a²+5b² form
For the "SOME" prime residue class, as being mentioned above,
being congruent to {2, 3, 7} mod 20.

first prime = 2; second prime = 3; third prime = 7; discriminant (-4×5 = -20);
X = 2; Z = 3; Y = √(4XZ - 4k) = √(24 - 20) = 2;

The companion polynomial to a²+5b² form is being, thereby, 2a² + 2ab + 3b² form, by itself.

thereby,
We will certainly have that with

Companion polynomial to a²+5b² form is 2a² + 2ab + 3b² form, with discriminant = -20.
Companion polynomial to a²+6b² form is 2a² + 3b² form, with discriminant = -24.
Companion polynomial to a²+10b² form is 2a² + 5b² form, with discriminant = -40.
Companion polynomial to a²+13b² form is 2a² + 2ab + 7b² form, with discriminant = -52.
Companion polynomial to a²+22b² form is 2a² + 11b² form, with discriminant = -88.
Companion polynomial to a²+37b² form is 2a² + 2ab + 19b² form, with discriminant = -148.
Companion polynomial to a²+58b² form is 2a² + 29b² form, with discriminant = -232.

[Raman]

I wrote with my own implementation to generate with the companion polynomials
for the a²+kb² forms, for the sixty-five "convenient numbers" of k, they are being much easier to generate with, automatically, list them out
off proper format, automated!

They are being much easier to be generated with - which generate with all the primes within a
certain set of residue classes (mod 4k), they are being the only primes that can be generated
by using the polynomial, quadratic form - by itself!

These companion polynomials are being much easier to be generated with, rather than testing out
the condition whenever N can be generated in the form a²+kb², as most of them behave differently
with - from each other - from among the interaction among these ideal classes being generated by
using these polynomials, to be being put altogether with.

The companion polynomials of the quadratic form a²+kb², as well as the different ideal classes of them,
classifying the different primes of them being generated by using the various residue classes, and then
testing out whether a prime number can be generated by them - are all being very much easier enough,
all at once!

The main polynomial a²+kb², along with all their own companion polynomials - have discriminant value
of -4k, which is the modulo of the generated primes / prime factors of N, in general being taken to classify them.

So, if the class number of Z[√-k] is being x, then there will be x-1 companion polynomials to the main
polynomial a²+kb² form, by itself. Thereby, signifying with all the possibly available ideal classes with,
themselves, with potentially, by itself.

Originally Posted by Raman

It is being good enough to observe that the companion polynomial for a²+kb² form of a certain prime
residue class (mod 4k), can be given by using
f(a,b) = Xa² + Yab + Zb², whereby
X = first prime of the residue class (mod 4k),
Z = second prime of the residue class (mod 4k),
Y² - 4XZ = discriminant (-4k),
thereby, Y = √(4XZ - 4k).

We will certainly have that with
f(1,0) = first prime
f(0,1) = second prime

Oh, this algorithm is being wrong again! The first failure occurs for k = 24.

My first guess at this was that Y = (third prime of the residue class (mod 4k)) - X - Z,
but that was being even more terribly wrong. Fails for k = 8.

If the algorithm is to be corrected, then Z need not to be the (second prime of the residue class (mod 4k)),
if in case that 4XZ - 4k is not being a perfect square. But Z = lower than this (second prime of the residue class (mod 4k))
number, that which would make with the value of 4XZ - 4k to be a perfect square, with case for the value of Y = √(4XZ - 4k),
by itself.



Notice that f(a,b) = 3a²+10ab+11b² is a companion form for the following main form a²+8b², thereby it generates with all primes
congruent to 3 (mod 8), they are the only primes, actually, this polynomial can generate with.
but consider with g(a,b) = 3a²+4ab+4b² form, this again generates with the same numbers that the polynomial 3a²+10ab+11b² can
generate with - Notice that again both having with the same discriminant -32.
But both these forms are being equivalent! We can get each other, onto with, the transformation, with, f(a,b) = g(a+b, b),
and then, with, g(a,b) = f(a-b, b).

Notice that the polynomial Xa²+Yab+Zb² can take with zero, or, with negative values, for the variables of a, b respectively, as such,
as since, the term "Yab" is not being symmetric with respect to negation operation, that I do realize with.

