 mersenneforum.org rational points on the unit circle
 Register FAQ Search Today's Posts Mark Forums Read 2020-12-20, 03:31 #1 bhelmes   Mar 2016 32×41 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   2020-12-20, 09:15   #2
Nick

Dec 2012
The Netherlands

3·587 Posts Quote:
 Originally Posted by bhelmes 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.   2020-12-20, 17:34 #3 Dr Sardonicus   Feb 2017 Nowhere 22·32·11·13 Posts 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.   2020-12-26, 01:48 #4 bhelmes   Mar 2016 36910 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.       2021-01-25, 23:59 #5 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 24×61 Posts The circle x^2 + y^2 = 3 passes through no rational points.   2021-01-26, 00:55   #6
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2·23·137 Posts Quote:
 Originally Posted by MattcAnderson 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   2021-01-26, 01:29   #7
dcheuk

Jan 2019
Tallahassee, FL

5·72 Posts Quote:
 Originally Posted by retina 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.   2021-01-26, 01:37   #8
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2×23×137 Posts Quote:
 Originally Posted by dcheuk 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.   2021-01-26, 01:41   #9
dcheuk

Jan 2019
Tallahassee, FL

5×72 Posts Quote:
 Originally Posted by dcheuk 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   2021-05-23, 01:22 #10 bhelmes   Mar 2016 36910 Posts A peaceful and pleasant night, I have added a pyramid like webapplication for primes, regarding gaussian numbers on the unit circle: http://devalco.de/unit_circle/system_unit_circle.php Perhaps someone likes it.      2021-06-02, 20:25 #11 bhelmes   Mar 2016 32×41 Posts You can visualize gaussian numers in two different ways: 1. Reducing it to the unit circle, so that the norm (a,b)=a²+b²= 1 mod p http://devalco.de/unit_circle/system...e.php?prim=127 2. Reducing it to the tangens = (a*c)/(b*c) = a/b http://www.devalco.de/unit_circle/sy...s.php?prim=127 I like writing graphical explications, but not every one will be astonished. Some ideas which kind of complex calculation is suitable for a prime test ?     Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post bhelmes Math 0 2018-03-04 18:50 MattcAnderson MattcAnderson 2 2016-12-14 16:58 wildrabbitt Math 17 2015-06-15 08:47 Unregistered Homework Help 6 2010-08-15 14:43 Mini-Geek Puzzles 1 2006-11-27 03:20

All times are UTC. The time now is 14:13.

Wed Dec 8 14:13:36 UTC 2021 up 138 days, 8:42, 1 user, load averages: 1.54, 1.47, 1.52