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Oct 8, 2002, 4:29:29 PM10/8/02

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A fellow from Monash University in Australia proves the Goldbach Conjecture:

http://arxiv.org/pdf/math.GM/0206033

I haven't found any discussions on this proof - don't know when it was posted.

-Han

Oct 8, 2002, 4:41:48 PM10/8/02

to

In article <580fae16.02100...@posting.google.com>,

The code indicates it was uploaded on June of 2002.

The manuscript does not seem very promising. Page 2 says:

"The method we use in our approach, relies on the following facts (or,

interpretations):

1. Controversial points with Golbach's conjecture are:

1.1 It seems as if there must be a kind of formula that can

produce prime numbers."

The rest of the manuscript seems likewise confusing. The attempt at a

formal definition of "row" seems hopelessly confused, and betrays an

unfamiliarity of the author with standard mathematical usage.

This does not mean that whatever it is the author is doing is

certainly wrong, but it does not inspire much confidence...

(The author lists his affiliation as "School of Physics and Materials

Engineering")

A more interesting manuscript with a proposed conditional proof of

Goldbach can be found in the same arxiv,

http://front.math.ucdavis.edu/math.GM/0209232

where the author states that he will prove that if every odd integer >

5 is a sum of three primes, then every even integer greater than 2 is

a sum of two primes.

======================================================================

"It's not denial. I'm just very selective about

what I accept as reality."

--- Calvin ("Calvin and Hobbes")

======================================================================

Arturo Magidin

mag...@math.berkeley.edu

Oct 8, 2002, 4:45:29 PM10/8/02

to

I get a little skeptical when I read "It seems as if there must be

a kind of formula that can produce prime numbers."

After browsing through the whole thing, I wonder where he uses any

property of the set of prime numbers. If he doesn't (or only very

trivial ones) then it is easy to prove the argument wrong by

creating a similar conjecture with a set thin enough (for example)

that it fails.

--

J K Haugland

http://www.neutreeko.com

Oct 8, 2002, 5:18:36 PM10/8/02

to

I'm avoiding writing up solutions to related rates problems, so I took

a look.

The author's notation is truly atrocious. But here is what I have been

able to gleam:

A "row" is a finite tuple of positive integers, listed from smallest

to largest.

A "range" is a finite tuple of positive integers, listed from smallest

to largest, with the property that if x,y,z are positive integers,

x<y<z, and x,z are in the range, then so is y.

In particular, a range it totally determined by its first and last

elements; so we can write R(a,b). Alas, the author prefers R(ab), so

the range from 1 to 100 is R(1100); it is hard to see how we will

distinguish between the range determined by 1 and 11000, and the range

determined by 11 and 1000.

Theorem 1 says that if you take two rows all of whose elements are

contained in some range, then their first elements differ by a

constant; it is unclear why "contained in some range" is necessary

here, but whatever.

Then he says "We define a function f(x) so that for any x=a, f(a) is a

prime number". Fair enough; let f(n) be the "n-th prime".

Then he attempts to define "degree of complexity" as "the number of

primes to be added to each other to yield an even number greater than

2". Obviously, a good definition would be:

DC(n) = least positive integer m such that n can be expressed as a sum

of n primes.

Since every even integer is a multiple of 2, DC(n) exists. However,

the author asserts that DC(8)=4, since 8 = 2+2+2+2, yet clearly DC(8)

should be 2, since 8=5+3. Which he notes later. So DC(8) is both 4 and

2. Very, very bad.

The proof then proceeds:

Assume that you have a finite ordered tuple of integers in a given

range, which contains at least 1 prime, has the same number of even

and odd positive integers, and has more odd integers than primes.

At this point the author says (freely translating into math):

Take an even number A greater than 2. Its degree of complexity is at

least 2; since A is even greater than 2, the number of even positive

integers required to express A as a sum is less than the number of

even number in the given row.

This is false in general. You need two even positive integers to

express A. The row is assumed to have at least one prime, hence at

least two odd integers, hence at least two even integers. But that is

all; for example, the row

(4, 5, 6, 9)

satisfies the hypothesis: it has at least one prime (1), has 2 even

and 2 positive integers, and has 2 odd numbers, more than the number

of primes you need. Yet 2 is not less than 2.

Fortunately for the author, he means "less than or equal" when he says

"less than". Since the row contains at least 2 even numbers, then the

assertion is true.

Unfortunately, the author asserts that DC(A) equals the least number

of even positive integers required to express A as a sum (pp. 7, line

20, equation labeled (4) following the discussion on the paragraph),

so the author is in fact ASSUMING that DC(A)<=2, which is, of course,

Goldbach's conjecture.

Oct 8, 2002, 5:28:11 PM10/8/02

to

In article <580fae16.02100...@posting.google.com>,

hanth...@hotmail.com (hantheman) wrote:

hanth...@hotmail.com (hantheman) wrote:

The usual method is to read the proof carefully and stop at the first

error. The mathematics in the paper doesn't look particularly difficult,

although the language does, so it is a bit hard to understand what the

author means.

However, the first bit of the proof is the definition of a "row" which

looks like complete nonsense to me. (A "row" is supposed to be a set of

integers with a smallest and a largest element such that every element

of the set is greater than the smallest element and less than the

largest). Doesn't exactly fill the reader with confidence.

Oct 8, 2002, 5:31:21 PM10/8/02

to

Oh my god - in his very first definition he writes that a set of

integers is a "row" if it has properties I, II, III and IV and then the

guy writes down only three of the properties!

I wouldn't be surprised if this paper was published on April, 1st.

Oct 8, 2002, 5:58:15 PM10/8/02

to

On Tue, 8 Oct 2002 21:18:36 +0000 (UTC), mag...@math.berkeley.edu

(Arturo Magidin) wrote something.

(Arturo Magidin) wrote something.

A typo?

