Self-gratification by a²+kb² forms "with ~10% FALSE STATEMENTS" (by author) - Page 14

Raman · 2013-04-02, 15:56

$\underline{Claim\ I} \\ If\ p\ and\ q\ are\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\ and\ N\ is\ some\ non-negative\ integer,\ then\ q\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)\ \\ if\ and\ only\ if\ p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)$

$\underline{Claim\ II} \\ If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\ p^2\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$

$and\ then\ if\ N\ =\ g^2\ +\ kh^2 \\ then\ certainly\ that\ the\ only\ solutions\ for\ p^2N\ =\ (pg)^2\ +\ k(ph)^2\ actually\ derives\ from\ N\ =\ g^2\ +\ kh^2.$

$\underline{Claim\ III} \\ If\ p\ is\ a\ prime\ number\ of\ the\ form\ a^2+kb^2,\ then\ \\ p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ IV} \\ If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\ there\ is\ no\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ V} \\ The\ product\ of\ any\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\ can\ always\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ VI} \\ No\ prime\ number\ p\ can\ be\ generated\ by\ using\ two\ different\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$

$\underline{Claim\ VII} \\ Cunningham\ numbers\ \frac{x^n-1}{x-1}\ numbers\ of\ this\ form\ are\ divisible\ only\ by\ numbers\ that\ are\ of\ form\ 2kn+1,\ n\ is\ prime,\ x\ \ge\ 2,\ x,\ n\ \in\ \mathbb{N},\ \\ except\ when\ n\ is\ a\ factor\ of\ (x-1).\ In\ such\ cases\ \frac{x^n-1}{x-1}\ is\ divisible\ by\ n\ itself$

$\underline{Claim\ VIII} \\ Any\ prime\ number\ representation\ by\ using\ a^2+kb^2\ form\ is\ always\ being\ unique.\ a\ \ge\ 0,\ b\ \ge\ 0.\ For\ the\ general\ quadratic\ form\ \\ polynomial\ of\ discriminant\ =\ -4k,\ there\ may\ be\ exactly\ four\ solutions\ for\ some\ prime\ number\ solution,\ whereby\ a,\ b\ may\ be\ even\ negative$

$\underline{Claim\ IX} \\ All\ prime\ numbers\ p\ for\ which\ -k\ is\ a\ quadratic\ residue\ (mod\ p)\ is\ generated\ by\ using\ exactly\ one\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ \\ and\ there\ is\ some\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ X} \\ If\ prime\ number\ p\ =\ a^2+kb^2,\ then\ certainly\ that\ always\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)$
$For\ all\ the\ other\ prime\ numbers\ p,\ for\ which\ -k\ is\ being\ a\ quadratic\ residue\ (mod\ p),\ including\ those\ such\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)\ \\\$
$there\ does\ certainly\ exist\ a\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2\ form\ \\\$
$in\ contrast\ to\ which,\ but\ that\ prime\ number\ p,\ \\\$ by making use for them a²+kb² form it itself, by a²+kb² form alone, the prime number p
$cannot\ be\ quite\ directly\ \\\$ being certainly be being
$written\ as\ a^2+kb^2\ form\ individually$

$\underline{Claim\ XI} \\ The\ following\ equation\ ma^2\ +\ nb^2\ =\ a^2\ +\ mnb^2\ will\ have\ a\ non-trivial\ non-zero\ integer\ solution,\ m\ \le\ n\ \\ if\ and\ only\ if\ m\ is\ being\ a\ quadratic\ residue\ (mod\ n)$

$\underline{Claim\ XII} \\ No\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ can\ be\ able\ to\ generate\ a\ prime\ number\ p,\ for\ which\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p)$
$Certain\ set\ of\ residue\ classes\ (mod\ 4k)\ is\ being\ the\ modulo\ usually\ being\ used\ in\ order\ to\ classify\ /\ categorize\ the\ any\ prime\ numbers\ p,\ \\ but\ that\ which\ are\ being\ generated\ by\ using\ some\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$

[COLOR=White]and then certainly that always that
but that which[/COLOR]

Raman · 2013-04-02, 15:57

$\underline{Claim\ I} \\ If\ p\ and\ q\ are\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\ and\ N\ is\ some\ non-negative\ integer,\ then\ q\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)\ \\ if\ and\ only\ if\ p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)$

$\underline{Claim\ II} \\ If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\ p^2\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$

$and\ then\ if\ N\ =\ g^2\ +\ kh^2 \\ then\ certainly\ that\ the\ only\ solutions\ for\ p^2N\ =\ (pg)^2\ +\ k(ph)^2\ actually\ derives\ from\ N\ =\ g^2\ +\ kh^2.$

$\underline{Claim\ III} \\ If\ p\ is\ a\ prime\ number\ of\ the\ form\ a^2+kb^2,\ then\ \\ p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ IV} \\ If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\ there\ is\ no\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ V} \\ The\ product\ of\ any\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\ can\ always\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ VI} \\ No\ prime\ number\ p\ can\ be\ generated\ by\ using\ two\ different\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$

$\underline{Claim\ VII} \\ Cunningham\ numbers\ \frac{x^n-1}{x-1}\ numbers\ of\ this\ form\ are\ divisible\ only\ by\ numbers\ that\ are\ of\ form\ 2kn+1,\ n\ is\ prime,\ x\ \ge\ 2,\ x,\ n\ \in\ \mathbb{N},\ \\ except\ when\ n\ is\ a\ factor\ of\ (x-1).\ In\ such\ cases\ \frac{x^n-1}{x-1}\ is\ divisible\ by\ n\ itself$

