- A beautifull evening,

I try to give you a sufficent proof for primes p:=x^2+x+1

p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time as

divisor of p-1

p mod 3 = 1 or in other words 3 | p-1

p mod 9 > 1 => 3 appears only one time as divisor of p-1

2 ^ ((p-1)/2) = p-1 mod p, must be verified

then p is prime

proof:

x and x^2 are 3.roots of 1 because of the construction of the prime

p

t1 | x and t2 | x+1 t1, t2 are odd t1=/=t2 t1>3 t2>3

2 ^ ((p-1)/t1)=1 and 2^ ((p-1)/t2) = 1

then 2 ^ {[(p-1)/t1 - (p-1)/t2)] / t1*t2} = 1

<=> 2 ^ [(t2-t1) *(p-1) / (t1*t2)] = 1 with t2-t1 = 2*r

<=> 2 ^ [(2*r) *(p-1) / (t1*t2)] = 1

A)<=> 4 ^ [ r *(p-1) / (t1*t2)] = 1

2^ [(p-1)/2] = p-1 with squaring

B) 4 ^[(p-1)/2] = 1

A) and B) are contradicted because (p-1)/2 is odd and r*(p-1) is

even

=> with the theorem of Pocklington follows that p is a prime.

A factorisation of p-1 is not needed.

Please let me know if the proof is o.k.

Friendly greetings from the primes

Bernhard Helmes

www.devalco.de - --- In primenumbers@yahoogroups.com, "Bernhard Helmes" <bhelmes@...>

wrote:>

Hi, Bernhard.

> A beautifull evening,

>

> I try to give you a sufficent proof for primes p:=x^2+x+1

>

> p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time as

> divisor of p-1

> p mod 3 = 1 or in other words 3 | p-1

> p mod 9 > 1 => 3 appears only one time as divisor of p-1

>

> 2 ^ ((p-1)/2) = p-1 mod p, must be verified

>

> then p is prime

>

I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241.

Is it? Bill

> proof:

>

> x and x^2 are 3.roots of 1 because of the construction of the prime

> p

>

> t1 | x and t2 | x+1 t1, t2 are odd t1=/=t2 t1>3 t2>3

>

> 2 ^ ((p-1)/t1)=1 and 2^ ((p-1)/t2) = 1

> then 2 ^ {[(p-1)/t1 - (p-1)/t2)] / t1*t2} = 1

> <=> 2 ^ [(t2-t1) *(p-1) / (t1*t2)] = 1 with t2-t1 = 2*r

> <=> 2 ^ [(2*r) *(p-1) / (t1*t2)] = 1

> A)<=> 4 ^ [ r *(p-1) / (t1*t2)] = 1

>

> 2^ [(p-1)/2] = p-1 with squaring

> B) 4 ^[(p-1)/2] = 1

>

> A) and B) are contradicted because (p-1)/2 is odd and r*(p-1) is

> even

>

> => with the theorem of Pocklington follows that p is a prime.

>

> A factorisation of p-1 is not needed.

>

> Please let me know if the proof is o.k.

>

> Friendly greetings from the primes

>

> Bernhard Helmes

>

> www.devalco.de

> - --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:>

as

> --- In primenumbers@yahoogroups.com, "Bernhard Helmes" <bhelmes@>

> wrote:

> >

> > A beautifull evening,

> >

> > I try to give you a sufficent proof for primes p:=x^2+x+1

> >

> > p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time

> > divisor of p-1

But p=241 is not congruent to 3 (mod 4). So it is not a counter-

> > p mod 3 = 1 or in other words 3 | p-1

> > p mod 9 > 1 => 3 appears only one time as divisor of p-1

> >

> > 2 ^ ((p-1)/2) = p-1 mod p, must be verified

> >

> > then p is prime

> >

>

> Hi, Bernhard.

> I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241.

> Is it? Bill

>

example.

Best regards,

Dario Alpern

Buenos Aires - Argentina

http://www.alpertron.com.ar/ENGLISH.HTM