Development of
Algorithmic Constructions

01:46:14

20.Apr 2024

1. Abstract

There is a relationship between the Pythagorean triples
resp. the corresponding rational points of the circle with unit 1 in the complex field
and the group of the complex elements (x+yI) with norm equal 1 mod p where x, y are elements of Np.
This relationship is one explication for the system of primes.

Primes are perfect circles generated by the Pythagorean triples.

2. Transformation

Pythagorean
triples

--->

complex vector
in the circle of unit 1

--->

group of complex elements
|a+bI|=1 mod p with a,b element Np

3. Mathematical proofs

    p an odd number > 1
    Let a,b,c a Pythagorean triple with c < p

    a²     + b²         = c²            with division by c²
<=> (a/c)² +(b/c)²      = 1             with I²=-1
<=> (a/c)² -(bI/c)²     = 1

    x, y element Np
    x:=(a+rp)/c with r element N, r is choosen, so that a+rp = 0 mod c
    r=-a*p^(-1) mod c       with gcd (p,c)=1

    y:=(b+sp)/c with s element N, s is choosen, so that b+sp = 0 mod c
    s=-b*p^(-1) mod c       with gcd (p,c)=1

    |x+yI|= (x+yI)(x-yI)=x²+y²      mod p
          = [(a+rp)/c]²+[(b+sp)/c]² mod p
          = (a/c)²+(b/c)²           mod p
          =       1                 mod p

4. Calculation

Let p be an odd number greater 1
Let a, b, c be all Pythagorean triples with a²+b²=c² and c < p

Dividing by c² gives the rational point in the circle of unit 1
(a/c)²+(b/c)²=1

With I²=-1 the rational point is transformed int the complex field.
(a/c)²-(bI/c)²=1

The rational point a/c is transformed in x element Np
(a+k*p) = 0 mod c with k element of Z
r=-a*p^(-1) mod c
x=(a+r*p)/c

(if p is divisible by c then the calculation is aborted, because p^(-1) mod c cannot be calculated.

Analog: The rational point b/c is transformed in y element Np
(b+s*p) = 0 mod c with k element of Z
s=-b*p^(-1) mod c
y=(a+k*p)/c

As a result you get x+yI with |x+yI|=(x+yI)(x-yI)=1 mod p

5. Symmetry

The symmetry of the resulting group of the complex elements with norm=1 based on the fact that
a²+b²=b²+a²=c² resp.
(a/c)²-(bI/c)²=(b/c)²-(aI/c)²=1
and that x²+y²=(p-x)²+y²=x²+(p-y)²=(p-x)²+(p-y)² mod p;

Therefore one Pythagorean triple gives always 4 or 8 solutions with |x+YI|=1 mod p

The first trivial Pythagorean triple (0,1,1) gives the 4 solution (0,1), (1,0), (0,p-1) and (p-1,0).

If a Pythagorean triple has as result (x, y) with x=y then there are the following 4 solutions:
(x, x), (x, p-x), (p-x, x) and (p-x, p-x)

Otherwise there are 8 solutions for (x, y):
(x, y), (x, p-y), (p-x, y) and (p-x, p-y) and
(y, x), (y, p-x), (p-y, x) and (p-y, p-x)

6. Program in pseudocode

This is not a practical test for prime numbers, but a short description, how the webprogram works.

Let p an odd number > 1;
prime=True;
for i from all pyth. triples (a, b, c) with c < p
    if (gcd (c, prim)>1) then prime=False; break;
    calculation (a,b,c)-> (x, y);
    if (gcd (x, prim)>1) then prime=False; break;
    if (gcd (y, prim)>1) then prime=False; break;
    output of all elements of (x,y);
end_for;
if prime=True then p is prime;

8. Links

Pythagoraen Triples by Chris K. Caldwell
Pythagoraen Triples, Wikipedia
Calculation of Pythagoraen Triples by Arndt Bruenner
Weisstein, Eric W. "Pythagorean Triple." From MathWorld--A Wolfram Web Resource.
List of Pythagoraen Triples with c < 10000, M. Somos,
A020882, Ordered hypotenuses (with multiplicity) of primitive Pythagorean triangles, Oeis

9. Webprogram for the demonstration

p < 200 table of Pythagorean triples + graphic is shown + table with elements
p < 500 table of Pythagorean triples + graphic is shown
p < 10000 table of Pythagorean triples is shown

29 31 33

p = 31

a, b, c is the Pythagorean triple with a²+b²=c²

nr.	a	b	c	->	(x,y) with \|x+yI\|=1	number of elements
0.	0	1	1	->	(0,1), (1,0), (0,30), (30,0)	4
1.	3	4	5	->	(7,13), (24,13), (7,18), (24,18) (13,7), (13,24), (18,7), (18,24)	8
2.	5	12	13	->	(2,11), (29,11), (2,20), (29,20) (11,2), (11,29), (20,2), (20,29)	8
3.	15	8	17	->	(5,10), (26,10), (5,21), (26,21) (10,5), (10,26), (21,5), (21,26)	8
4.	7	24	25	->	(4,4), (27,4), (4,27), (27,27)	4
						group order (31)=32