$\underline{Claim\ VIII} \\ Any\ prime\ number\ representation\ by\ using\ a^2+kb^2\ form\ is\ always\ being\ unique.\ a\ \ge\ 0,\ b\ \ge\ 0.\ For\ the\ general\ quadratic\ form\ \\ polynomial\ of\ discriminant\ =\ -4k,\ there\ may\ be\ exactly\ four\ solutions\ for\ some\ prime\ number\ solution,\ whereby\ a,\ b\ may\ be\ even\ negative$

$\underline{Claim\ IX} \\ All\ prime\ numbers\ p\ for\ which\ -k\ is\ a\ quadratic\ residue\ (mod\ p)\ is\ generated\ by\ using\ exactly\ one\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ \\ and\ there\ is\ some\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$

$\underline{Claim\ X} \\ If\ prime\ number\ p\ =\ a^2+kb^2,\ then\ certainly\ that\ always\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)$
$For\ all\ the\ other\ prime\ numbers\ p,\ for\ which\ -k\ is\ being\ a\ quadratic\ residue\ (mod\ p),\ including\ those\ such\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)\ \\\$
$there\ does\ certainly\ exist\ a\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2\ form\ \\\$
$in\ contrast\ to\ which,\ but\ that\ prime\ number\ p,\ \\\$ by making use for them a²+kb² form it itself, by a²+kb² form alone, the prime number p
$cannot\ be\ quite\ directly\ \\\$ being certainly be being
$written\ as\ a^2+kb^2\ form\ individually$

$\underline{Claim\ XI} \\ The\ following\ equation\ ma^2\ +\ nb^2\ =\ c^2\ +\ mnd^2\ will\ have\ a\ non-trivial\ non-zero\ integer\ solution,\ for\ the\ (a,\ b,\ c,\ d),\ m\ \le\ n\ \\ if\ and\ only\ if\ m\ is\ being\ a\ quadratic\ residue\ (mod\ n)$

$\underline{Claim\ XII} \\ No\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ can\ be\ able\ to\ generate\ a\ prime\ number\ p,\ for\ which\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p)$
$Certain\ set\ of\ residue\ classes\ (mod\ 4k)\ is\ being\ the\ modulo\ usually\ being\ used\ in\ order\ to\ classify\ /\ categorize\ the\ any\ prime\ numbers\ p,\ \\ but\ that\ which\ are\ being\ generated\ by\ using\ some\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$

[COLOR=White]and then certainly that always that
but that which[/COLOR]

Raman · 2013-04-14, 20:24

Code:

For a given k,
there is a lowest square m²
such that for every prime number p ≡ x² (mod 4k) and p ≡ x² + k (mod 4k)
then m² × p can be written as a²+kb² form.

This m is being constant enough
for all the inequivalent quadratic polynomial forms of discriminant = -4k that generates some prime numbers p such that p ≡ x² (mod 4k) and then p ≡ x² + k (mod 4k)
for a given fixed value of k.

For an example,
Consider k = 59
There are being five inequivalent quadratic polynomial forms of discriminant = -236
that whenever everything is being combined together
generates all prime numbers p such that p ≡ {0, 1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57} (mod 59)

p(a,b) = a²+59b²
q(a,b) = 3a²+2ab+20b²
r(a,b) = 5a²+2ab+12b²
s(a,b) = 7a²+4ab+9b²
t(a,b) = 4a²+2ab+15b²

Each of these polynomials have a lowest square n²
such that for every prime number P being generated by using this polynomial n² × P is being of the following form, a²+59b² form

For p(a,b) = a²+59b², n² = 1
For q(a,b) = 3a²+2ab+20b², n² = 25
For r(a,b) = 5a²+2ab+12b², n² = 36, 49
For s(a,b) = 7a²+4ab+9b², n² = 9
For t(a,b) = 4a²+2ab+15b², n² = 4

Notice that each of the above polynomials can generate with a perfect square, that generated square number will be (most probably?) be the lowest square n²
such that for every prime number P being generated by using this polynomial n² × P is being of the form a²+kb² form ...

The polynomial a²+kb² form for which coefficient of ab term is being zero, can generate with ALL squares, n² = 1, being lowest possible square satisfying this condition...
as since 1 × (a²+kb²) is atleast being already of the form a²+kb² form
but that for the other polynomials, not ALL squares can be generated, but some square can be generated, this will serve as a square l² such that
for which if l² can be multiplied by using this polynomial, it can be written as form a²+kb² form

as since product of any two prime numbers / composite numbers being generated by using same polynomial of discriminant = -4k
can be everytime written as form a²+kb² form...

Observe that each of the identities for them can be obtained by using a substitution of square generating values for the given fixed polynomial onto this following equation, alone
$ (Xa^2+Yab+Zb^2) \times (Xc^2+Ycd+Zd^2) = |acX+bdZ+ad\frac{Y}{2}+bc\frac{Y}{2}|^2 + (XZ-\frac{Y^2}{4})|ad-bc|^2 \\ (Xa^2+Yab+Zb^2) \times (Xc^2+Ycd+Zd^2) = |acX+bd(\frac{Y^2}{2X}-Z)+ad\frac{Y}{2}+bc\frac{Y}{2}|^2 + (XZ-\frac{Y^2}{4})|bd\frac{Y}{X}+ad+bc|^2 $

X=4, Y=2, Z=15, c=1, d=0
$ 4 \times (4a^2+2ab+15b^2) = (4a+b)^2 + 59b^2 $

X=7, Y=4, Z=9, c=0, d=1
$ 9 \times (7a^2+4ab+9b^2) = (2a+9b)^2 + 59a^2 $

X=3, Y=2, Z=20, c=1, d=1
$ 25 \times (3a^2+2ab+20b^2) = (4a+21b)^2 + 59(a-b)^2 $

X=5, Y=2, Z=12, c=2, d=1
$ 36 \times (5a^2+2ab+12b^2) = (11a+14b)^2 + 59(a-2b)^2 $

