rational points on the unit circle

[bhelmes]

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

[Nick]

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.

[Dr Sardonicus]

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

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 Z-linear 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 Q-linearly independent. This is a much stronger statement than saying that the individual numbers θ_p/π are irrational.

[bhelmes]

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.

[MattcAnderson]

The circle x^2 + y^2 = 3 passes through no rational points.

[retina]

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.

[dcheuk]

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.

[retina]

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.

[dcheuk]

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