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From: Archimedes Plutonium on 12 Jul 2010 15:13


Archimedes Plutonium wrote:
> Archimedes Plutonium wrote:
> > Archimedes Plutonium wrote:
> > (snipped)
> > >
> > > There maybe a sticking point about this proof procedure for large even
> > > numbered
> > > prime pairs such as say p_n and p_n+1000, that we may have to have
> > > some
> > > more qualifiers to impose the Square Root Patch, due to the large
> > > spread between
> > > the n and even numbered n_
> > >
> >
>
> Let me try this on a proof of the Infinitude of N+6 Primes:
>
> Let me write a list of the first few primes:
>
>
> 2, 3, 5, 7, 11, 13, 17, 19, 23, . .
>
>
> The first twin primes is 3, 5
>
>
> The first quad primes (n+4) is 3, 7
>
>
> The first (n + 6) primes is 5, 11
>
>
> The first (n +8) primes is 3, 11
>
>
> etc etc
>
>
> And so it matters not, how large the N+2k becomes, because the
> square-
> root
> patch on the last prime in the successive prime list captures all the
> primes
> of that succession, that finite succession, for example, doing the
> N +6 primes proof would go like this:
>
>
> Proof of the Infinitude of N+6 Primes:
>
>
> (1) Definition of prime
>  (2) Hypothetical Assumption, suppose set of all N+6 Primes is finite
>  with
>  5,11 being the last two N+6 primes of the sequence set S = 5, 7, 11
>  where
>  11 is the last and largest of the N+6 primes. Note we have to
> eliminate
> 3 from the S succession list the prime factors of the 6/2 in N+6 so we
> delete
> 3.
>  (3) Multiply the lot and add 3 and subtract 3, yielding  W+3 =
> (2x5x7x11) +3 = 773 and
>  W-3 = (2x5x7x11) -3 = 767
>  (4) Take the square root of W+3 and I get 27.8.... I do not need
> square root of W-3, and there cannot be any
>  regular primes
>  for consideration of being a prime factor of this S secession that is
> greater
> than 27
>  (5) Successively divide all the primes in the sequence S into W+3
> and
>  W-3 and
>  they all leave a remainder but I also need to consider primes from
> 11 out to 27 and they are 13, 17, 19, and 23
>
> And this is where I may get into new trouble, in that I cannot
> eliminate them
> and that the method would then only be a proof of the Infinitude of
> Twin Primes
> and that the proof of N +2k Primes has to await another method.
>
> And the only reason that Twin Primes is amenable to this method is
> because of the
> separation of just 2
>
> So it is an open question at this moment whether the method is
> restricted to a proof of
> Twin Primes but no larger N+2k primes.
>

Now what could save the above:

(5) Successively divide all the primes in the sequence S into W+3
and W-3 and they all leave a remainder but I also need to consider
primes from
11 out to 27 and they are 13, 17, 19, and 23

Is that somehow, if 13, 17, 19, and 23 were prime factors of W+3 = 773
and
W-3 = 767 then 2, 3, 5, 7, 11 would also be a prime factor of W+3 and
W-3

So there is still hope that the Euclid Indirect of Euclid's Number
necessarily prime
captures not only the Twin Primes infinitude proof but captures all
even numbered Primes
infinitude proof.

I am still hopeful that another patch can wrangle out the pesky
regular primes that look
to interfer

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
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