>

> Then he attempts to define "degree of complexity" as "the number of

> primes to be added to each other to yield an even number greater than

> 2". Obviously, a good definition would be:

>

> DC(n) = least positive integer m such that n can be expressed as a sum

> of _m_ primes. <------?

>

> Since every even integer is a multiple of 2, DC(n) exists. However,

> the author asserts that DC(8)=4, since 8 = 2+2+2+2, yet clearly DC(8)

> should be 2, since 8=5+3. Which he notes later. So DC(8) is both 4 and

> 2. Very, very bad.

>

:-)

F.

Oct 8, 2002, 6:21:07 PM10/8/02

to

In article <3da35404...@news.t-online.de>,

Franz Fritsche <in...@simple-line.de> wrote:

>On Tue, 8 Oct 2002 21:18:36 +0000 (UTC), mag...@math.berkeley.edu

>(Arturo Magidin) wrote something.

>

>A typo?

>

>>

>> Then he attempts to define "degree of complexity" as "the number of

>> primes to be added to each other to yield an even number greater than

>> 2". Obviously, a good definition would be:

>>

>> DC(n) = least positive integer m such that n can be expressed as a sum

>> of _m_ primes. <------?

Franz Fritsche <in...@simple-line.de> wrote:

>On Tue, 8 Oct 2002 21:18:36 +0000 (UTC), mag...@math.berkeley.edu

>(Arturo Magidin) wrote something.

>

>A typo?

>

>>

>> Then he attempts to define "degree of complexity" as "the number of

>> primes to be added to each other to yield an even number greater than

>> 2". Obviously, a good definition would be:

>>

>> DC(n) = least positive integer m such that n can be expressed as a sum

>> of _m_ primes. <------?

Don't see why.

DC(n)=2 if n can be expressed as a sum of 2 primes, and is not a prime.

DC(n)=7 if n can be expressed as a sum of 7 primes, but not of 6 or

fewer primes.

etc.

Goldbach's conjecture is equivalent to:

"For every positive even integer n>2, DC(n)=2."

Oct 8, 2002, 8:01:25 PM10/8/02

to

Theorem: A fool has more ways in which to be foolish than a wise man has to

be wise.

be wise.

Proof: "Arturo Magidin" agi...@math.berkeley.edu wrote in message

news:anvi3c$142v$1...@agate.berkeley.edu...

Re http://arxiv.org/pdf/math.GM/0206033

>The author's notation is truly atrocious.

Yes, and in so saying, you show you are a generous man. For example, page

4 refers repeatedly to the meeting of condition IV; no such condition is

defined. Page 5 condition [5] either confuses the variable with its index

(example H suggests what he might mean) or states a property of all reals.

The incidence of errors per page monotonically increases with page number.

>A "row" is a finite tuple of positive integers, listed from smallest

>to largest.

If you are correct about what is meant, please explain why at page 3 s1.1.1

D and E are not rows. You have as long as you wish, Arturo. :)

>A "range" is a finite tuple of positive integers, listed from smallest

>to largest, with the property that if x,y,z are positive integers,

>x<y<z, and x,z are in the range, then so is y.

Here you may have gone wrong, Arturo. While what he is saying is virtually

meaningless and wholly useless, he is not saying what you are saying he is

saying. You are stating that a *range* must have order and consecutivity.

But that is what make his examples D and E *rows*.

If you are implicitly assuming his examples D and E are "wrong", are you

claiming that all ranges are rows?

As an aside, how do you deal with his reference to "ordered rows"? :)

See to what use he puts them, to find what they might be. Better to ignore

all definitions and examples. :)

>In particular, a range it totally determined by its first and last

>elements; so we can write R(a,b). Alas, the author prefers R(ab), so

>the range from 1 to 100 is R(1100); it is hard to see how we will

>distinguish between the range determined by 1 and 11000, and the range

>determined by 11 and 1000.

This could give the proof some considerable "fluidity". Alas, the maroon

does not avail himself thereof.

>Theorem 1 says that if you take two rows all of whose elements are

>contained in some range, then their first elements differ by a

>constant; it is unclear why "contained in some range" is necessary

>here, but whatever.

. and which constant cannot be zero, you forgot he "proved" - i.e. the

word "distinct" is also to be scattered around the "proof" with liberality.

Simplest is to think of a "row" as being some non-empty subset of a

"range", and both being ordered n-tuples of consecutive naturals,

irrespective of anything to the contrary contained in any of the

"definitions". I know your definition of a row does not require

consecutivity.

>Then he says "We define a function f(x) so that for any x=a, f(a) is a

>prime number". Fair enough; let f(n) be the "n-th prime".

No, no, you are restricting his function enormously. Your one yields

consecutive primes. His one yields primes, period. Your one has the

naturals as an implied domain. His one probably has a much wider domain!

For example, his f(x) might well be defined over (say) R as

f(x) = 2 for all x<3

f(x) = f(2) * f(3) * .. * f(int(x)-1) + 1 otherwise (so f(pi)=3,f(2*e)=43))

:)

>Then he attempts to define "degree of complexity" as "the number of

>primes to be added to each other to yield an even number greater than

>2". Obviously, a good definition would be:

>

>DC(n) = least positive integer m such that n can be expressed as a sum

> of n primes.

No. That way lies lunacy, conceivably even beyond what the wanton wanted. A

good definition would instead be:

DC(n) = least positive integer m such that n can be expressed as a sum

of m primes.

Is it catching? :)

Should I be surprised that you have lept away from page 5, and failed to

comment on his assertion that there are [but] ten subsets (rows) of [range]

R(1100)? Taking, as is clear from context and definition, R(1,100) as what

is being referred to, there are a great deal more than 10 subsets thereof

that qualify as rows (by either his definitionS or your definition of

"row"). :)

>Since every even integer is a multiple of 2, DC(n) exists. However,

>the author asserts that DC(8)=4, since 8 = 2+2+2+2, yet clearly DC(8)

>should be 2, since 8=5+3. Which he notes later. So DC(8) is both 4 and

>2. Very, very bad.