X=5, Y=2, Z=12, c=-1, d=2
$ 49 \times (5a^2+2ab+12b^2) = (3a-23b)^2 + 59(2a+b)^2 $

Lowest square m²
such that for every prime number p such that p ≡ {0, 1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57} (mod 59)
m² × p is being of the following form, a²+59b² form
is being equal to
m² = LCM(1, 4, 9, 25, 36) = 900.

for the r r r r done enough for the r r r r

Raman · 2013-04-14, 20:25

SURPRISE! Really nice sequence! HOT

enough!
whenever if it is being as submitted sequence onto for the OEIS. for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
These are being the
values of k such that
there are no numbers N such that if N cannot be written as a²+kb² form,
then N³ can be written as a²+kb² form.
(Or apparently any odd power of N):

k =
1, 2, 3, 5, 6, 9, 10, 12, 13, 14, 17, 21, 22, 30, 33, 34, 37, 42, 46, 49, 57, 58,
65, 66, 69, 70, 73, 77, 78, 82, 85, 90, 93, 97, 102, 105, 114, 117, 130, 133, 138,
141, 142, 145, 154, 165, 177, 190, 193, 198, 205, 210, 213, 217, 238, 253, 258, 265,
273, 282, 285, 301, 310, 322, 330, 333, 345, 357, 385, 390, 418, 429, 438, 442, 445,
462, 465, 498, 505, 510, 522, 553, 561, 570, 598, 609, 630, 645, 658, 690, 697, 742,
765, 777, 793, 798, 805, 858, 870, 897, 910, 957, 1005, 1045, 1065, 1105, 1110, 1113,
1122, 1170, 1185, 1197, 1290, 1302, 1353, 1365, 1513, 1582, 1590, 1605, 1645, 1653,
1677, 1705, 1710, 1785, 1870, 1885, 1918, 2002, 2013, 2145, 2170, 2233, 2277, 2310,
2385, 2542, 2730, 2737, 2790, 3045, 3465, 3502, 3570, 3705, 4305, 4810, 4830, 4845,
5005, 5037, 5910, 6402, 6678, 7137, 7245, 7585, 8710, 10605, 11713, 13398, 14637,
16830, 18018, 18285, 20097, 23142, 49665, 51870, 57057, 66045

If a lowest odd power of N, N^r can be written as a²+kb² form, then so can be N^r+2, N^r+4, ...,
there will exist a lowest power of three: 3^x, 3^x ≥ r, such that N^3[sup]x[/sup] can be written as a²+kb² form ...

k = 11 is not being a member into this sequence as since as as follows as
3 is not being of form a²+11b² form, but that 3³ = 27 = 4² + 11 × 1².
5 is not being of form a²+11b² form, but that 5³ = 125 = 9² + 11 × 2².
23 is not being of form a²+11b² form, but that 23³ = 12167 = 54² + 11 × 29².

If for a given fixed value of k, suppose integer N violates this condition: N is not being of form a²+kb² form; but that N³ is being of form a²+kb² form
if such an integer N exists; and then
will it be always be ever be N < k?

Is this sequence being finite?
Are there being furthermore terms beyond 66045?

It is being worth noticing that if k ≡ {0, 4, 7} (mod 8), as since as as follows as
some number N with prime factor of 2 to be
being raised on to an odd power violates this following condition; no numbers k ≡ {0, 4, 7} (mod 8) is being
belonging into this sequence.

It is being worth noticing that if square of odd / even prime number p² | k, that if p ≡ {1, 2} (mod 4), as since as as follows as
some number N with prime factor of p to be
being raised on to an odd power violates this following condition; for no numbers that if square of odd / even prime number p² | k, that if p ≡ {1, 2} (mod 4), is being
belonging into this sequence.

Values of k such that
there exist no (p, N)
prime number p, non-negative integer N
such that N is being of form a²+kb², p × N is being of form a²+kb², but that p is not being of form a²+kb²

AND THEN

Values of k such that
there exist no (p, N)
prime number p, non-negative integer N
such that p² × N is being of form a²+kb², but that N is not being of form a²+kb²

are being proper subsets of above mentioned sequence.

Values of k satisfying above two mentioned conditions are being
square-free ideonal numbers ≡ {1, 2, 3, 5, 6} (mod 8) / or aka square-free convenient numbers ≡ {1, 2, 3, 5, 6} (mod 8)

k =
1, 2, 3, 5, 6, 10, 13, 21, 22, 30, 33, 37, 42, 57, 58, 70, 78, 85, 93, 102, 105, 130, 133,
165, 177, 190, 210, 253, 273, 330, 345, 357, 385, 462, 1365

k = 14 is not being a member into this sequence as since as as follows as
7 is not being of form a²+14b² form, but that 7 × 3² = 63 = 7² + 14 × 1².

k = 14 is not being a member into this sequence as since as as follows as
2² + 14 × 1² = 18 is being of form a²+14b² form, 6² + 14 × 0² = 36 is being of form a²+14b² form, but that 36 ÷ 18 = 2 is not being of form a²+14b² form.
2² + 14 × 1² = 18 is being of form a²+14b² form, 0² + 14 × 3² = 126 is being of form a²+14b² form, but that 126 ÷ 18 = 7 is not being of form a²+14b² form.

