Bernhard Helmes and X^2 - Free people check

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math, x^2 = y (mod p)

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Bernhard Helmes. Wed, 12 May :34:04 GMT. William Ellio # x^2 = y (mod p). It's not easy. As a matter of fact, it's used in crytology. Otherwise study ...

Google Groups: Suche zwei 100 stellige Primzahlen - Google Groupsgroups.google.com › de.sci.mathematik

Bernhard Helmes: > Ein möglicher einfacher Primzahltest ist folgender: > Sei p=x^2+1, dann ist ein hinreichendes Kriterium:

Polynomzerlegung in komplexe Linearfaktorende.sci.mathematik.narkive.com › WkIrfT55 › polyno...

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Post by Bernhard Helmes Einen erfreulichen guten Morgen, wie läßt sich das Polynom f(x)=x^4+x^3+x^2+x+1 in komplexe. Linearfaktoren zerlegen ?

Google Groups: Primzahlalgorithmus p:=x^2+1

: Bernhard Helmes de sci mathematik ... der Form p=x^2+

Google Groups: x^3 =2^y (mod p) or x^2 = 1 (mod p) //primes

: Bernhard Helmes sci math Hello What is easier to calculate a) x

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Primzahlalgorithmus fuer Primzahlen der Form p=4x^2+1 ...www.devalco.de › ...

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Primzahlalgorithmus fuer Primzahlen der Form p=4x^2+ Bernhard Helmes Aachen, www.devalco.de

Prime numbers and primality testing - Yahoo Groups

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sufficent proof for primes p:=x^2+x+1. A beautifull day, I have examine the prime numbers which lay on the polynom f(x)=x^2+x+1 My hypothese was : if 2^[(p-1)/3] = x or 2^[(p-1)/3] =x^2, then p is. Bernhard Helmes. 06 Mar, #

sufficent proof for primes of the kind p:=x^2+x+1 - Yahoo! Groups

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2^((240/2)) == 1 mod Is it? Bill. Message 2 of 3 , Nov 14, View Source In , "Bernhard Helmes" <bhelmes@...>

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Bernhard Helmes and X^2 Free People Check