So, what was the significance of "to be added to each other with a + sign

between each" as distinct from "to be added to each other"?

>The proof then proceeds:

>Assume that you have a finite ordered tuple of integers in a given

>range, which contains at least 1 prime, has the same number of even

>and odd positive integers, and has more odd integers than primes.

Consecutivity? Page 7 2) says "the number of even and odd numbers are equal

to each [..] So in any Row r(ij) there exists odd and even numbers

alternatively" - the "so" must imply that the numbers were consecutive in

the first place, else (31,33,37,38) would qualify. And if consecutivity is

a given (as all his examples of what stops things from being rows or ranges

state), all this means is that the row must have even cardinality.

And on page 7 3) "number of odd numbers be more than prime numbers" must

either imply "more than includes equal to" or that he is placing quite a

restriction on row selection.

>At this point the author says (freely translating into math):

Your translation is too free, as is shown below.

>Take an even number A greater than 2. Its degree of complexity is at

>least 2; since A is even greater than 2, the number of even positive

>integers required to express A as a sum is less than the number of

>even number in the given row.

>

>This is false in general. You need two even positive integers to

>express A. The row is assumed to have at least one prime, hence at

>least two odd integers, hence at least two even integers. But that is

>all; for example, the row

>

>(4, 5, 6, 9)

Wearily, I requote: "{49,51,53,55} .. is NOT a Row"

And also his "assumption" (i.e. selection criterion) 2) on page 7

explicitly excludes rows where evens and odds did not alternate.

>satisfies the hypothesis: it has at least one prime (1), has 2 even

>and 2 positive integers, and has 2 odd numbers, more than the number

>of primes you need. Yet 2 is not less than 2.

Irrelevant in the light of my earlier statement. Your row does not qualify.

>Fortunately for the author, he means "less than or equal" when he says

>"less than". Since the row contains at least 2 even numbers, then the

>assertion is true.

>

>Unfortunately, the author asserts that DC(A) equals the least number

>of even positive integers required to express A as a sum (pp. 7, line

>20, equation labeled (4) following the discussion on the paragraph),

>so the author is in fact ASSUMING that DC(A)<=2, which is, of course,

>Goldbach's conjecture.

:)

I could be cruel and point out that the learned author is not saying this

at all. May I?

Read it again.

I cannot tell you exactly what he is saying or implying, but I can tell you

he is not saying or implying what you are saying he is saying or implying.

And you have abandoned your post much too early. Delicious material follows

over pages 8-10, wherein he attempts to extrapolate properties "within" his

specially chosen row(s) to ... something else.

Summary:

He is wrong in ways you (and I) are incapable of comprehending.

So, have I proved 'it'?

>======================================================================

>"It's not denial. I'm just very selective about

> what I accept as reality."

> --- Calvin ("Calvin and Hobbes")

>======================================================================

>

>Arturo Magidin

>mag...@math.berkeley.edu

Sirrah Elnoostu Semaj

Oct 9, 2002, 5:46:18 AM10/9/02

to

Well, by lack of having to do anything better, maybe you want

to have a look at my 2 cents, re Goldbach:

http://home.iae.nl/users/benschop/ng-abstr.htm

Unlike the OP, I do not have the advantage anymore (since dec01) of

posting on arXiv, since my ISP is not an academic one. I did provide

an academic sponsor on their suggestion, which they appear to ignore.

Let this not discourage you of pointing out my first error. Thanks.

BTW: RC is welcome too: his earlier remarks, on the previous MS being

too long (16 pgs), prompted the present reduced version 7oct02 (7 pgs).

Oct 9, 2002, 8:42:39 AM10/9/02

to

On Tue, 08 Oct 2002 23:41:48 +0300, Arturo Magidin wrote:

> A more interesting manuscript with a proposed conditional proof of

> Goldbach can be found in the same arxiv,

>

> http://front.math.ucdavis.edu/math.GM/0209232

>

> where the author states that he will prove that if every odd integer >

> 5 is a sum of three primes, then every even integer greater than 2 is

> a sum of two primes.

> A more interesting manuscript with a proposed conditional proof of

> Goldbach can be found in the same arxiv,

>

> http://front.math.ucdavis.edu/math.GM/0209232

>

> where the author states that he will prove that if every odd integer >

> 5 is a sum of three primes, then every even integer greater than 2 is

> a sum of two primes.

Someone I shared that with reported back:

# "Considering the alternative (i) and comparing (4) with (2) and (3) we obtain

# a contradiction."

#

# (2) and (3) don't rule out (4) for ALL values of a,b,c, only for those such

# that (j+k) is PRIME and q is PRIME.

which I agree with. There were too many things to follow for simple me,

and it appears sleight of hand has been used.

Phil

Oct 9, 2002, 3:08:40 PM10/9/02

to

In article <3DA3FAEA...@chello.nl>,

Nico Benschop <n.ben...@chello.nl> wrote:

Nico Benschop <n.ben...@chello.nl> wrote:

> Well, by lack of having to do anything better, maybe you want

> to have a look at my 2 cents, re Goldbach:

> http://home.iae.nl/users/benschop/ng-abstr.htm

Sorry, Internet Explorer says "Requested server cannot be found". I'll

try again tomorrow.

Oct 10, 2002, 5:22:33 AM10/10/02

to

Nico Benschop <n.ben...@chello.nl> wrote in message news:<3DA3FAEA...@chello.nl>...

> Christian Bau wrote:

> >

> > The usual method is to read the proof carefully and stop at the

> > first error. The mathematics in the paper doesn't look particularly

> > difficult, although the language does, so it is a bit hard to

> > understand what the author means.

> Christian Bau wrote:

> >

> > The usual method is to read the proof carefully and stop at the

> > first error. The mathematics in the paper doesn't look particularly

> > difficult, although the language does, so it is a bit hard to

> > understand what the author means.

