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 2020-12-20, 03:31 #1 bhelmes     Mar 2016 1010000102 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

11·151 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 23·13·43 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 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.
 2021-01-25, 23:59 #5 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 22·179 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

22×1,531 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

257 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

22·1,531 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

257 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

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