25² + 14 × 1² = 639 is being of form a²+14b² form, 135² + 14 × 48² = 50481 = 9² + 14 × 60² is being of form a²+14b² form, but that 50481 ÷ 639 = 79 = 2 × 6² + 7 × 1² is not being of form a²+14b² form.
19² + 14 × 5² = 711 is being of form a²+14b² form, 9² + 14 × 60² = 50481 = 135² + 14 × 48² is being of form a²+14b² form, but that 50481 ÷ 711 = 71 = 2 × 2² + 7 × 3² is not being of form a²+14b² form.

It is being worth noticing that if k ≡ {0, 4, 7} (mod 8), as since as as follows as
some number N with prime factor of 2 to be
being raised on to an odd power violates this following condition; no numbers k ≡ {0, 4, 7} (mod 8) is being
belonging into this sequence.

It is being worth noticing that if square of odd / even prime number p² | k, as since as as follows as
some number N with prime factor of p to be
being raised on to an odd power violates this following condition; for no numbers that if square of odd / even prime number p² | k, is being
belonging into this sequence.

for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
These are being the
values of k such that
there are no numbers N such that if N cannot be written as a²+kb² form,
then N³ can be written as a²+kb² form.
(Or apparently any odd power of N):
for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
This is being indeed exceptionally
really nice sequence for which
such that I could be able in order to be being in to in
in
within into from among within onto to be being
to add / submit them onto for the OEIS.
and then but
that there are being a lot furthermore
within on to
a lump of plenty matters
for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
to be being as submitted sequences
from among with in in to
for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
If in case whether if
OEIS editors are being less stringent, furthermore lesser onerous onto me towards limits
for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
Limits are being done
proven with in within into from among within onto to be being
on to be
being ridiculous enough, quite directly, it is being all dedicated hard-work / taking precious preparation time process at all
/ efforts of authors / contributors...

Please do support me,
kindly enough! HELP!

I will be blocked onto OEIS till until April 28, 2013,
and then afterwards let's see
look at all
away that way
but that which way
later onto me
onwards towards

for the r r r rdone enough for the r r r r

There does exist no
value of k such that
there exist no (p, N)
prime number p, non-negative integer N
such that N is not being of form a²+kb², p is not being of form a²+kb², but that p × N is not being of form a²+kb²

Raman · 2013-04-14, 21:06

Quote:

Originally Posted by Raman

Representation by quadratic forms

After research by writing with my own program, I got with this following result,
great to be had them proved, been!

$\begin{tabular}{|l|l|l|l|l|} \hline {If\ N\ can\ be\\ written\ as\\ (with\ a\ \ge\ 0,\ b\ \ge\ 0)} & {Prime\ factors\ of\ N\\ which\ have\ no\\ restriction} & {Sum\ of\ exponents\ of\\ all\ these\ prime\ factors\\ of\ N\ should\ be\ even\\ when\ combined\ together} & {Prime\ factors\ of\ N\\ whose\ exponent\\ should\ be\ \ge \ 2} & {Each\ of\ these\\ prime\ factors\ of\ N\\ can\ only\ occur\\ in\ pairs\ individually} \\ \hline a^2+b^2 & {2\\ 1\ mod\ 4} & ~ & ~ & 3\ mod\ 4 \\ \hline a^2+2b^2 & {2\\ 1\ mod\ 8\\ 3\ mod\ 8} & ~ & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline a^2+3b^2 & {3\\ 1\ mod\ 3} & ~ & ~ & 2\ mod\ 3 \\ \hline a^2+4b^2 & 1\ mod\ 4 & ~ & 2 & 3\ mod\ 4 \\ \hline a^2+5b^2 & {5\\ 1\ mod\ 20\\ 9\ mod\ 20} & {2\\ 3\ mod\ 20\\ 7\ mod\ 20} & ~ & {11\ mod\ 20\\ 13\ mod\ 20\\ 17\ mod\ 20\\ 19\ mod\ 20} \\ \hline a^2+6b^2 & {1\ mod\ 24\\ 7\ mod\ 24} & {2\\ 3\\ 5\ mod\ 24\\ 11\ mod\ 24} & ~ & {13\ mod\ 24\\ 17\ mod\ 24\\ 19\ mod\ 24\\ 23\ mod\ 24} \\ \hline a^2+7b^2 & {7\\ 1\ mod\ 14\\ 9\ mod\ 14\\ 11\ mod\ 14} & ~ & 2 & {3\ mod\ 14\\ 5\ mod\ 14\\ 13\ mod\ 14} \\ \hline {a^2+8b^2\\ (for\ odd\ numbers)} & 1\ mod\ 8 & 3\ mod\ 8 & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+8b^2\\ (for\ even\ numbers)} & {1\ mod\ 8\\ 3\ mod\ 8} & ~ & 2 & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+9b^2\\ (for\ non-multiples\ of\ 3)} & 1\ mod\ 12 & {2\\ 5\ mod\ 12} & ~ & {7\ mod\ 12\\ 11\ mod\ 12} \\ \hline {a^2+9b^2\\ (for\ multiples\ of\ 3)} & {2\\ 1\ mod\ 12\\ 5\ mod\ 12} & ~ & ~ & {3\\ 7\ mod\ 12\\ 11\ mod\ 12} \\ \hline a^2+10b^2 & {1\ mod\ 40\\ 9\ mod\ 40\\ 11\ mod\ 40\\ 19\ mod\ 40} & {2\\ 5\\ 7\ mod\ 40\\ 13\ mod\ 40\\ 23\ mod\ 40\\ 37\ mod\ 40} & ~ & {3\ mod\ 40\\ 17\ mod\ 40\\ 21\ mod\ 40\\ 27\ mod\ 40\\ 29\ mod\ 40\\ 31\ mod\ 40\\ 33\ mod\ 40\\ 39\ mod\ 40} \\ \hline \end{tabular}$

Is it possible in general to give by an explicit formula, for the general representation of the quadratic form a²+kb² form, like this?
The necessary condition for a prime p to be written as a²+kb² form, is being that (-k/p) = 1.