Anyone who thinks that this is the "usual method" can never have read

a crank MS. While mathematicians may make errors, cranks don't

make errors, they simply don't make sense. When reading a crank paper

one comes across something which doesn't make sense. With a great deal

of effort one may manage to extract some meaning from it, and when one

has done that, it may even be correct. Then this happens again,

and again, and again, and again, ... each time it's harder and harder

to extract any meaning from it. Eventually one gives up, usually without

identifying anything which the author would acknowledge as a mistake.

>

> Well, by lack of having to do anything better, maybe you want

> to have a look at my 2 cents, re Goldbach:

> http://home.iae.nl/users/benschop/ng-abstr.htm

According to the website, this is "submitted for publication".

Where, Mr Benschop?

I quote

" Theorem 4.1 (Goldbach conjecture 'GC') Each 2n > 4 is the

sum of two odd primes.

Proof. Apply Euclidean prime sieve (5) by induction over k,

with base values k <= 3 (lem3.2), using sums of *initial* prime

units I(k) subset G(k) for k >3. If GC fails for some

2n in I(k) + I(k) subset So(k), then all extensions 2n + cm_k

(c > 0) would be missing as well (lem3.1), not restoring residue

2n mod m_k for any extension pair sum a + b = c > 0, which

contradicts GR(k) [thm2.1]. Hence no 2n is missing from

successive prime sum intervals [2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k),

which overlap by Bertrand's postulate (7), proving Goldbach's conjecture."

This is a fine example of my above thesis. Is there a first mistake?

I contend that it's just nonsense.

Some of Benschop's notation: p_k is the k-th prime, m_k is the

product of the first k primes, G(k) is the set of integers r

with 0 < r < m_k and r coprime to m_k, I(k) is the set of primes

p with P_{k+1}<= p < p_{k+1}^2, So(k) = G(k) + G(k) (the set of

sums of pairs of elements of G(k)).

Earlier results: Lemma 3.1 is the triviality that the least positive

residue of r in G(k+1) modulo m_k lies in G(k),

Lemma 3.2 states that each even integer btween 6 and 90 inclusive

is the sum of two primes between 3 and 47 inclusive,

Theorem 2.1 states that each even integer is congruent modulo m_k

to a sum of two elements of G(k).

References: There is no equation (5), there are (5a), (5b) and (5c)

which are trivial consequences of defintions and (7) is a standard

statement of Bertrand's postulate (proved by Chebyshev).

So with this background, does this "proof" make sense? Has it even

a first mistake?

For a start Benschop uses "induction over k". It's not clear what

the inductive hypothesis is (the first mistake?). But in the last

sentence he blithely asserts that

[2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k) (one presumes that the left

side of this inclusion is the set of even m with

2_p{k+1} <= m <= 4p_{k+1}). Certainly this assertion, with this

interpretation, together with Bertrand's postulate would suffice to

prove Goldbach. I'll assume this is the inductive hypothesis.

Alas, it isn't true. The set I(5) = {13, 17, ...} and so

I(5) + I(5) = {26, 30, ...} does not contain all even numbers

between 26 and 52. Was that the first mistake, or were there earlier ones?

Benschop's second sentence states that "If GC fails for some

2n in I(k) + I(k) subset So(k) ...". What does that mean?

Presumably "GC fails for some 2n" means that 2n is not the sum of

two primes. But Benschop also asserts that 2n in I(k) + I(k);

as I(k) is a set of primes then 2n is a sum of two primes. So is

Benschop saying "If some 2n which is not a sum of 2 primes is

a sum of two primes ..."? Is this his first mistake? If 2n wasn't

a sum of two primes, then 2n is not in I(k) + I(k). Let's

ignore this assertion and pretend that Benschop actually meant to

say that "If GC fails for some 2n, then all extensions 2n + cm_k

(c > 0) would be missing as well (lem3.1)". Well, what does "missing" mean

here? Not a sum of two primes? If so there is a non sequitur here,

that if 2n is not a sume of two primes then neither is 2n + cm_k,

as lemma 3.1 does not refer to sums of two primes. Is a non sequitur

a mistake? A non sequitur is a gap, which may or may not be fillable.

What is certain is that their non sequiturs are rarely acknowledged as

mistakes by cranks. But perhaps "missing" means something else here.

If so is the failure to explain to the reader the meaning of

"missing" a mistake?

In any case, this paragraph is falsely labelled: it is not a proof of

Goldbach's conjecture.

> Unlike the OP, I do not have the advantage anymore (since dec01) of

> posting on arXiv, since my ISP is not an academic one.

It's regrettable that the arXiv now makes this restriction. It is

also regrettable that the arXiv has been abused by various contributors,

including Mr Benschop, who have uploaded worthless trash to the server.

The restriction to academic users, while reducing this problem,

has not eliminated it.

> BTW: RC is welcome too: his earlier remarks, on the previous MS being

> too long (16 pgs), prompted the present reduced version 7oct02 (7 pgs).

Well, the previous version was too long, by 16 pages.

Mr Benschop, this MS does not contain, as claimed,

a proof of Goldbach's conjecture, and the remainder is of no

other interest. Please withdraw this MS from submission, and save

a referee from the wasted effort of reading it.

Robin Chapman

Oct 10, 2002, 6:08:23 AM10/10/02

to

Indeed, my ISP (iae.nl) had trouble yesterday, but it does work now.

-- NB

Oct 10, 2002, 11:46:11 AM10/10/02

to

In article <pan.2002.10.09.15....@yahoo.co.uk>,

Fair enough; I didn't hold much hope given that it was posted to the

GM section rather than the NT section, but you must agree the

manuscript looks much better than the one that started the thread. At

least it read in coherent sentences.

