mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math > Number Theory Discussion Group

Reply
 
Thread Tools
Old 2020-12-20, 03:31   #1
bhelmes
 
bhelmes's Avatar
 
Mar 2016

1010000102 Posts
Default 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
bhelmes is offline   Reply With Quote
Old 2020-12-20, 09:15   #2
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

11·151 Posts
Default

Quote:
Originally Posted by bhelmes View Post
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 ?
Lin Tan's article referenced at the bottom of the Wikipedia page you linked to gives more details.
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.
Nick is offline   Reply With Quote
Old 2020-12-20, 17:34   #3
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

23·13·43 Posts
Default

I note that if p is prime, p = a2 + b2, and tan(θp) = b/a, then θ2 = π/4, and for p > 2, tan(2*θp) = 2*a*b/( a2 - b2)

Interchanging the order of a and b changes θp to π/2 - θp , hence 2*θp to π - 2*θp.

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 nth 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, θp/π is irrational. Therefore, the additive group {n*θp (mod 2*π)} is infinite cyclic, and

2) For any finite set of primes p == 1 (mod 4) the only Z-linear combination of the θp which is 0 (mod 2*π) is the one where all the coefficients are 0.

Another way of stating this is that the numbers θp/π, p == 1 (mod 4), are Q-linearly independent. This is a much stronger statement than saying that the individual numbers θp/π are irrational.
Dr Sardonicus is offline   Reply With Quote
Old 2020-12-26, 01:48   #4
bhelmes
 
bhelmes's Avatar
 
Mar 2016

2×7×23 Posts
Default

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.


bhelmes is offline   Reply With Quote
Old 2021-01-25, 23:59   #5
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

22·179 Posts
Default

The circle x^2 + y^2 = 3 passes through no rational points.
MattcAnderson is offline   Reply With Quote
Old 2021-01-26, 00:55   #6
retina
Undefined
 
retina's Avatar
 
"The unspeakable one"
Jun 2006
My evil lair

22×1,531 Posts
Default

Quote:
Originally Posted by MattcAnderson View Post
The circle x^2 + y^2 = 3 passes through no rational points.
Yes, if we ignore [(9/5), (12/5)], and an infinity of other points.


Last fiddled with by retina on 2021-01-26 at 00:58
retina is online now   Reply With Quote
Old 2021-01-26, 01:29   #7
dcheuk
 
dcheuk's Avatar
 
Jan 2019
Tallahassee, FL

257 Posts
Default

Quote:
Originally Posted by retina View Post
Yes, if we ignore [(9/5), (12/5)], and an infinity of other points.

You might be thinking about x^2+y^2=9. You'll have to multiply both sides of (3/5)^2+(4/5)^2=1 by sqrt(3) to get x=9/5 and y=12/5. There aren't any rational roots to this equation X^2+y^2=p for positive integer p if p mod 4 =3 I believe.

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.
dcheuk is offline   Reply With Quote
Old 2021-01-26, 01:37   #8
retina
Undefined
 
retina's Avatar
 
"The unspeakable one"
Jun 2006
My evil lair

22·1,531 Posts
Default

Quote:
Originally Posted by dcheuk View Post
You might be thinking about x^2+y^2=9. You'll have to multiply both sides of (3/5)^2+(4/5)^2=1 by sqrt(3) to get x=9/5 and y=12/5.
yes, you are correct. I was somehow seeing x^2+y^2=3^2.
retina is online now   Reply With Quote
Old 2021-01-26, 01:41   #9
dcheuk
 
dcheuk's Avatar
 
Jan 2019
Tallahassee, FL

257 Posts
Default

Quote:
Originally Posted by dcheuk View Post
You'll have to multiply both sides of (3/5)^2+(4/5)^2=1 by sqrt(3) to get x=9/5 and y=12/5
oops I meant by 9
dcheuk is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
the unit circle, tangens and the complex plane bhelmes Math 0 2018-03-04 18:50
a rational sequence MattcAnderson MattcAnderson 2 2016-12-14 16:58
rational integers wildrabbitt Math 17 2015-06-15 08:47
Two Questions about Unit Circle Unregistered Homework Help 6 2010-08-15 14:43
Circle intersection points Mini-Geek Puzzles 1 2006-11-27 03:20

All times are UTC. The time now is 13:54.

Tue Apr 20 13:54:41 UTC 2021 up 12 days, 8:35, 0 users, load averages: 2.79, 3.30, 3.30

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.