sufficent proof for primes of the kind p:=x^2+x+1

Expand Messages
• 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
Message 1 of 3 , Nov 10, 2006
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
• ... Hi, Bernhard. I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241. Is it? Bill
Message 2 of 3 , Nov 14, 2006
--- 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 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
>

Hi, Bernhard.
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
>
• ... as ... But p=241 is not congruent to 3 (mod 4). So it is not a counter- example. Best regards, Dario Alpern Buenos Aires - Argentina
Message 3 of 3 , Dec 13, 2006
wrote:
>
> --- 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
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
> >
>
> Hi, Bernhard.
> I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241.
> Is it? Bill
>

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

Best regards,

Dario Alpern
Buenos Aires - Argentina
http://www.alpertron.com.ar/ENGLISH.HTM
Your message has been successfully submitted and would be delivered to recipients shortly.