- **Number Theory Discussion Group**
(*https://www.mersenneforum.org/forumdisplay.php?f=132*)

- - **a suggestion for a fast probablistic prime test**
(*https://www.mersenneforum.org/showthread.php?t=28016*)

a suggestion for a fast probablistic prime testA peaceful night for you,
Imho this is a nice piece of math, a suggestion for a fast probablistic prime test, perhaps the finding of two suitable bases with the same norm can be improved, nevertheless the calculation is easy and there is an added example. Mathematical feedback is welcome and the next Mp not far away. (The first part is written on english, the second part on german.) [URL]http://devalco.de/fast_prob_prime_test.pdf[/URL] :anonymous: |

Two months later, 270 views and 0 reply for a at least nice idea for a mersenne probabilistic prime test,
Is it worth to make an implementation or should I save my time for more interesting work ? Is there a mathematical or a calculative reason, why the described algorithm should be not o.k. ? Mathematical feedback would be nice ..... :brian-e::uncwilly: :rakes: :judge: |

That sounds like an interesting approach.
However, I cannot follow the practical example:[QUOTE]A:=16α=15,724699572=0,702849249836+6*pi B:=16β=19,5524051714=0,702849249836+6*pi[/QUOTE]19.5524051714 = 0.702849249836+6*Pi but 15.724699572 <> 0.702849249836+6*Pi Can you please describe the example in a little more detail? |

There is no way that a/b mod p can possibly be defined as an element of Q/Z. Assuming a/b mod p is defined, there is an integer r in [0,p-1] such that a/b == r (mod p). And, of course, r is 0 in Q/Z.
The group Q/Z is merely an additive group. Multiplication is not defined in Q/Z. |

A peaceful and pleasant night for all members, especially for Dr Sardonicus,
I appreciate really your mathematical explications. [QUOTE=Dr Sardonicus;616464]There is no way that a/b mod p can possibly be defined as an element of Q/Z. Assuming a/b mod p is defined, there is an integer r in [0,p-1] such that a/b == r (mod p). And, of course, r is 0 in Q/Z.[/QUOTE] If I calculate a/b=a*b^(-1) mod p. Is this an advantage, because of changing the group ? [QUOTE=Dr Sardonicus;616464] The group Q/Z is merely an additive group. Multiplication is not defined in Q/Z.[/QUOTE] I only need a scalar multiplication which is granted if the additive group is abel and where the scalar multiplication does not hurt the seclusion. Q/Z is Z-linear for the scalar. Besides: Is there a connection between primitive Pythagorean triples and rotation matrices, because a²+b²=c² <=> (a/c)²+(b/c)²=1 <=> [a*c^(-1)]²+[b*c^(-1)]²=1 <=> x²+y²=1 where the rotation matrix M of (x; -y) (y; x) has the determinant x²+y²=1 Best regards :crazy: :loco::hello::uncwilly: |

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