So, a prime number can be written into, by using four different ways by using a quadratic form Xa²+Yab+Zb², by itself.
For h(a,b) = a²+kb², for example, if in case h(a,b) = prime, then h(a,b) = h(a,-b) = h(-a,b) = h(-a,-b).

Similarly, consider with the f(a,b) = 3a²+10ab+11b² form; 19 is being congruent to the 3 (mod 8) form of the residue class, by itself.
Thereby, 19 = f(4,-1) = f(-4,1) = f(5,-2) = f(-5,2)

of course, with the f(a,b) = f(-a,-b) property; f(a,-b) = f(-a,b) property holding out
off, by itself.

if in case g(a,b) = 3a²+4ab+4b² form,
then certainly,
we will have with,
Thereby, 19 = f(3,-1) = f(-3,1) = f(3,-2) = f(-3,2).

Thereby, any number being writable in the form 3a²+10ab+11b² or the form 3a²+4ab+4b² should have with
sum of exponents of 3 (mod 8) prime factors being odd.



For an other example, the companion polynomial for the a²+9b² form is being 2a²+2ab+5b² form, by itself.

For an other example, the companion polynomial for the a²+15b² form is being 3a²+5b² form, by itself.

It is not being necessary that X = (first prime of the residue class (mod 4k)), by itself,
but there is being no harm in doing so with going ahead in that same way,

which by itself, by some side

It is always being true, again that,
For the quadratic form a²+kb² form - if in case k = p × q, with (p,q) being co-prime to each other, but
and then necessarily that,

A companion polynomial for the a²+kb² form is being pa²+qb² form, by itself. but

For this following example, consider with, u(a,b) = 8a²+9b² form, this form is being a hidden companion polynomial to the visible
main polynomial a²+72b² form, by itself - thereby this polynomial generates with all primes
congruent to 17 (mod 24), they are the only primes, actually, it can generate with, by using that same way.

By using the mentioned algorithm, as being quoted as above,
and then we will certainly get with the companion polynomial for the main quadratic form
a²+72b² form - to be v(a,b) = 17a²+50ab+41b² form - notice that this form is being again,
being equivalent to the form, 8a²+9b² form - just simply - by going ahead with the same numbers that both of these two polynomials certainly can be able to
generate with

Notice that again both of these two forms having with the same discriminant -288.

We can certainly be able to get each other, with, onto applying with, the following transformation, with, v(a,b) = u(a+2b, a+b), by itself, by using that same way.

Thereby, any number being writable in the form 8a²+9b² or the form 17a²+50ab+41b² should have with
sum of exponents of 17 (mod 24) prime factors being odd.

[henryzz]

p=x^2+5*y^2 iff p = 1 or 9 (mod 20)
2p=x^2+5*y^2 iff p = 3 or 7 (mod 20)

iff stands for "if and only if"

As far as I can tell a lot of primes not representable in a form can be represented if they are doubled.

a*x^2+b*x*y+c*y^2
The discriminant is b^2 -4ac.

The class number refers to the number of polynomials in each group. There are also polynomials with class numbers not of the form 2^n. Each group is usually refered to by disciminant rather than k as the groups with class numbers not of the form 2^n don't have ks.
There are 3 polynomials with a discriminant of -23 for example.

[Raman]

For the record, this is what my own program source code / implementation - had generated with. - Companion polynomials for the main polynomials.

