## Primes in the field of |a+b*sqrt(3)*I|=1

Expand Messages
• A peaceful day to all group members and especially for Kermit and David, i would like to start a discussion about primes and some special theory concerning the
Message 1 of 1 , Feb 24, 2014
A peaceful day to all group members and especially for Kermit and David,

i would like to start a discussion about primes and some special theory
concerning the imaginary quadratic field a+b*sqrt(3)*I mod  p
with the reduction to the norm = 1 mod p of these elements

I hope that this field could be useful for developing a sufficient prime test.
It would be nice if some more mathematicians could participate in the discussion.

As far as i see there exists always a trivial third root for all odd numbers:

Let A=sqrt (3)

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
=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 | subtraction 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. Choosing an element of this group

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)

If someone likes more an illustrate picture of the different number in this field,
there is a web application in order to see the different kind of fields
http://devalco.de 4. k) with adjoined √3 in the complex field with norm=1 and a metric

Some questions for those you had enough patience to follow the text:
1. Is it possible to distinguish the elements in quadratic and non quadratic residues
2. For primes p = 1 mod 6 there seems to be always p-1 elements and
for primes  p = 5 mod 6 there seems to be always p+1 elements
Kermit, can you transmit your proof for the group elements ?
3. Is it sensful to develop a kind of rabin-miller test in order to check the 2. and 3. root of 1

The best greetings from the primes
Bernhard

Your message has been successfully submitted and would be delivered to recipients shortly.