How to determine the class number, in general, but what it is being, first of all, in properly defined sense? It seems to me that it is being an abstract concept...

For this example, consider with the representation by using the next quadratic form, namely, a²+11b² form, in general.

The necessary condition for a prime p to be written as a²+11b² form, is being that (-11/p) = (p/11) = 1.
i.e. p ≡ 0, 1, 3, 4, 5, 9 (mod 11).
It is not being a sufficient condition as since every prime of these residue classes cannot be written, in general, in the form, a²+11b² form.
But if we extend it to include numbers of the form (a/2)² + 11(b/2)², whereby a, b are being integers ≥ 0,
then certainly every prime of these residue classes can be uniquely written of the form (a/2)² + 11(b/2)².
But how do we know that, what primes of the form 0, 1, 3, 4, 5, 9 (mod 11) do require with odd a, b values, for the representation of the form (a/2)² + 11(b/2)²?

In general, the square-free numbers 1, 2, 3, 7, 11, 19, 43, 67, 163 have got with class number 1. Excluding thereby the non-square-free numbers 4, 8, 12, 27.
Of course, for these k values, in a²+kb² form, (or thereby (a/2)²+k(b/2)² form - if k ≡ 3 mod 4), if a composite number C can be written in that form, then
every prime factor P of it can be written in that form, thereby

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

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Raman · 2013-04-15, 05:16

Quote:

Originally Posted by Raman

Representation by quadratic forms

After research by writing with my own program, I got with this following result,
great to be had them proved, been!

$\begin{tabular}{|l|l|l|l|l|} \hline {If\ N\ can\ be\\ written\ as\\ (with\ a\ \ge\ 0,\ b\ \ge\ 0)} & {Prime\ factors\ of\ N\\ which\ have\ no\\ restriction} & {Sum\ of\ exponents\ of\\ all\ these\ prime\ factors\\ of\ N\ should\ be\ even\\ when\ combined\ together} & {Prime\ factors\ of\ N\\ whose\ exponent\\ should\ be\ \ge \ 2} & {Each\ of\ these\\ prime\ factors\ of\ N\\ can\ only\ occur\\ in\ pairs\ individually} \\ \hline a^2+b^2 & {2\\ 1\ mod\ 4} & ~ & ~ & 3\ mod\ 4 \\ \hline a^2+2b^2 & {2\\ 1\ mod\ 8\\ 3\ mod\ 8} & ~ & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline a^2+3b^2 & {3\\ 1\ mod\ 3} & ~ & ~ & 2\ mod\ 3 \\ \hline a^2+4b^2 & 1\ mod\ 4 & ~ & 2 & 3\ mod\ 4 \\ \hline a^2+5b^2 & {5\\ 1\ mod\ 20\\ 9\ mod\ 20} & {2\\ 3\ mod\ 20\\ 7\ mod\ 20} & ~ & {11\ mod\ 20\\ 13\ mod\ 20\\ 17\ mod\ 20\\ 19\ mod\ 20} \\ \hline a^2+6b^2 & {1\ mod\ 24\\ 7\ mod\ 24} & {2\\ 3\\ 5\ mod\ 24\\ 11\ mod\ 24} & ~ & {13\ mod\ 24\\ 17\ mod\ 24\\ 19\ mod\ 24\\ 23\ mod\ 24} \\ \hline a^2+7b^2 & {7\\ 1\ mod\ 14\\ 9\ mod\ 14\\ 11\ mod\ 14} & ~ & 2 & {3\ mod\ 14\\ 5\ mod\ 14\\ 13\ mod\ 14} \\ \hline {a^2+8b^2\\ (for\ odd\ numbers)} & 1\ mod\ 8 & 3\ mod\ 8 & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+8b^2\\ (for\ even\ numbers)} & {1\ mod\ 8\\ 3\ mod\ 8} & ~ & 2 & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+9b^2\\ (for\ non-multiples\ of\ 3)} & 1\ mod\ 12 & {2\\ 5\ mod\ 12} & ~ & {7\ mod\ 12\\ 11\ mod\ 12} \\ \hline {a^2+9b^2\\ (for\ multiples\ of\ 3)} & {2\\ 1\ mod\ 12\\ 5\ mod\ 12} & ~ & ~ & {3\\ 7\ mod\ 12\\ 11\ mod\ 12} \\ \hline a^2+10b^2 & {1\ mod\ 40\\ 9\ mod\ 40\\ 11\ mod\ 40\\ 19\ mod\ 40} & {2\\ 5\\ 7\ mod\ 40\\ 13\ mod\ 40\\ 23\ mod\ 40\\ 37\ mod\ 40} & ~ & {3\ mod\ 40\\ 17\ mod\ 40\\ 21\ mod\ 40\\ 27\ mod\ 40\\ 29\ mod\ 40\\ 31\ mod\ 40\\ 33\ mod\ 40\\ 39\ mod\ 40} \\ \hline \end{tabular}$

Is it possible in general to give by an explicit formula, for the general representation of the quadratic form a²+kb² form, like this?
The necessary condition for a prime p to be written as a²+kb² form, is being that (-k/p) = 1.

How to determine the class number, in general, but what it is being, first of all, in properly defined sense? It seems to me that it is being an abstract concept...

For this example, consider with the representation by using the next quadratic form, namely, a²+11b² form, in general.