k = 1

Main polynomial
a² + b²

Discriminant = -4

k = 2

Main polynomial
a² + 2b²

Discriminant = -8

k = 3

Main polynomial
a² + 3b²

Discriminant = -12

k = 4

Main polynomial
a² + 4b²

Discriminant = -16

k = 5

Main polynomial
a² + 5b²

Companion polynomials
2a² + 2ab + 3b²

Discriminant = -20

k = 6

Main polynomial
a² + 6b²

Companion polynomials
2a² + 3b²

Discriminant = -24

k = 7

Main polynomial
a² + 7b²

Discriminant = -28

k = 8

Main polynomial
a² + 8b²

Companion polynomials
3a² + 4ab + 4b²

Discriminant = -32

k = 9

Main polynomial
a² + 9b²

Companion polynomials
2a² + 2ab + 5b²

Discriminant = -36

k = 10

Main polynomial
a² + 10b²

Companion polynomials
2a² + 5b²

Discriminant = -40

k = 12

Main polynomial
a² + 12b²

Companion polynomials
3a² + 4b²

Discriminant = -48

k = 13

Main polynomial
a² + 13b²

Companion polynomials
2a² + 2ab + 7b²

Discriminant = -52

k = 15

Main polynomial
a² + 15b²

Companion polynomials
3a² + 5b²

Discriminant = -60

k = 16

Main polynomial
a² + 16b²

Companion polynomials
4a² + 4ab + 5b²

Discriminant = -64

k = 18

Main polynomial
a² + 18b²

Companion polynomials
2a² + 9b²

Discriminant = -72

k = 21

Main polynomial
a² + 21b²

Companion polynomials
2a² + 2ab + 11b²
3a² + 7b²
5a² + 6ab + 6b²

Discriminant = -84

k = 22

Main polynomial
a² + 22b²

Companion polynomials
2a² + 11b²

Discriminant = -88

k = 24

Main polynomial
a² + 24b²

Companion polynomials
3a² + 8b²
5a² + 2ab + 5b²
7a² + 10ab + 7b²

Discriminant = -96

k = 25

Main polynomial
a² + 25b²

Companion polynomials
2a² + 2ab + 13b²

Discriminant = -100

k = 28

Main polynomial
a² + 28b²

Companion polynomials
4a² + 7b²

Discriminant = -112

k = 30

Main polynomial
a² + 30b²

Companion polynomials
2a² + 15b²
3a² + 10b²
5a² + 6b²

Discriminant = -120

k = 33

Main polynomial
a² + 33b²

Companion polynomials
2a² + 2ab + 17b²
3a² + 11b²
6a² + 6ab + 7b²

Discriminant = -132

k = 37

Main polynomial
a² + 37b²

Companion polynomials
2a² + 2ab + 19b²

Discriminant = -148

k = 40

Main polynomial
a² + 40b²

Companion polynomials
5a² + 8b²
7a² + 8ab + 8b²
4a² + 4ab + 11b²

Discriminant = -160

k = 42

Main polynomial
a² + 42b²

Companion polynomials
2a² + 21b²
3a² + 14b²
6a² + 7b²

Discriminant = -168

k = 45

Main polynomial
a² + 45b²

Companion polynomials
2a² + 2ab + 23b²
5a² + 9b²
7a² + 4ab + 7b²

Discriminant = -180

k = 48

Main polynomial
a² + 48b²

Companion polynomials
3a² + 16b²
7a² + 12ab + 12b²
4a² + 4ab + 13b²

Discriminant = -192

k = 57

Main polynomial
a² + 57b²

Companion polynomials
2a² + 2ab + 29b²
3a² + 19b²
6a² + 6ab + 11b²

Discriminant = -228

k = 58

Main polynomial
a² + 58b²

Companion polynomials
2a² + 29b²

Discriminant = -232

k = 60

Main polynomial
a² + 60b²

Companion polynomials
3a² + 20b²
5a² + 12b²
4a² + 15b²

Discriminant = -240

k = 70

Main polynomial
a² + 70b²

Companion polynomials
2a² + 35b²
5a² + 14b²
7a² + 10b²

Discriminant = -280

k = 72

Main polynomial
a² + 72b²

Companion polynomials