Oct 10, 2002, 5:58:47 PM10/10/02

to

On Thu, 10 Oct 2002 18:46:11 +0300, Arturo Magidin wrote:

> In article <pan.2002.10.09.15....@yahoo.co.uk>,

> Phil Carmody <thefatphi...@yahoo.co.uk> wrote:

>>On Tue, 08 Oct 2002 23:41:48 +0300, Arturo Magidin wrote:

>>> A more interesting manuscript with a proposed conditional proof of

>>> Goldbach can be found in the same arxiv,

>>>

>>> http://front.math.ucdavis.edu/math.GM/0209232

>>>

>>> where the author states that he will prove that if every odd integer >

>>> 5 is a sum of three primes, then every even integer greater than 2 is

>>> a sum of two primes.

>>

>>Someone I shared that with reported back:

>>

>># "Considering the alternative (i) and comparing (4) with (2) and (3) we obtain

>># a contradiction."

>>#

>># (2) and (3) don't rule out (4) for ALL values of a,b,c, only for those such

>># that (j+k) is PRIME and q is PRIME.

>>

>>which I agree with. There were too many things to follow for simple me,

>>and it appears sleight of hand has been used.

>

> Fair enough; I didn't hold much hope given that it was posted to the

> GM section rather than the NT section, but you must agree the

> manuscript looks much better than the one that started the thread. At

> least it read in coherent sentences.

Absolutely, and the background information it starts with is all spot on,

and also more detailed than most summaries I've seen.

Phil

Oct 11, 2002, 6:00:35 AM10/11/02

to

Robin Chapman wrote:

>

> Nico Benschop <n.ben...@chello.nl> wrote :

> > Christian Bau wrote:

> > >

> > > The usual method is to read the proof carefully and stop at

> > > the first error. The mathematics in the paper doesn't look

> > > particularly difficult, although the language does, so it is a

> > > bit hard to understand what the author means.

>

> Anyone who thinks that this is the "usual method" can never have

> read a crank MS. While mathematicians may make errors, cranks don't

> make errors, they simply don't make sense. When reading a crank paper

> one comes across something which doesn't make sense. With a great

> deal of effort one may manage to extract some meaning from it, and

> when one has done that, it may even be correct. Then this happens

> again, and again, and again, and again, ... each time it's harder

> and harder to extract any meaning from it. Eventually one gives up,

> usually without identifying anything which the author would

> acknowledge as a mistake.

>

> >

> > Well, by lack of having to do anything better, maybe you want

> > to have a look at my 2 cents, re Goldbach:

> > http://home.iae.nl/users/benschop/ng-abstr.htm ..[1]

>

> Nico Benschop <n.ben...@chello.nl> wrote :

> > Christian Bau wrote:

> > >

> > > The usual method is to read the proof carefully and stop at

> > > the first error. The mathematics in the paper doesn't look

> > > particularly difficult, although the language does, so it is a

> > > bit hard to understand what the author means.

>

> Anyone who thinks that this is the "usual method" can never have

> read a crank MS. While mathematicians may make errors, cranks don't

> make errors, they simply don't make sense. When reading a crank paper

> one comes across something which doesn't make sense. With a great

> deal of effort one may manage to extract some meaning from it, and

> when one has done that, it may even be correct. Then this happens

> again, and again, and again, and again, ... each time it's harder

> and harder to extract any meaning from it. Eventually one gives up,

> usually without identifying anything which the author would

> acknowledge as a mistake.

>

> >

> > Well, by lack of having to do anything better, maybe you want

> > to have a look at my 2 cents, re Goldbach:

So far so good, apparently not all "just nonsense".

> References: There is no equation (5), there are (5a), (5b) and (5c)

Indeed a major mistake: (5) was meant to include its three sub items.

This, in my opinion, is in the order of a typo, readily corrected.

> which are trivial consequences of defintions and (7) is a

> standard statement of Bertrand's postulate (proved by Chebyshev).

> So with this background, does this "proof" make sense?

> Has it even a first mistake?

>

> For a start Benschop uses "induction over k". It's not clear what

> the inductive hypothesis is (the first mistake?). But in the last

> sentence he blithely asserts that

> [2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k) (one presumes that the

> left side of this inclusion is the set of even m with

> 2_p{k+1} <= m <= 4p_{k+1}).

Indeed, good guess, and as such defined in this MS.

> Certainly this assertion, with this interpretation, together with

> Bertrand's postulate would suffice to prove Goldbach.

> I'll assume this is the inductive hypothesis.

> Alas, it isn't true. The set I(5) = {13, 17, ...} and so

> I(5) + I(5) = {26, 30, ...} does not contain all even numbers

> between 26 and 52.

> [...]

Good point, but again: readily corrected. In fact we're doing

induction on k, and the missing 28 from I(5)+I(5), caused by the

larger than 2 gap 17 - 13 = 4 as 'edge effect', is already in

I(4)+I(4), namely 28 = 11 + 17, where prime 11 not in I(5) with

base_primes 2,3,5,7,11 for G(5).

In fact a smaller example of such edge effect is 16 = 5 + 11, which

is in I(2)+ I(2) with base primes 2,3 - while 16 not in I(3)+I(3)

since 5 is not 'anymore' in I(3) with base primes 2,3,5.

> > Unlike the OP, I do not have the advantage anymore (since dec01)

> > of posting on arXiv, since my ISP is not an academic one.

>

> It's regrettable that the arXiv now makes this restriction.

> It is also regrettable that the arXiv has been abused by

> various contributors, including Mr Benschop, who have uploaded

> worthless trash to the server.

> The restriction to academic users, while reducing this problem,

> has not eliminated it.

>

> > BTW: RC is welcome too: his earlier remarks, on the previous MS

> > being too long (16 pgs), prompted the present reduced version

> > 7oct02 (7 pgs).

>

> Well, the previous version was too long, by 16 pages. Mr Benschop,

> this MS does not contain, as claimed, a proof of Goldbach's

> conjecture, and the remainder is of no other interest.

