20201220, 03:31  #1 
Mar 2016
101000010_{2} Posts 
rational points on the unit circle
A peaceful and pleasant night for you,
I found the text in the english Wikipedia: https://en.wikipedia.org/wiki/Group_...roup_structure I did not understand, why "the point (a² − b²)/p + (2ab/p)i is a generator of Gp" Is this the same as a primitve root of p ? Is there a mathematical proof availible ? Greetings from the unit circle and tangens Bernhard 
20201220, 09:15  #2  
Dec 2012
The Netherlands
11·151 Posts 
Quote:
If you are interested in this topic, the book "Rational Points on Elliptic Curves" by Joseph Silverman and John Tate is a good place to start. 

20201220, 17:34  #3 
Feb 2017
Nowhere
2^{3}·13·43 Posts 
I note that if p is prime, p = a^{2} + b^{2}, and tan(θ_{p}) = b/a, then θ_{2} = π/4, and for p > 2, tan(2*θ_{p}) = 2*a*b/( a^{2}  b^{2})
Interchanging the order of a and b changes θ_{p} to π/2  θ_{p} , hence 2*θ_{p} to π  2*θ_{p}. Furthermore, if (A, B, C) is a primitive Pythagorean triple, cos(θ) = A/C, and sin(θ) = B/C, then if C > 2, θ is not a rational multiple of π. For if it were, n*θ would be an integer multiple of 2*π for some n, and z = cos(θ) + i*sin(θ) and its conjugate cos(θ)  i*sin(θ) would be n^{th} roots of unity, hence algebraic integers. Their sum 2*cos(θ) would then also be an algebraic integer, and, being rational, a rational integer. The only possible values for cos(θ) would then be 1, 1/2, 0, 1/2, and 1. It follows easily from this that, 1) Except for p = 2, θ_{p}/π is irrational. Therefore, the additive group {n*θ_{p} (mod 2*π)} is infinite cyclic, and 2) For any finite set of primes p == 1 (mod 4) the only Zlinear combination of the θ_{p} which is 0 (mod 2*π) is the one where all the coefficients are 0. Another way of stating this is that the numbers θ_{p}/π, p == 1 (mod 4), are Qlinearly independent. This is a much stronger statement than saying that the individual numbers θ_{p}/π are irrational. 
20201226, 01:48  #4 
Mar 2016
2×7×23 Posts 
It is always a pleasure to read your clear explications, many thanks for your support.
I am still missing the next Mersenne Prime number, by all the amazing amount of work in 2020 there will be sooner or later the next Mp. A peaceful and pleasant Christmas for you and your family. 
20210125, 23:59  #5 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}·179 Posts 
The circle x^2 + y^2 = 3 passes through no rational points.

20210126, 00:55  #6  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×1,531 Posts 
Quote:
Last fiddled with by retina on 20210126 at 00:58 

20210126, 01:29  #7  
Jan 2019
Tallahassee, FL
257 Posts 
Quote:
Second the recommendation on "Rational Points on Elliptic Curves" by Silverman And Tate. For a more introductory and algebraic geometry approach I also recommend "Algebraic Geometry: A Problem Solving Approach" by Garrity et al. 

20210126, 01:37  #8 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,531 Posts 

20210126, 01:41  #9 
Jan 2019
Tallahassee, FL
257 Posts 

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