11a² + 14ab + 11b²
8a² + 9b²
4a² + 4ab + 19b²

Discriminant = -288

k = 78

Main polynomial
a² + 78b²

Companion polynomials
2a² + 39b²
3a² + 26b²
6a² + 13b²

Discriminant = -312

k = 85

Main polynomial
a² + 85b²

Companion polynomials
2a² + 2ab + 43b²
5a² + 17b²
10a² + 10ab + 11b²

Discriminant = -340

k = 88

Main polynomial
a² + 88b²

Companion polynomials
8a² + 11b²
8a² + 8ab + 13b²
4a² + 4ab + 23b²

Discriminant = -352

k = 93

Main polynomial
a² + 93b²

Companion polynomials
2a² + 2ab + 47b²
3a² + 31b²
6a² + 6ab + 17b²

Discriminant = -372

k = 102

Main polynomial
a² + 102b²

Companion polynomials
2a² + 51b²
3a² + 34b²
6a² + 17b²

Discriminant = -408

k = 105

Main polynomial
a² + 105b²

Companion polynomials
2a² + 2ab + 53b²
3a² + 35b²
5a² + 21b²
7a² + 15b²
11a² + 8ab + 11b²
13a² + 16ab + 13b²
6a² + 6ab + 19b²

Discriminant = -420

k = 112

Main polynomial
a² + 112b²

Companion polynomials
7a² + 16b²
11a² + 16ab + 16b²
4a² + 4ab + 29b²

Discriminant = -448

k = 120

Main polynomial
a² + 120b²

Companion polynomials
3a² + 40b²
5a² + 24b²
11a² + 2ab + 11b²
12a² + 12ab + 13b²
8a² + 8ab + 17b²
8a² + 15b²
4a² + 4ab + 31b²

Discriminant = -480

k = 130

Main polynomial
a² + 130b²

Companion polynomials
2a² + 65b²
5a² + 26b²
10a² + 13b²

Discriminant = -520

k = 133

Main polynomial
a² + 133b²

Companion polynomials
2a² + 2ab + 67b²
7a² + 19b²
13a² + 14ab + 14b²

Discriminant = -532

k = 165

Main polynomial
a² + 165b²

Companion polynomials
2a² + 2ab + 83b²
3a² + 55b²
5a² + 33b²
11a² + 15b²
13a² + 4ab + 13b²
19a² + 28ab + 19b²
6a² + 6ab + 29b²

Discriminant = -660

k = 168

Main polynomial
a² + 168b²

Companion polynomials
3a² + 56b²
7a² + 24b²
13a² + 24ab + 24b²
12a² + 12ab + 17b²
8a² + 8ab + 23b²
8a² + 21b²
4a² + 4ab + 43b²

Discriminant = -672

k = 177

Main polynomial
a² + 177b²

Companion polynomials
2a² + 2ab + 89b²
3a² + 59b²
6a² + 6ab + 31b²

Discriminant = -708

k = 190

Main polynomial
a² + 190b²

Companion polynomials
2a² + 95b²
5a² + 38b²
10a² + 19b²

Discriminant = -760

k = 210

Main polynomial
a² + 210b²

Companion polynomials
2a² + 105b²
3a² + 70b²
5a² + 42b²
7a² + 30b²
14a² + 15b²
10a² + 21b²
6a² + 35b²

Discriminant = -840

k = 232

Main polynomial
a² + 232b²

Companion polynomials
8a² + 29b²
8a² + 8ab + 31b²
4a² + 4ab + 59b²

Discriminant = -928

k = 240

Main polynomial
a² + 240b²

Companion polynomials
3a² + 80b²
5a² + 48b²
17a² + 14ab + 17b²
19a² + 22ab + 19b²
12a² + 12ab + 23b²
15a² + 16b²
4a² + 4ab + 61b²

Discriminant = -960

k = 253

Main polynomial
a² + 253b²

Companion polynomials
2a² + 2ab + 127b²
11a² + 23b²
17a² + 22ab + 22b²

Discriminant = -1012

k = 273

Main polynomial
a² + 273b²

Companion polynomials
2a² + 2ab + 137b²
3a² + 91b²
7a² + 39b²
13a² + 21b²
17a² + 8ab + 17b²
23a² + 32ab + 23b²
6a² + 6ab + 47b²

Discriminant = -1092

k = 280

Main polynomial
a² + 280b²

Companion polynomials
5a² + 56b²
7a² + 40b²
17a² + 28ab + 28b²
19a² + 20ab + 20b²
8a² + 8ab + 37b²
8a² + 35b²
4a² + 4ab + 71b²

