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• This is directed to the entire group: From: Paul Jobling Date: Thu Dec 9, 2004 9:22 am Subject: RE: [PrimeNumbers] GC proof complete?
Message 1 of 31 , Dec 9, 2004
This is directed to the entire group:

From: "Paul Jobling" <Paul.Jobling@W...>
Date: Thu Dec 9, 2004 9:22 am
Subject: RE: [PrimeNumbers] GC proof complete?

Jens,

As has often been pointed out, what people forget in their 'proofs' of GC is
that the proof must depend somehow on numbers being prime, rather than
simply having the same density as the prime numbers in some range. Proofs
that take that into account will, necessarily IMHO, be difficult.

There is the other point that says that any 100 line elementary proof of GC
that was missed by Euler, let alone Erdos, is necessarily nonsense.

Right. Any proof which relies on the density argument alone is
necessarily flawed, because it is possible to construct a sequence
which is "statistically identical" to the prime numbers (same
asymptotic density, same distribution of gaps, etc.) but for which
GC is false.

Now for my own (Greg's) comments:

My posted examples "depend somehow on numbers being prime."
I extracted one of those examples below to demostrate that I must count 22
primes < (88 + 2).
q = 22.
Lets look at density.
(88 + 2) can have 90 number pairs that sum to 90.
(0,90), (1,89), (2,88) ...(45,45), (46,44), (47,43), (48,42), (49,41) ...
(89,1)
Note that pairs after (45,45) are mirror images of those that came before 45.
The 2nd half are equivalent and are not unique pairs but instead are
duplicates.
Therefore I discard the 2nd half leaving 45 unique number pairs.
Of the 45 pairs, 23 of them are odd number pairs.
Since I incorporate all 22 primes in the form (prime + 1) even numbers,
I now discard all odd number pairs.
Of 90 sequential numbers we are left with ONLY 22 even numbers to pair.

Even if ALL 22 were (prime + 1), a density of 100%, then that leaves a
maximum of ONLY
22 candidates that must be eliminated for the counter-example to be true.
It turns out that the density is not 100% since only 13 of the 22 numbers
are type 1 (prime + 1): 4,6,8,12,14,18,20,24,30,32,38,42,44.
Again, I "depend somehow on numbers being prime."
In my "Stepladder Proof" we just scaled 13 of 22 rungs on the ladder.
After 44, we reach the top and are headed down again.
Since I know there NUST be 22 type 1 (prime + 1) even numbers,
non-type 1 are simply every other even number < 90.
That calculation is E1 = A/2 - q - 1 = 44 - 22 - 1 = 21.
Its simple elimination.
Now my "counter-example assumption" plays a very very prominent role:

What happens if we "assume" that every type 1 MUST pair with a non-type 1?
Given that false assumption we nevertheless MUST end up with a SURPLUS
of "unique" type 1 numbers that can't be paired with anything but themselves.
That is the calculation:
Q1 = q - E1 + 1 = 22 - 21 + 1 = 2.
So, if everything got paired to favor the counter-example,
we still end up with two unpaired type 1's (prime + 1, prime + 1) = 90.
I don't attempt to predict which ones they are -- only that they MUST exist.
Next, my algorithm ties up some loose ends to quantify why there must be
more prime pairs than the minimum number of pairs calculated above.
My method of proof: Assume a counter-example exists until inconsistencies
result.

EXTRACT FROM MY POST:

Class 2: Q1 Example With A + 1 Prime, A/2 Type 1 and A/2 + 1 Odd Composite:

A = 88, A + 1 = prime, A + 2 = 90, q = 22, E1 = A/2 - q - 1 = 44 - 22 - 1 =
21
Q1 = q - E1 + 1 = 22 - 21 + 1 = 2.
Also
Q1 = 2q + 2 - A/2 = 44 + 2 - 44 = 2.
Step 1: A + 2 is type 1; Q1 - 1 = 1
Step 2: A = 88 is not type 1; add 1 to step 1: Q1 = 2.
Step 3: A/2 + 1 = 45 is not type 1; go to step 4 with Q1 = 2.
Step 4: A/2 + 1 = 45 is not type 2 or 3; retain Q1 = 2.
Step 5: Type 2 {26,64}, {34,56}, {40,50} permit 3 more type 1 pairs: 2 +
(3*2) = 8.

Four type 1 pairs {6,84}, {18,72}, {30,60} and {42,48} are predicted by 8
excess type 1 numbers that must pair with each other. There are (a) 14 mixed
{type 1, type 2 pairs}, (b) 1 pair with 2, (c) 3 type 2 pairs and (d) 4 type 1
pairs or 22 even number pairs in A + 2.

A and A + 2 both have A/4 = 22 unique even number pairs. A + 2 replaces {44,
44} with {45, 45}. But {44,46} and {46,44} in A + 2 remain unchanged subsets
from A that are merely paired differently in the 22 unique even number pairs of
A + 2.

So do you believe as I do that my method of proof "must depend somehow on
numbers being prime."

Greg

[Non-text portions of this message have been removed]
• In a message dated 12/10/04 6:13:13 AM Eastern Standard Time, ... In my proof I make the outrageously false assumption that every sequential prime from p_1
Message 31 of 31 , Dec 10, 2004
In a message dated 12/10/04 6:13:13 AM Eastern Standard Time,
Paul.Jobling@... writes:
> Odd, because your 'proof' contains the following as its crux:
>
> "Therefore we find a logical inconsistency:
> The frequency and distribution of primes declines after p_n/2 for larger
> composites, C.
> Therefore, we must expect the relation
> r_((n/2)+1) >p_((n/2)+1) as opposed to the calculated result above."
>
> So you *are* simply claiming that your proof is true because the number of
> primes decreases - ie it is based entirely on the distribution of primes. I
> tried to follow your logic, by the way, but I couldn't because you leave so
> many things undefined (what is s_2, in fact why bother with those at all?).
>

In my "proof " I make the outrageously false assumption that
every sequential prime from p_1 = 3 thru p_(n/2) MUST pair with an odd
composite.
The difference from my examples is that I am pairing odd numbers only. I
arrive at an arbitrary
stopping point, the median prime since I felt the inconsistency is more acute
at that median prime.

The 10 classes of examples make the same outrageously false assumption --
that ALL even numbers of the form (prime + 1) MUST pair with even numbers
of the form (largest prime + (x > 1)).

Effectively, I see it as the same assumption and the same approach.
The advantage of the examples is that they leave a visible
trail of quantified yet unmatched "unique" type 1 (prime + 1) numbers that
must pair with
"something." That is, 22 primes populate the range 0 thru 90.
They must all get squeezed into the 1st 22 even numbers where they can't
possibly fit.
Only 13 fit. So 9 become candidates for pairing with the other 13.
But I assume non-type 1 numbers all get paired first to favor the
counter-example assumption.
That leaves unpaired (prime + 1, prime + 1) sets.

Now if my examples in some way don't mimic the proof approach on my URL,
then I suppose they should be used as a basis for a separate proof.

Regarding s_2, I had to invent a reason why the counter-example assumption
can proceed down to the median prime. Its a "padding" number and in reality
may be zero.
It too is based on a false assumption
that the counter-example exists. I agree the format is lacking in rigor and
formalism. In fact, I agreed that a formalistic approach would likely break
the
proof since this proof would have been done already.

Greg

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