Development of |
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29 31 33 |
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 | |
0 | 30√3 |
|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)