The necessary condition for a prime p to be written as a²+11b² form, is being that (-11/p) = (p/11) = 1.
i.e. p ≡ 0, 1, 3, 4, 5, 9 (mod 11).
It is not being a sufficient condition as since every prime of these residue classes cannot be written, in general, in the form, a²+11b² form.
But if we extend it to include numbers of the form (a/2)² + 11(b/2)², whereby a, b are being integers ≥ 0,
then certainly every prime of these residue classes can be uniquely written of the form (a/2)² + 11(b/2)².
But how do we know that, what primes of the form 0, 1, 3, 4, 5, 9 (mod 11) do require with odd a, b values, for the representation of the form (a/2)² + 11(b/2)²?

In general, the square-free numbers 1, 2, 3, 7, 11, 19, 43, 67, 163 have got with class number 1. Excluding thereby the non-square-free numbers 4, 8, 12, 27.
Of course, for these k values, in a²+kb² form, (or over thereby (a/2)²+k(b/2)² form - if k ≡ 3 mod 4), if a composite number C can be written in that form, then
every prime factor P of it can be written in that form, other than 27 over thereby

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

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Raman · 2013-04-15, 05:21

Quote:

Originally Posted by Raman

Representation by quadratic forms

After research by writing with my own program, I got with this following result,
great to be had them proved, been!

$\begin{tabular}{|l|l|l|l|l|} \hline {If\ N\ can\ be\\ written\ as\\ (with\ a\ \ge\ 0,\ b\ \ge\ 0)} & {Prime\ factors\ of\ N\\ which\ have\ no\\ restriction} & {Sum\ of\ exponents\ of\\ all\ these\ prime\ factors\\ of\ N\ should\ be\ even\\ when\ combined\ together} & {Prime\ factors\ of\ N\\ whose\ exponent\\ should\ be\ \ge \ 2} & {Each\ of\ these\\ prime\ factors\ of\ N\\ can\ only\ occur\\ in\ pairs\ individually} \\ \hline a^2+b^2 & {2\\ 1\ mod\ 4} & ~ & ~ & 3\ mod\ 4 \\ \hline a^2+2b^2 & {2\\ 1\ mod\ 8\\ 3\ mod\ 8} & ~ & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline a^2+3b^2 & {3\\ 1\ mod\ 3} & ~ & ~ & 2\ mod\ 3 \\ \hline a^2+4b^2 & 1\ mod\ 4 & ~ & 2 & 3\ mod\ 4 \\ \hline a^2+5b^2 & {5\\ 1\ mod\ 20\\ 9\ mod\ 20} & {2\\ 3\ mod\ 20\\ 7\ mod\ 20} & ~ & {11\ mod\ 20\\ 13\ mod\ 20\\ 17\ mod\ 20\\ 19\ mod\ 20} \\ \hline a^2+6b^2 & {1\ mod\ 24\\ 7\ mod\ 24} & {2\\ 3\\ 5\ mod\ 24\\ 11\ mod\ 24} & ~ & {13\ mod\ 24\\ 17\ mod\ 24\\ 19\ mod\ 24\\ 23\ mod\ 24} \\ \hline a^2+7b^2 & {7\\ 1\ mod\ 14\\ 9\ mod\ 14\\ 11\ mod\ 14} & ~ & 2 & {3\ mod\ 14\\ 5\ mod\ 14\\ 13\ mod\ 14} \\ \hline {a^2+8b^2\\ (for\ odd\ numbers)} & 1\ mod\ 8 & 3\ mod\ 8 & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+8b^2\\ (for\ even\ numbers)} & {1\ mod\ 8\\ 3\ mod\ 8} & ~ & 2 & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+9b^2\\ (for\ non-multiples\ of\ 3)} & 1\ mod\ 12 & {2\\ 5\ mod\ 12} & ~ & {7\ mod\ 12\\ 11\ mod\ 12} \\ \hline {a^2+9b^2\\ (for\ multiples\ of\ 3)} & {2\\ 1\ mod\ 12\\ 5\ mod\ 12} & ~ & ~ & {3\\ 7\ mod\ 12\\ 11\ mod\ 12} \\ \hline a^2+10b^2 & {1\ mod\ 40\\ 9\ mod\ 40\\ 11\ mod\ 40\\ 19\ mod\ 40} & {2\\ 5\\ 7\ mod\ 40\\ 13\ mod\ 40\\ 23\ mod\ 40\\ 37\ mod\ 40} & ~ & {3\ mod\ 40\\ 17\ mod\ 40\\ 21\ mod\ 40\\ 27\ mod\ 40\\ 29\ mod\ 40\\ 31\ mod\ 40\\ 33\ mod\ 40\\ 39\ mod\ 40} \\ \hline \end{tabular}$

Is it possible in general to give by an explicit formula, for the general representation of the quadratic form a²+kb² form, like this?
The necessary condition for a prime p to be written as a²+kb² form, is being that (-k/p) = 1.

How to determine the class number, in general, but what it is being, first of all, in properly defined sense? It seems to me that it is being an abstract concept...

For this example, consider with the representation by using the next quadratic form, namely, a²+11b² form, in general.

The necessary condition for a prime p to be written as a²+11b² form, is being that (-11/p) = (p/11) = 1.
i.e. p ≡ 0, 1, 3, 4, 5, 9 (mod 11).
It is not being a sufficient condition as since every prime of these residue classes cannot be written, in general, in the form, a²+11b² form.
But if we extend it to include numbers of the form (a/2)² + 11(b/2)², whereby a, b are being integers ≥ 0,
then certainly every prime of these residue classes can be uniquely written of the form (a/2)² + 11(b/2)².
But how do we know that, what primes of the form 0, 1, 3, 4, 5, 9 (mod 11) do require with odd a, b values, for the representation of the form (a/2)² + 11(b/2)²?

