- View SourceAs Norman has intimated, here is what is occurring:

Choose consecutive odd primes p and q. Let m be the number exactly between them, at a distance x from p and q. Thus m-x = p and m+x = q. So of course m +/- x are primes, because they are the p and q we originally chose! So we must mind our p's and q's !

(2x represents the gap between p and q. Depending on the p and q chosen, x can be any number > 1. For instance if p is 89 and q is 97, we have m = 93, and 93 +/- 4 is prime. )

Mark

--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

>

> m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others?Â

>

>

> ________________________________

> De: Maximilian Hasler <maximilian.hasler@...>

> Para: Sebastian Martin Ruiz <s_m_ruiz@...>

> Enviado: SÃ¡bado 16 de junio de 2012 23:21

> Asunto: Re: [PrimeNumbers] Prime equivalence theorem

>

>

> Don't you have the same also for m +/- 1 Â and m +/- 6 (p>7)?

>

> Maximilian

>

>

>

>

> On Sat, Jun 16, 2012 at 11:37 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

>

>

> >Â

> >Hello all:

> >

> >Theorem:

> >

> >Let p and q consecutive prime numbers >=5

> >

> >Let m=(p+q)/2

> >

> >m+2 is prime if and only if m-2 is prime.

> >

> >Sincerely

> >

> >Sebastian Martin Ruiz

> >

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

> >

> >

> >

>

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

> - View SourceI'm not sure if I understand completely what you meant to say, because you

use "the same x" for (q-p)/2 (which varies with each pair (p,q))

and for the m +/- x (which is constant for the whole procedure /

"theorem", e.g. x=2 in the theorem of this thread).

But indeed these "theorems" can easily be proved by analysing the possible

values in a similar way :

q-p > 2 d => m +/- d are both composite, since strictly comprised between

the two consecutive primes.

q-p = 2 d => m +/- d are the two primes p & q.

Thereafter, there may be remaining cases (for d>1) :

If d=2,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

If d=3,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)

q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

If d=4 or d=5, exceptions are possible.

If d=6,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)

q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)

q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)

q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)

q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

This shows that the Theorem holds for d=1, 2, 3 and 6.

I think that for all other d there are infinitely many counter-examples.

Maximilian

[Non-text portions of this message have been removed] - View SourceAnother argument is:

m-k must be p and m+k must be q, otherwise m-k is the consecutive prime to p .. and that is q

(q-p)=2k

________________________________

Von: Maximilian Hasler <maximilian.hasler@...>

An: Mark <mark.underwood@...>

CC: primenumbers@yahoogroups.com

Gesendet: 8:19 Sonntag, 17.Juni 2012

Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem

I'm not sure if I understand completely what you meant to say, because you

use "the same x" for (q-p)/2 (which varies with each pair (p,q))

and for the m +/- x (which is constant for the whole procedure /

"theorem", e.g. x=2 in the theorem of this thread).

But indeed these "theorems" can easily be proved by analysing the possible

values in a similar way :

q-p > 2 d => m +/- d are both composite, since strictly comprised between

the two consecutive primes.

q-p = 2 d => m +/- d are the two primes p & q.

Thereafter, there may be remaining cases (for d>1) :

If d=2,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

If d=3,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)

q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

If d=4 or d=5, exceptions are possible.

If d=6,

q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)

q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)

q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)

q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)

q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

This shows that the Theorem holds for d=1, 2, 3 and 6.

I think that for all other d there are infinitely many counter-examples.

Maximilian

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

[Non-text portions of this message have been removed] - View SourceMaximilian is correct in that he actually shows why m-k must be p and m+k must be q. It really has to do with admissible 3 - prime constellations of a given length.

Mark

--- In primenumbers@yahoogroups.com, Norman Luhn <n.luhn@...> wrote:

>

> Another argument is:

>

> m-k must be p and m+k must be q, Â otherwise Â m-k is the consecutive prime to p .. and that is q

>

> (q-p)=2k

>

>

> ________________________________

> Von: Maximilian Hasler <maximilian.hasler@...>

> An: Mark <mark.underwood@...>

> CC: primenumbers@yahoogroups.com

> Gesendet: 8:19 Sonntag, 17.Juni 2012

> Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem

>

>

> Â

> I'm not sure if I understand completely what you meant to say, because you

> use "the same x" for (q-p)/2 (which varies with each pair (p,q))

> and for the m +/- x (which is constant for the whole procedure /

> "theorem", e.g. x=2 in the theorem of this thread).

>

> But indeed these "theorems" can easily be proved by analysing the possible

> values in a similar way :

>

> q-p > 2 d => m +/- d are both composite, since strictly comprised between

> the two consecutive primes.

>

> q-p = 2 d => m +/- d are the two primes p & q.

>

> Thereafter, there may be remaining cases (for d>1) :

>

> If d=2,

> q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

>

> If d=3,

> q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)

> q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

>

> If d=4 or d=5, exceptions are possible.

>

> If d=6,

> q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)

> q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)

> q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)

> q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)

> q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

>

> This shows that the Theorem holds for d=1, 2, 3 and 6.

>

> I think that for all other d there are infinitely many counter-examples.

>

> Maximilian

>

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

>

>

>

>

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

>