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@...>

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

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