Discriminant = -1120

k = 312

Main polynomial
a² + 312b²

Companion polynomials
3a² + 104b²
13a² + 24b²
19a² + 24ab + 24b²
12a² + 12ab + 29b²
8a² + 8ab + 41b²
8a² + 39b²
4a² + 4ab + 79b²

Discriminant = -1248

k = 330

Main polynomial
a² + 330b²

Companion polynomials
2a² + 165b²
3a² + 110b²
5a² + 66b²
11a² + 30b²
15a² + 22b²
10a² + 33b²
6a² + 55b²

Discriminant = -1320

k = 345

Main polynomial
a² + 345b²

Companion polynomials
2a² + 2ab + 173b²
3a² + 115b²
5a² + 69b²
19a² + 30ab + 30b²
15a² + 23b²
10a² + 10ab + 37b²
6a² + 6ab + 59b²

Discriminant = -1380

k = 357

Main polynomial
a² + 357b²

Companion polynomials
2a² + 2ab + 179b²
3a² + 119b²
7a² + 51b²
17a² + 21b²
19a² + 4ab + 19b²
29a² + 44ab + 29b²
6a² + 6ab + 61b²

Discriminant = -1428

k = 385

Main polynomial
a² + 385b²

Companion polynomials
2a² + 2ab + 193b²
5a² + 77b²
7a² + 55b²
11a² + 35b²
22a² + 22ab + 23b²
14a² + 14ab + 31b²
10a² + 10ab + 41b²

Discriminant = -1540

k = 408

Main polynomial
a² + 408b²

Companion polynomials
3a² + 136b²
17a² + 24b²
23a² + 24ab + 24b²
12a² + 12ab + 37b²
8a² + 8ab + 53b²
8a² + 51b²
4a² + 4ab + 103b²

Discriminant = -1632

k = 462

Main polynomial
a² + 462b²

Companion polynomials
2a² + 231b²
3a² + 154b²
7a² + 66b²
11a² + 42b²
21a² + 22b²
14a² + 33b²
6a² + 77b²

Discriminant = -1848

k = 520

Main polynomial
a² + 520b²

Companion polynomials
5a² + 104b²
13a² + 40b²
23a² + 40ab + 40b²
20a² + 20ab + 31b²
8a² + 8ab + 67b²
8a² + 65b²
4a² + 4ab + 131b²

Discriminant = -2080

k = 760

Main polynomial
a² + 760b²

Companion polynomials
5a² + 152b²
19a² + 40b²
29a² + 40ab + 40b²
20a² + 20ab + 43b²
8a² + 8ab + 97b²
8a² + 95b²
4a² + 4ab + 191b²

Discriminant = -3040

k = 840

Main polynomial
a² + 840b²

Companion polynomials
3a² + 280b²
5a² + 168b²
7a² + 120b²
29a² + 2ab + 29b²
31a² + 22ab + 31b²
37a² + 46ab + 37b²
24a² + 24ab + 41b²
20a² + 20ab + 47b²
24a² + 35b²
21a² + 40b²
15a² + 56b²
12a² + 12ab + 73b²
8a² + 8ab + 107b²
8a² + 105b²
4a² + 4ab + 211b²

Discriminant = -3360

k = 1320

Main polynomial
a² + 1320b²

Companion polynomials
3a² + 440b²
5a² + 264b²
11a² + 120b²
37a² + 60ab + 60b²
41a² + 38ab + 41b²
43a² + 46ab + 43b²
24a² + 24ab + 61b²
20a² + 20ab + 71b²
33a² + 40b²
24a² + 55b²
15a² + 88b²
12a² + 12ab + 113b²
8a² + 8ab + 167b²
8a² + 165b²
4a² + 4ab + 331b²

Discriminant = -5280

k = 1365

Main polynomial
a² + 1365b²

Companion polynomials
2a² + 2ab + 683b²
3a² + 455b²
5a² + 273b²
7a² + 195b²
13a² + 105b²
37a² + 4ab + 37b²
42a² + 42ab + 43b²
30a² + 30ab + 53b²
59a² + 92ab + 59b²
14a² + 14ab + 101b²
10a² + 10ab + 139b²
21a² + 65b²
15a² + 91b²
179a² + 218ab + 74b²
6a² + 6ab + 229b²

Discriminant = -5460

k = 1848

Main polynomial
a² + 1848b²

Companion polynomials
3a² + 616b²
7a² + 264b²
11a² + 168b²
43a² + 84ab + 84b²
47a² + 38ab + 47b²
53a² + 62ab + 53b²
28a² + 28ab + 73b²
24a² + 24ab + 83b²
33a² + 56b²
24a² + 77b²
21a² + 88b²
12a² + 12ab + 157b²
8a² + 8ab + 233b²
8a² + 231b²
4a² + 4ab + 463b²

Discriminant = -7392



Are the other people reading / listening to this? Or rather I am talking to myself?