In general, the square-free numbers 1, 2, 3, 7, 11, 19, 43, 67, 163 have got with class number 1. Excluding thereby the non-square-free numbers 4, 8, 12, 27.
Of course, for these k values, in a²+kb² form, (or thereby (a/2)²+k(b/2)² form - if k ≡ 3 mod 4), if a composite number C can be written in that form, then
every prime factor P of it can be written in that form, thereby

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

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ID: 9673

kar_bon · 2013-04-15, 07:32

Is it possible in gerneral you don't know what you're doing here?

Read about the "code"-tag and try preview your post first.

And

<color=white>
for the r r r r done enough for the r r r r
for the r r r r done enough for the r r r r
<color=black>

is really good for trying to post 'new' things and keep youself serious.

NBtarheel_33 · 2013-04-15, 18:39

Certainly nice to see that Raman learned some better behavior whilst on enforced break...NOT.

2013-04-02, 15:56	#144
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3·419 Posts	$\underline{Claim\ I} \\ <br /> If\ p\ and\ q\ are\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\<br /> and\ N\ is\ some\ non-negative\ integer,\ then\ q\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)\ \\<br /> if\ and\ only\ if\ p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ (or\ some\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k)$ $\underline{Claim\ II} \\ <br /> If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\<br /> p^2\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$ $and\ then\ if\ N\ =\ g^2\ +\ kh^2 \\ <br /> then\ certainly\ that\ the\ only\ solutions\ for\ p^2N\ =\ (pg)^2\ +\ k(ph)^2\ actually\ derives\ from\ N\ =\ g^2\ +\ kh^2.$ $\underline{Claim\ III} \\ <br /> If\ p\ is\ a\ prime\ number\ of\ the\ form\ a^2+kb^2,\ then\ \\<br /> p\ \times\ N\ can\ be\ written\ as\ a^2+kb^2\ if\ and\ only\ if\ N\ can\ be\ written\ as\ a^2+kb^2$ $\underline{Claim\ IV} \\ <br /> If\ for\ a\ prime\ number\ p,\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p),\ then\ \\<br /> there\ is\ no\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$ $\underline{Claim\ V} \\ <br /> The\ product\ of\ any\ two\ numbers\ (prime\ /\ composite)\ generated\ by\ using\ the\ same\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k,\ \\<br /> can\ always\ be\ written\ as\ a^2+kb^2$ $\underline{Claim\ VI} \\ <br /> No\ prime\ number\ p\ can\ be\ generated\ by\ using\ two\ different\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$ $\underline{Claim\ VII} \\ <br /> Cunningham\ numbers\ \frac{x^n-1}{x-1}\ numbers\ of\ this\ form\ are\ divisible\ only\ by\ numbers\ that\ are\ of\ form\ 2kn+1,\ n\ is\ prime,\ x\ \ge\ 2,\ x,\ n\ \in\ \mathbb{N},\ \\<br /> except\ when\ n\ is\ a\ factor\ of\ (x-1).\ In\ such\ cases\ \frac{x^n-1}{x-1}\ is\ divisible\ by\ n\ itself$ $\underline{Claim\ VIII} \\ <br /> Any\ prime\ number\ representation\ by\ using\ a^2+kb^2\ form\ is\ always\ being\ unique.\ a\ \ge\ 0,\ b\ \ge\ 0.\ For\ the\ general\ quadratic\ form\ \\<br /> polynomial\ of\ discriminant\ =\ -4k,\ there\ may\ be\ exactly\ four\ solutions\ for\ some\ prime\ number\ solution,\ whereby\ a,\ b\ may\ be\ even\ negative$ $\underline{Claim\ IX} \\ <br /> All\ prime\ numbers\ p\ for\ which\ -k\ is\ a\ quadratic\ residue\ (mod\ p)\ is\ generated\ by\ using\ exactly\ one\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ \\<br /> and\ there\ is\ some\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2$ $\underline{Claim\ X} \\ <br /> If\ prime\ number\ p\ =\ a^2+kb^2,\ then\ certainly\ that\ always\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)$ $For\ all\ the\ other\ prime\ numbers\ p,\ for\ which\ -k\ is\ being\ a\ quadratic\ residue\ (mod\ p),\ including\ those\ such\ that\ p\ \equiv\ x^2\ (mod\ 4k)\ or\ p\ \equiv\ x^2\ +\ k\ (mod\ 4k)\ \\\$ $there\ does\ certainly\ exist\ a\ multiple\ of\ p,\ to\ which\ p\ is\ being\ raised\ to\ an\ odd\ power,\ that\ can\ be\ written\ as\ a^2+kb^2\ form\ \\\$ $in\ contrast\ to\ which,\ but\ that\ prime\ number\ p,\ \\\$ by making use for them a²+kb² form it itself, by a²+kb² form alone, the prime number p $cannot\ be\ quite\ directly\ \\\$ being certainly be being $written\ as\ a^2+kb^2\ form\ individually$ $\underline{Claim\ XI} \\ <br /> The\ following\ equation\ ma^2\ +\ nb^2\ =\ a^2\ +\ mnb^2\ will\ have\ a\ non-trivial\ non-zero\ integer\ solution,\ m\ \le\ n\ \\<br /> if\ and\ only\ if\ m\ is\ being\ a\ quadratic\ residue\ (mod\ n)$ $\underline{Claim\ XII} \\ <br /> No\ quadratic\ form\ polynomial\ of\ discriminant\ =\ -4k\ can\ be\ able\ to\ generate\ a\ prime\ number\ p,\ for\ which\ -k\ is\ a\ quadratic\ non-residue\ (mod\ p)$ $Certain\ set\ of\ residue\ classes\ (mod\ 4k)\ is\ being\ the\ modulo\ usually\ being\ used\ in\ order\ to\ classify\ /\ categorize\ the\ any\ prime\ numbers\ p,\ \\<br /> but\ that\ which\ are\ being\ generated\ by\ using\ some\ quadratic\ form\ polynomials\ of\ same\ discriminant\ =\ -4k$ [COLOR=White]and then certainly that always that but that which[/COLOR]