The rather large number of visits yesterday to the given ref ..[1]

tells me otherwise. But then, your opinion 'worthless trash' is only

_your_ opinion, slightly biased I presume by your inability to

adjust it or any other opinion previously put forward too strongly

(... keep barking, to His Master's Voice:

http://home.iae.nl/users/benschop/nr-th.htm )

Your increasingly abusive style reminds me of John Cleese in

"Faulty Towers": once taking a wrong position/assumption he

escalates in ridiculous extreme behaviour (though he's funny,

you're not) - until he almost literally goes over the edge.

NB: Just be sure, dear Robin, you don't act this out in public.

I'd advise you to 'cover your retreat', just in case you may need to.

Oct 11, 2002, 12:24:38 PM10/11/02

to

Nico Benschop wrote:

> Robin Chapman wrote:

>>

>>

>> Proof. Apply Euclidean prime sieve (5) by induction over k,

>> with base values k <= 3 (lem3.2), using sums of *initial* prime

>> units I(k) subset G(k) for k >3. If GC fails for some

>> 2n in I(k) + I(k) subset So(k), then all extensions 2n + cm_k

>> (c > 0) would be missing as well (lem3.1), not restoring residue

>> 2n mod m_k for any extension pair sum a + b = c > 0, which

>> contradicts GR(k) [thm2.1]. Hence no 2n is missing from

>> successive prime sum intervals [2p_{k+1} ... 4p_{k+1}] subset

>> I(k) + I(k), which overlap by Bertrand's postulate (7), proving

>> Goldbach's conjecture."

>>

>> This is a fine example of my above thesis. Is there a first mistake?

>> I contend that it's just nonsense.

>>

>

> So far so good, apparently not all "just nonsense".

Up to the penultimate paragraph of page 5 most of the paper is fairly

sound. It's a catalogue of eccentric notation and trivial results, and

there are some errors (in the abstract there's a m^k which should be m_k,

and on page 5 is the false assertion that p_{i+j} > p_i + p_j),

mind you.

>>

>> For a start Benschop uses "induction over k". It's not clear what

>> the inductive hypothesis is (the first mistake?). But in the last

>> sentence he blithely asserts that

>> [2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k) (one presumes that the

>> left side of this inclusion is the set of even m with

>> 2_p{k+1} <= m <= 4p_{k+1}).

>

> Indeed, good guess, and as such defined in this MS.

I cannot see such a definition, but as I pointed out the statement

"[2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k)" is false. In any case

the main difficulty I identified here remains: what is the inductive

hypothesis?

>> Certainly this assertion, with this interpretation, together with

>> Bertrand's postulate would suffice to prove Goldbach.

>> I'll assume this is the inductive hypothesis.

>> Alas, it isn't true. The set I(5) = {13, 17, ...} and so

>> I(5) + I(5) = {26, 30, ...} does not contain all even numbers

>> between 26 and 52.

>> [...]

>

> Good point, but again: readily corrected. In fact we're doing

> induction on k, and the missing 28 from I(5)+I(5), caused by the

> larger than 2 gap 17 - 13 = 4 as 'edge effect', is already in

> I(4)+I(4), namely 28 = 11 + 17, where prime 11 not in I(5) with

> base_primes 2,3,5,7,11 for G(5).

> In fact a smaller example of such edge effect is 16 = 5 + 11, which

> is in I(2)+ I(2) with base primes 2,3 - while 16 not in I(3)+I(3)

> since 5 is not 'anymore' in I(3) with base primes 2,3,5.

But {26, 28, ..., 52} is not a subset of I(5) + I(5). Exactly

what do you wish to assert here, Mr Benschop?

Now, Mr Benschop, you did not address the most important point

(why am I not surprised?); let me repeat a section of my post that

you snipped, Mr Benschop:

Benschop's second sentence states that "If GC fails for some

2n in I(k) + I(k) subset So(k) ...". What does that mean?

Presumably "GC fails for some 2n" means that 2n is not the sum of

two primes. But Benschop also asserts that 2n in I(k) + I(k);

as I(k) is a set of primes then 2n is a sum of two primes. So is

Benschop saying "If some 2n which is not a sum of 2 primes is

a sum of two primes ..."? Is this his first mistake? If 2n wasn't

a sum of two primes, then 2n is not in I(k) + I(k). Let's

ignore this assertion and pretend that Benschop actually meant to

say that "If GC fails for some 2n, then all extensions 2n + cm_k

(c > 0) would be missing as well (lem3.1)". Well, what does "missing" mean

here? Not a sum of two primes? If so there is a non sequitur here,

that if 2n is not a sume of two primes then neither is 2n + cm_k,

as lemma 3.1 does not refer to sums of two primes. Is a non sequitur

a mistake? A non sequitur is a gap, which may or may not be fillable.

What is certain is that their non sequiturs are rarely acknowledged as

mistakes by cranks. But perhaps "missing" means something else here.

If so is the failure to explain to the reader the meaning of

"missing" a mistake?

What about this non-sequitur Mr Benschop? Are you ready to explain

why 2n being "missing" implies that 2n + c m_k must be "missing" too

(and perhaps you might start by explaining your usage of the word

"missing")?

>> Well, the previous version was too long, by 16 pages. Mr Benschop,

>> this MS does not contain, as claimed, a proof of Goldbach's

>> conjecture, and the remainder is of no other interest.

>

> The rather large number of visits yesterday to the given ref ..[1]

It's really a bit sad to see someone count the number of vists to their

site. In case you are interested, I confess several were from me.

These visits are no evidence that you have produced anything of value.

> tells me otherwise. But then, your opinion 'worthless trash' is only

> _your_ opinion, slightly biased I presume by your inability to

> adjust it or any other opinion previously put forward too strongly

Biased? No, "worthless trash" is a simple description of some of

the crank rubbish uploaded to the arXiv.

> (... keep barking, to His Master's Voice:

> http://home.iae.nl/users/benschop/nr-th.htm )

Oh dear. Still whingeing about your contribution being rejected by

the NMBRTHY list. It's a moderated (some would say censored) list

you know; that means the moderator decided what to accept. Don't you

realize this? The only surprise is that the moderator accepted your first

posting.