[Raman]

Notice with the correction being made!

Originally Posted by henryzz

p=x²+5*y² iff p ≡ 1 or 5 or 9 (mod 20)
2*p=x²+5*y² iff p ≡ 2 or 3 or 7 (mod 20)

iff stands for "if and only if"

This is only a small example, by itself - a tiny miniscule of the whole idea in detail, in my own humble opinion.

A prime p is of the form a²+5b² if and only if it is congruent to {1, 5, 9} mod 20.
A prime p is of the form 2a²+2ab+3b² if and only if it is congruent to {2, 3, 7} mod 20.

In general,
a number can be written in the form a²+5b² form if and only if
- it has got with no prime factors congruent to {11, 13, 17, 19} mod 20 raised to an odd power.
- sum of exponents of {2, 3, 7} mod 20 prime factors is even.

In general,
a number can be written in the form 2a²+2ab+3b² form if and only if
- it has got with no prime factors congruent to {11, 13, 17, 19} mod 20 raised to an odd power.
- sum of exponents of {2, 3, 7} mod 20 prime factors is odd.

Any product of two numbers of the form 2a²+2ab+3b² is being a number of the form a²+5b² form, by itself.
And then, in general, the product of a number of the form 2a²+2ab+3b², and then a number of the form a²+5b² form - is being, again, a number of the form 2a²+2ab+3b² form, by itself.

Originally Posted by henryzz

As far as I can tell a lot of primes not representable in a form can be represented if they are doubled.

This is because Z[√-5] is not being a unique factorization domain.
It does not have with class number of 1, let alone with class number of 2.

Originally Posted by henryzz

The class number refers to the number of polynomials in each group. There are also polynomials with class numbers not of the form 2ⁿ. Each group is usually refered to by disciminant rather than k as the groups with class numbers not of the form 2ⁿ don't have k's.
There are 3 polynomials with a discriminant of -23 for example.

Thanks to you, for this useful information.
This concept is being much more concrete to visualize with, rather than going on
with - by using the classification of ideals - abstractly inside the ring Z[√-k].

I was getting ready, next, to experiment with that out,
off.

But actually

only for values of k - cases, with the ring Z[√-k] - thereby having with the class number of 2ⁿ, for these sixty-five values of k,
as being mentioned as above

we can expect with the all the primes / only primes being generated by using the main polynomial / the companion polynomials
to be categorized by using a certain set of the residue classes (mod 4k), by itself.

For this example, consider this with,

A prime number p is of the form a² + 105b² form - if and only if - p is being congruent to [1, 109, 121, 169, 289, 361] mod 420.
A prime number p is of the form 2a² + 2ab + 53b² form - if and only if - p is being congruent to [2, 53, 113, 137, 197, 233, 317] mod 420.
A prime number p is of the form 3a² + 35b² form - if and only if - p is being congruent to [3, 47, 83, 143, 167, 227, 383] mod 420.
A prime number p is of the form 5a² + 21b² form - if and only if - p is being congruent to [5, 41, 89, 101, 209, 269, 341] mod 420.
A prime number p is of the form 7a² + 15b² form - if and only if - p is being congruent to [7, 43, 67, 127, 163, 247, 403] mod 420.
A prime number p is of the form 11a² + 8ab + 11b² form - if and only if - p is being congruent to [11, 71, 179, 191, 239, 359] mod 420.
A prime number p is of the form 13a² + 16ab + 13b² form - if and only if - p is being congruent to [13, 73, 97, 157, 313, 397] mod 420.
A prime number p is of the form 6a² + 6ab + 19b² form - if and only if - p is being congruent to [19, 31, 139, 199, 271, 391] mod 420.