2013-04-14, 20:24	#146
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 10011101001₂ Posts	Code: $4 \times (3a^2 + 2ab + 4b^2) = (a + 4b)^2 + 11a^2 \\ 9 \times (2a^2 + 7b^2) = (2a + 7b)^2 + 14(a - b)^2 = (2a - 7b)^2 + 14(a + b)^2\\ 9 \times (2a^2 + 2ab + 9b^2) = (a + 9b)^2 + 17a^2 \\ 4 \times (4a^2 + 2ab + 5b^2) = (4a + b)^2 + 19b^2 \\ 4 \times (4a^2 + 5b^2) = (4a)^2 + 20b^2$ For a given k, there is a lowest square m² such that for every prime number p ≡ x² (mod 4k) and p ≡ x² + k (mod 4k) then m² × p can be written as a²+kb² form. This m is being constant enough for all the inequivalent quadratic polynomial forms of discriminant = -4k that generates some prime numbers p such that p ≡ x² (mod 4k) and then p ≡ x² + k (mod 4k) for a given fixed value of k. For an example, Consider k = 59 There are being five inequivalent quadratic polynomial forms of discriminant = -236 that whenever everything is being combined together generates all prime numbers p such that p ≡ {0, 1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57} (mod 59) p(a,b) = a²+59b² q(a,b) = 3a²+2ab+20b² r(a,b) = 5a²+2ab+12b² s(a,b) = 7a²+4ab+9b² t(a,b) = 4a²+2ab+15b² Each of these polynomials have a lowest square n² such that for every prime number P being generated by using this polynomial n² × P is being of the following form, a²+59b² form For p(a,b) = a²+59b², n² = 1 For q(a,b) = 3a²+2ab+20b², n² = 25 For r(a,b) = 5a²+2ab+12b², n² = 36, 49 For s(a,b) = 7a²+4ab+9b², n² = 9 For t(a,b) = 4a²+2ab+15b², n² = 4 Notice that each of the above polynomials can generate with a perfect square, that generated square number will be (most probably?) be the lowest square n² such that for every prime number P being generated by using this polynomial n² × P is being of the form a²+kb² form ... The polynomial a²+kb² form for which coefficient of ab term is being zero, can generate with ALL squares, n² = 1, being lowest possible square satisfying this condition... as since 1 × (a²+kb²) is atleast being already of the form a²+kb² form but that for the other polynomials, not ALL squares can be generated, but some square can be generated, this will serve as a square l² such that for which if l² can be multiplied by using this polynomial, it can be written as form a²+kb² form as since product of any two prime numbers / composite numbers being generated by using same polynomial of discriminant = -4k can be everytime written as form a²+kb² form... Observe that each of the identities for them can be obtained by using a substitution of square generating values for the given fixed polynomial onto this following equation, alone $<br /> (Xa^2+Yab+Zb^2) \times (Xc^2+Ycd+Zd^2) = \|acX+bdZ+ad\frac{Y}{2}+bc\frac{Y}{2}\|^2 + (XZ-\frac{Y^2}{4})\|ad-bc\|^2 \\<br /> (Xa^2+Yab+Zb^2) \times (Xc^2+Ycd+Zd^2) = \|acX+bd(\frac{Y^2}{2X}-Z)+ad\frac{Y}{2}+bc\frac{Y}{2}\|^2 + (XZ-\frac{Y^2}{4})\|bd\frac{Y}{X}+ad+bc\|^2<br />$ X=4, Y=2, Z=15, c=1, d=0 $<br /> 4 \times (4a^2+2ab+15b^2) = (4a+b)^2 + 59b^2<br />$ X=7, Y=4, Z=9, c=0, d=1 $<br /> 9 \times (7a^2+4ab+9b^2) = (2a+9b)^2 + 59a^2<br />$ X=3, Y=2, Z=20, c=1, d=1 $<br /> 25 \times (3a^2+2ab+20b^2) = (4a+21b)^2 + 59(a-b)^2<br />$ X=5, Y=2, Z=12, c=2, d=1 $<br /> 36 \times (5a^2+2ab+12b^2) = (11a+14b)^2 + 59(a-2b)^2<br />$ X=5, Y=2, Z=12, c=-1, d=2 $<br /> 49 \times (5a^2+2ab+12b^2) = (3a-23b)^2 + 59(2a+b)^2<br />$ Lowest square m² such that for every prime number p such that p ≡ {0, 1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57} (mod 59) m² × p is being of the following form, a²+59b² form is being equal to m² = LCM(1, 4, 9, 25, 36) = 900. for the r r r r done enough for the r r r r

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2013-04-15, 07:32	#151
kar_bon Mar 2006 Germany B58₁₆ Posts	Is it possible in gerneral you don't know what you're doing here? Read about the "code"-tag and try preview your post first. And <color=white> for the r r r r done enough for the r r r r for the r r r r done enough for the r r r r <color=black> is really good for trying to post 'new' things and keep youself serious.

2013-04-15, 18:39	#152
NBtarheel_33 "Nathan" Jul 2008 Maryland, USA 1115₁₀ Posts	Certainly nice to see that Raman learned some better behavior whilst on enforced break...NOT.