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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
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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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