The product of any two primes from among any one single residue class can be written into the form a² + 105b² form, by itself.



If the class number of the underlying ring Z[√-k] is not being 2ⁿ, then you cannot categorize the different primes
being writable into the form a²+kb² forms, such cases, as by using the residue classes technique (mod 4k), by itself. Can you do so with it?

Can you be able to give with a simple / easy condition for the stating of with - thereby what are being the different possible prime numbers that can be written into
the form a²+23b² form, by itself? Or with the form a²+31b² form? Thereby, these underlying rings Z[√-k] have got with class number of 3.

Even, if in case, the class number of the underlying ring Z[(1 + √-k)/2] for the some of the values of the k = 3 (mod 4) cases, or the k = 0 (mod 4) cases, is being equal to 2ⁿ, even then,
you cannot classify with all the writable primes of the form a²+kb² form - like these certain set of residue classes (mod 4k), technique, by itself. Can you be able to do so with it?
Especially for with these following seventy-four values of k, as, such cases.

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

Will you be able to classify with the all the writable prime numbers of the form a²+kb² form - such as like with these following residue classes (mod 4k),
techniques, for within these above mentioned seventy-four values of k - cases, as such? You will be actually, only be able to be classify with the primes, p
that which can be written into the form 4 × p = a²+kb² form, if in case I am not being mistaken with - that which I do realize with, appropriately, by itself.

And then, thereby noticing that for the following rings
that which are being nothing but unique factorization domains, appropriately, by itself
Z[i], Z[√-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]
values of k - cases, as such
k = 1, 2, 3, (4), 7
k = (8), 11, (12), 19, 43, 67, 163
It does have got with class number of 1, okay alright and then
crazily enough
crazy enough
rather from among with in
by going onto using itself with, in to be, being from among, again
k being square-free again
(k) being non-square-free again
by using that some same way again, that which
by making use ofr, again
anyway, somehow
noticing that, nothing but

[henryzz]

A prime p can be represented in the form a^2+k*b^2 if p doesn't divide k and p=x^2 (mod 4k) or p=x^2+n (mod 4k).

I got this from the end of section 2 in http://www.math.harvard.edu/~chaoli/...2/SethNeel.pdf

In general I believe it is more normal to exclude special cases like 2 and 5. It helps with clarity.
By showing that 2*p=x2+5*y2 iff p ≡ 2 or 3 or 7 (mod 20) I hoped to show that the other forms can represent a multiple of p. According to what I have read I think it is p, -p, 2p, -2p are possiblities.

[ewmayer]

Originally Posted by Raman

Are the other people reading / listening to this? Or rather I am talking to myself?

Assuredly so.

[Raman]

Can you please take more care with your own posts, into the future?
There is no n anywhere within the statement.
But, I shall assume that you meant with k.

Originally Posted by henryzz

A prime p can be represented in the form a²+k*b² if p doesn't divide k and
p ≡ x² (mod 4k) or p ≡ x²+k (mod 4k).

After the correction, I'm afraid that this statement is FALSE.

For example, consider the case with value of k = 11.

Let p = 89 ≡ 1 (mod 44). p ≡ 1² (mod 4*11), but can 89 be written into the form - a²+11b² form, by itself?
Let p = 97 ≡ 9 (mod 44). p ≡ 3² (mod 4*11), but can 97 be written into the form - a²+11b² form, by itself?
Let p = 67 ≡ 23 (mod 44). p ≡ 10² + 11 (mod 4*11), but can 67 be written into the form - a²+11b² form, by itself?

Please clarify with these cases!



(Although I believe that this statement is being essentially true enough
for the sixty-five values of k - such as cases like - as they are being mentioned above
over thereby!)


53 = (3)² + 11×(2)²
89 = (4½)² + 11×(2½)²
97 = (8½)² + 11×(1½)²

67 = (6½)² + 11×(1½)²
199 = (10)² + 11×(3)²


25² ≡ -11 (mod 53); the GCD of 53 with 25+√-11 inside the ring Z[√-11], by itself, will certainly give with 3-2√-11.
such that 53 = (3)² + 11×(2)².

over thereby

Likewise, 38² ≡ -11 (mod 97); but, why does the GCD of 97 with 38+√-11 fail with?
again!