> Your increasingly abusive style reminds me of John Cleese in

> "Faulty Towers": once taking a wrong position/assumption he

"Fawlty"

> escalates in ridiculous extreme behaviour (though he's funny,

> you're not) - until he almost literally goes over the edge.

The only extreme behaviour evidenced here is your continued

promotion of your nonsensical theses. My pointing out their errors

is really moderate, not extreme.

Incidentally, you failed to respond to my first question.

Where exactly did you submit this paper?

--

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

"His mind has been corrupted by colours, sounds and shapes."

The League of Gentlemen

Oct 12, 2002, 9:05:28 AM10/12/02

to

Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message news:<ao6ttm$h2n$1...@news5.svr.pol.co.uk>...

> Nico Benschop wrote:

>

> > Robin Chapman wrote:

> >> [...]

>

> Up to the penultimate paragraph of page 5 most of the paper is fairly

> sound. It's a catalogue of eccentric notation and trivial results, and

> there are some errors (in the abstract there's a m^k which should be m_k,

> and on page 5 is the false assertion that p_{i+j} > p_i + p_j),

> mind you.

> Nico Benschop wrote:

>

> > Robin Chapman wrote:

>

> Up to the penultimate paragraph of page 5 most of the paper is fairly

> sound. It's a catalogue of eccentric notation and trivial results, and

> there are some errors (in the abstract there's a m^k which should be m_k,

> and on page 5 is the false assertion that p_{i+j} > p_i + p_j),

> mind you.

Both corrected, see V-12oct02.

> >> For a start Benschop uses "induction over k". It's not clear what

> >> the inductive hypothesis is (the first mistake?). But in the last

> >> sentence he blithely asserts that

> >> [2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k) (one presumes that the

> >> left side of this inclusion is the set of even m with

> >> 2_p{k+1} <= m <= 4p_{k+1}).

> >

> > Indeed, good guess, and as such defined in this MS.

>

> I cannot see such a definition, but as I pointed out the statement

> "[2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k)" is false. In any case

> the main difficulty I identified here remains: what is the inductive

> hypothesis?

>

> >> Certainly this assertion, with this interpretation, together with

> >> Bertrand's postulate would suffice to prove Goldbach.

> >> I'll assume this is the inductive hypothesis.

> >> Alas, it isn't true. The set I(5) = {13, 17, ...} and so

> >> I(5) + I(5) = {26, 30, ...} does not contain all even numbers

> >> between 26 and 52.

> >> [...]

> >

> > Good point, but again: readily corrected:

... if I'm not mistaken (in thm4.1), by quantifiers

"for some 2n" and "for all k>3". The term "missing 2n"

and "if GC fails" is now (V 12oct02) described directly by

a property, not really requiring the concept of induction.

> [...]

> It's really a bit sad to see someone count the number of vists to their

> site. In case you are interested, I confess several were from me.

> These visits are no evidence that you have produced anything of value.

True, but is does indicate some interest from sci.math readers,

contrary to your opinion.

> > tells me otherwise. But then, your opinion 'worthless trash' is only

> > _your_ opinion, slightly biased I presume by your inability to

> > adjust it or any other opinion previously put forward too strongly

>

> Biased? No, "worthless trash" is a simple description of some of

> the crank rubbish uploaded to the arXiv.

>

> > (... keep barking, to His Master's Voice:

> > http://home.iae.nl/users/benschop/nr-th.htm )

>

> Oh dear. Still whingeing about your contribution being rejected by

> the NMBRTHY list. It's a moderated (some would say censored) list

> you know; that means the moderator decided what to accept. Don't you

> realize this? The only surprise is that the moderator accepted your

> first posting.

Reading the process described in */nr-th.htm , especially the

responses of Alf v.d. Poorten and Victor Miller (the moderator), they

obviously (& on purpose?-) refrain from any answer to the content of

my suggestion re FLT case_1 : for *any* residue solution mod p^2,

the exponent p distributes over a sum, with as special case the cubic

roots of 1 mod p^2 (for any prime p with 3 dividing p-1, and given

example for p=7) NOT extendable to integer p-th powers -> breaking

the Hensel lift, contrary to common opinion of (re quoted Bob S.):

that one cannot use a local [residue] property to derive a global

[integer] conclusion. This holds *in general* all right, but NOT

in the *special* FLT case_1 with the syntax property 'EDS'

(Exponent p Distributes over a Sum).

This point is, by both gentlemen, ignored - a despicable state of

affairs for a NBMRTH list - thus in fact blocking any discussion.

> [...]

> Incidentally, you failed to respond to my first question.

> Where exactly did you submit this paper?

None of your business.

Oct 12, 2002, 1:36:05 PM10/12/02

to

Nico Benschop wrote:

> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> wrote in message

> news:<ao6ttm$h2n$1...@news5.svr.pol.co.uk>...

>> Nico Benschop wrote:

>>

>> > Robin Chapman wrote:

>> >> [...]

>>

>

>> >> For a start Benschop uses "induction over k". It's not clear what

>> >> the inductive hypothesis is (the first mistake?). But in the last

>> >> sentence he blithely asserts that

>> >> [2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k) (one presumes that the

>> >> left side of this inclusion is the set of even m with

>> >> 2_p{k+1} <= m <= 4p_{k+1}).

>> >

>> > Indeed, good guess, and as such defined in this MS.

>>

>> I cannot see such a definition, but as I pointed out the statement

>> "[2p_{k+1} ... 4p_{k+1}] subset I(k) + I(k)" is false. In any case

>> the main difficulty I identified here remains: what is the inductive

>> hypothesis?

So, what IS the inductive hypothesis?

>> >> Certainly this assertion, with this interpretation, together with

>> >> Bertrand's postulate would suffice to prove Goldbach.

>> >> I'll assume this is the inductive hypothesis.

