## Prime equivalence theorem

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• 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
Message 1 of 6 , Jun 16, 2012
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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]
• m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others?  ________________________________ De: Maximilian Hasler
Message 2 of 6 , Jun 16, 2012
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m +/-1, m+/-2, m+/-3, m+/-6 are 4 values for the theorem. It is possible to find others?

________________________________
De: Maximilian Hasler <maximilian.hasler@...>
Para: Sebastian Martin Ruiz <s_m_ruiz@...>
Asunto: Re: [PrimeNumbers] Prime equivalence theorem

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

Maximilian

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

>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]
• As 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
Message 3 of 6 , Jun 16, 2012
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As 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@...>
> 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]
>
• 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
Message 4 of 6 , Jun 16, 2012
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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]
• Another 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
Message 5 of 6 , Jun 17, 2012
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Another 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@...>
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]
• Maximilian 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
Message 6 of 6 , Jun 17, 2012
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Maximilian 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@...>
> 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]
>
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