I will certainly come back returning over hereby, afterwards,
once I have got with, for the three inequivalent polynomial forms,
for value of the following cases,
for the rings Z[√-23], and then, Z[√-31] cases,
likewise, afterwards, towards, beforehand, by itself, as such, as since, in to be, being inside
over thereby of r

[henryzz]

Hmm, I checked one way not the other.


Looking harder at that link it looks like rather than a^2+k*b^2 this applies to the principle form. If you look at definition 3, then the principle form is defined as a^2-D/4*b^2 if D is 0 (mod 4) and a^2+a*b+(1-D)/4*b^2 if D is 1 (mod 4).
I haven't checked this but I think it is correct. I haven't a clue what you do for D=2,3 (mod 4)

[Raman]

I have got with the general IF AND ONLY IF conditions for a few other special cases of the k values.

k = 27

Primes p of the form a²+27b²: All primes congruent to {1} mod 3 for which 2^((p-1)/3) ≡ 1 (mod p).
if N is a non-negative integer that can be written as a²+27b², then p × N can be written as a²+27b²
if N is a non-negative integer that cannot be written as a²+27b², then p × N cannot be written as a²+27b²

N can be written as a²+27b² if and only if
- N is not congruent to {3, 6} mod 9.
- N has no prime factors congruent to {2} mod 3 to an odd power.
- If N is co-prime to 6, then the sum of exponents of {1} mod 3 prime factors P of N for which 2^((P-1)/3) ≠ 1 (mod P) is either zero or at least two.

k = 32

Primes p of the form a²+32b²: All primes congruent to {1} mod 8 for which (p-4)^((p-1)/8) ≡ 1 (mod p).
if N is a non-negative integer that can be written as a²+32b², then p × N can be written as a²+32b²
if N is a non-negative integer that cannot be written as a²+32b², then p × N cannot be written as a²+32b²

N can be written as a²+32b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16).
- N has no prime factors congruent to {5, 7} mod 8 to an odd power.
- If N is odd or N is congruent to 4 (mod 8), then the sum of exponents of {3} mod 8 prime factors of N is even.
- If N has no prime factors congruent to {2, 3} mod 8, then the sum of exponents of {1} mod 8 prime factors P of N for which (P-4)^((P-1)/8) ≠ 1 (mod P) is even.

k = 64

Primes p of the form a²+64b²: All primes congruent to {1} mod 8 for which 2^((p-1)/4) ≡ 1 (mod p).
if N is a non-negative integer that can be written as a²+64b², then p × N can be written as a²+64b²
if N is a non-negative integer that cannot be written as a²+64b², then p × N cannot be written as a²+64b²

N can be written as a²+64b² if and only if
- N is not congruent to 2 (mod 4) or 8 (mod 16) or 32 (mod 64).
- N has no prime factors congruent to {3} mod 4 to an odd power.
- If N is odd or N is congruent to 4 (mod 8), then the sum of exponents of {5} mod 8 prime factors of N is even.
- If N has no prime factors congruent to {2, 5} mod 8, then the sum of exponents of {1} mod 8 prime factors P of N for which 2^((P-1)/4) ≠ 1 (mod P) is even.



But, for the other values of k, - such of those cases - I am being unable to classify and then categorize the primes of the form
a²+kb² = p form, let alone prove that there exist a multiple of p containing with the prime factor of p raised to an odd power,

if p is a quadratic residue (mod 4k), or p-k is a quadratic residue (mod 4k),
i.e. if -k is a quadratic residue (mod p).

But, the method of infinite descent to prove that if in case that there exist values for (a, b) such that a²+kb² = n × p, for some value of n - not being a multiple of p,
then there exist values for (a, b) such that a²+kb² = p,
fails for k = 11, 14, 17, 19, 20, 23, 26, 27, 29, 31, 32, 34, 35, 36, 38, 39, 41, 43, 44, 46, 47, 49, 50, ...
etc.