## more equivalence theorem

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• Hello all: Also we have: Theorem: Let p and q consecutive prime numbers =5 Let m=(p+q)/2 m+6 is prime if and only if m-6 is prime. Sincerely Sebastian Martin
Message 1 of 2 , Jun 16, 2012
Hello all:

Also we have:

Theorem:

Let p and q consecutive prime numbers >=5

Let m=(p+q)/2

m+6 is prime if and only if m-6 is prime.

Sincerely

Sebastian Martin Ruiz

then for k=2 , 3 , and 6
m+k is prime if and only if m-k is prime.

I think is possible to found more k....

[Non-text portions of this message have been removed]
• Hello Sebastian,  I think it is trivial. m=(p+q)/2 z1=m-k and z2=m+k are prime. m-k+m+k=2m=p+q. So m-k =p and m+k=q , your primes to calculate m. Norman
Message 2 of 2 , Jun 16, 2012
Hello Sebastian,
I think it is trivial.

m=(p+q)/2

z1=m-k and z2=m+k are prime.

m-k+m+k=2m=p+q. So m-k =p and m+k=q , your primes to calculate m.

Norman

________________________________
Von: Sebastian Martin Ruiz <s_m_ruiz@...>
Gesendet: 20:19 Samstag, 16.Juni 2012

Hello all:

Also we have:

Theorem:

Let p and q consecutive prime numbers >=5

Let m=(p+q)/2

m+6 is prime if and only if m-6 is prime.

Sincerely

Sebastian Martin Ruiz

then for k=2 , 3 , and 6
m+k is prime if and only if m-k is prime.

I think is possible to found more k....

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed]
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