Inhaltsverzeichnis

Development of
Algorithmic Constructions

14:37:31
Deutsch
29.Mar 2024

29 31 33

Number p < 10000=

Calculation in the complex field with adjoined A=√3 with the reduction to the elements with |(a+bAI)|=1 mod p

A=√3 Let a,b element of N and p an odd number > 3 (a+bAI)² = a²-b²A² + 2abAI mod p |(a+bAI)| = (a+bAI)(a-bAI) = a²+b²A² mod p 1. |(p-1)/2+[(p-1)/2]AI|=1 mod p is always true. Proof: 2. {(p-1)/2+[(p-1)/2]AI}³=1 mod p is always true. Proof: 3. If a = (c^2 - 3 d^2) / (c^2 + 3 d^2) mod p and b = (2 c d) / (c^2 + 3 d^2) mod p, then a^2 + 3 b^2 = 1 mod p. Proof: 4. Metric: Calculation of the metric is the distance from the point x+yAI to the point (p-1)/2+[(p-1)/2]*AI in the complex field center:=(p-1)/2 if x>center then x:=p - x if y>center then y:=p - y diff_x:=center-x diff_y:=center-y metric:=diff_x²+diff_y²*A² mod p

Number = 31, Adjoined = √3, Exponent = 1, Jacobi (3, 31) = -1, 31 = 3 mod 4, 31 = 7 mod 8

30
030√3
Number of elements = 30

Number = 31, Adjoined = √3, Exponent = 1, Jacobi (3, 31) = -1, 31 = 3 mod 4, 31 = 7 mod 8

Potence 2 Potence 3 Potence 5 Potence 7 Potence 11 Potence 13 Potence 17 Potence 19 Potence 23 Potence 29

|30+0AI|
=1
|28+7AI|
=1
metric = 26
|28+24AI|
=1
metric = 26
|25+3AI|
=1
metric = 17
|25+28AI|
=1
metric = 17
|23+14AI|
=1
metric = 21
|23+17AI|
=1
metric = 21
|22+5AI|
=1
metric = 26
|22+26AI|
=1
metric = 26
|19+2AI|
=1
metric = 20
|19+29AI|
=1
metric = 20
|17+11AI|
=1
metric = 18
|17+20AI|
=1
metric = 18
|16+15AI|
=1
metric = 0
|16+16AI|
=1
metric = 0
|15+15AI|
=1
metric = 0
|15+16AI|
=1
metric = 0
|14+11AI|
=1
metric = 18
|14+20AI|
=1
metric = 18
|12+2AI|
=1
metric = 20
|12+29AI|
=1
metric = 20
|9+5AI|
=1
metric = 26
|9+26AI|
=1
metric = 26
|8+14AI|
=1
metric = 21
|8+17AI|
=1
metric = 21
|6+3AI|
=1
metric = 17
|6+28AI|
=1
metric = 17
|3+7AI|
=1
metric = 26
|3+24AI|
=1
metric = 26
|1+0AI|
=1


    1. |(p-1)/2+[(p-1)/2]AI|=1 mod p is always true

    Proof:

     |(p-1)/2+[(p-1)/2]AI|                         mod p
    =[(p-1)/2+[(p-1)/2]AI]*[(p-1)/2-[(p-1)/2]AI]   mod p | with A²=3
    =[(p-1)/2]²+3*[(p-1)/2]²                       mod p | adding
    =4[(p-1)/2]²                                   mod p | squaring
    =4[(p²-2p+1)/4]                                mod p | cancel down
    =  (p²-2p+1)                                   mod p | with p² = 0 mod p and 2p = 0 mod p
    = 1


    2. {(p-1)/2+[(p-1)/2]AI}³=1 mod p is always true

    Proof:

    {(p-1)/2+[(p-1)/2]AI}²  mod p
    =[(p-1)/2]²  -[(p-1)/2]²A²   +2[(p-1)/2]²     AI mod p | with A²=3
    =[(p-1)/2]²  -3[(p-1)/2]²     +2[(p-1)/2]²    AI mod p | substraction by the second term
    =-2[(p-1)/2]²                 +2[(p-1)/2]²    AI mod p | squaring
    =-2(p-1)(p-1)/4               +2[(p-1)(p-1)/4]AI mod p | cancel down
    = -(p-1)(p-1)/2               + [(p-1)(p-1)/2]AI mod p | p-1=-1 mod p
    =       (p-1)/2               - [     (p-1)/2]AI mod p


    Proof:

    =  { (p-1)/2 + [(p-1)/2]AI }³                             mod p |  with  { (p-1)/2 + [(p-1)/2]AI }²= { (p-1)/2 - [(p-1)/2]AI }  mod p
    =  { (p-1)/2 - [(p-1)/2]AI } * { (p-1)/2 + [(p-1)/2]*AI } mod p
    =  | (p-1)/2 + [(p-1)/2]AI |                              mod p |  with proof 1.
    =         1


    3. If

    a = (c^2 - 3 d^2) / (c^2 + 3 d^2) mod p

    and

    b = (2 c d) / (c^2 + 3 d^2) mod p,

    then

    a^2 + 3 b^2 = 1 mod p.

    Proof:

    a^2 + 3 b^2

    = [c^2 - 3 d^2)   / (c^2 + 3 d^2)]^2 + 3 [ (2 c d)   / (c^2 + 3 d^2)]^2  | squaring
    = (c^2 - 3 d^2)^2 / (c^2 + 3 d^2)^2  + 3 (4 c^2 d^2) / (c^2 + 3 d^2)^2
    = (1 / (c^2 + 3 d^2)^2 ) ( (c^2 - 3 d^2)^2 + 12 c^2 d^2 )
    = (1 / (c^2 + 3 d^2)^2 ) (c^4 - 6 c^2 d^2 + 9 d^4 + 12 c^2 d^2)
    = (1 / (c^2 + 3 d^2)^2 ) (c^4 + 6 c^2 d^2 + 9 d^4)
    = (1 / (c^2 + 3 d^2)^2 ) (c^2 + 3 d^2)^2
    = 1

    (Proof friendly transmitted by Kermit Rose)