>> >> Alas, it isn't true. The set I(5) = {13, 17, ...} and so

>> >> I(5) + I(5) = {26, 30, ...} does not contain all even numbers

>> >> between 26 and 52.

>> >> [...]

>> >

>> > Good point, but again: readily corrected:

>

> ... if I'm not mistaken (in thm4.1), by quantifiers

> "for some 2n" and "for all k>3". The term "missing 2n"

> and "if GC fails" is now (V 12oct02) described directly by

> a property, not really requiring the concept of induction.

The alleged "proof" now reads

"Proof. Apply Euclidean prime sieve (5a-c) by induction over k,

with base values k <= 3 (lem3.2), using sums of *initial* prime

units I(k) subset G(k) for k >3. If GC fails, viz. for some 2n and

all k >1: 2n notin I(k) + I(k) subset So(k), then all extensions

2n + cm_k

(c > 0) would be missing as well (lem3.1), not restoring residue

2n mod m_k (k >1) for any extension pair sum a + b = c > 0 (5b), which

contradicts GR(k) [thm2.1]. Hence each 2n > 4 is, for some k, in a

prime sum interval of even naturals

[2p_{k+1} . . 4p_{k+1}] subset I(k) + I(k),

which overlap by Bertrand's postulate (7), proving Goldbach's

Conjecture."

It still starts with a mention of induction, so I ask again:

what is the inductive hypothesis? Now the non-sequitur is even more

blatant. We are given a 2n held to falsify Goldbach, and it is stated

of this that "all extensions 2n + cm_k would be missing as well (lem3.1)".

The term "missing" is not defined, but one presumes that it means

that 2n + c m_k is not a sum of two primes. If so then lemma 3.1

(which doe not refer to primes at all) does not account for why this

might be true. The "proof" collapses at this point.

>> [...]

>> It's really a bit sad to see someone count the number of vists to their

>> site. In case you are interested, I confess several were from me.

>> These visits are no evidence that you have produced anything of value.

>

> True, but is does indicate some interest from sci.math readers,

> contrary to your opinion.

My opinion? I deal here with matters of fact, not opinion. Visits

to your site are no evidence of any value contained therein. Also

you are committing your favourite sin of non-sequiturization again.

Visits to your site are no evidence of interest from sci.math readers:

they may no even be from sci.math readers. I suggest a different index

of the interest in you on this group: the number of readers who reply

to your postings. According to this, the interest in you is

a tiny fraction of that afforded to Mr James Harris.

>>

>> > (... keep barking, to His Master's Voice:

>> > http://home.iae.nl/users/benschop/nr-th.htm )

>>

>> Oh dear. Still whingeing about your contribution being rejected by

>> the NMBRTHY list. It's a moderated (some would say censored) list

>> you know; that means the moderator decided what to accept. Don't you

>> realize this? The only surprise is that the moderator accepted your

>> first posting.

>

> Reading the process described in */nr-th.htm , especially the

> responses of Alf v.d. Poorten and Victor Miller (the moderator), they

> obviously (& on purpose?-) refrain from any answer to the content of

> my suggestion re FLT case_1 : for *any* residue solution mod p^2,

> the exponent p distributes over a sum, with as special case the cubic

> roots of 1 mod p^2 (for any prime p with 3 dividing p-1, and given

> example for p=7) NOT extendable to integer p-th powers -> breaking

> the Hensel lift, contrary to common opinion of (re quoted Bob S.):

Aha! The Benschop mantra "Breaking The Hensel Lift". Never

explained what that means, have you, Mr Benschop?

> that one cannot use a local [residue] property to derive a global

> [integer] conclusion. This holds *in general* all right, but NOT

> in the *special* FLT case_1 with the syntax property 'EDS'

> (Exponent p Distributes over a Sum).

I don't recall you ever presenting any evidence for this;

come to think of it, I don't ever recall your presenting a coherent

statement of what you are claiming to have proved here.

> This point is, by both gentlemen, ignored - a despicable state of

> affairs for a NBMRTH list - thus in fact blocking any discussion.

Oh dear, how sad, never mind.

Mr Benschop, if you post to a moderated group or list, you must accept

the fact that your article may be rejected. Live with it. Incidentally

yours was not the only article in that thread not to be accepted.

>> [...]

>> Incidentally, you failed to respond to my first question.

>> Where exactly did you submit this paper?

>

> None of your business.

Touchy!

I'll ask again: where have you submitted this paper, Mr Benschop?

Oct 13, 2002, 6:09:28 PM10/13/02

to

Robin Chapman wrote:

>

> Nico Benschop wrote:

>

> > [...]

> > Reading the process described in */nr-th.htm , especially the

> > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > they obviously (& on purpose?-) refrain from any answer to the

> > content of my suggestion re FLT case_1 : for *any* residue solution

> > mod p^2, the exponent p distributes over a sum, with as special

> > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > p-1, and given example for p=7) NOT extendable to integer p-th

> > powers -> breaking the Hensel lift, contrary to common opinion of

> > (re quoted Bob S.):

>

> Aha! The Benschop mantra "Breaking The Hensel Lift".

> Never explained what that means, have you, Mr Benschop?

^^^^^^^^^^^^^^^^??

>

> Nico Benschop wrote:

>

> > [...]

> > Reading the process described in */nr-th.htm , especially the

> > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > they obviously (& on purpose?-) refrain from any answer to the

> > content of my suggestion re FLT case_1 : for *any* residue solution

> > mod p^2, the exponent p distributes over a sum, with as special

> > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > p-1, and given example for p=7) NOT extendable to integer p-th

> > powers -> breaking the Hensel lift, contrary to common opinion of

> > (re quoted Bob S.):

>

> Aha! The Benschop mantra "Breaking The Hensel Lift".

> Never explained what that means, have you, Mr Benschop?

Short memory, mr. Chapman? ... (or just deaf, see further...)

See http://mathforum.org/discuss/sci.math/a/m/363273/363276

(to which you replied, so don't say you didn't see it).

And that is only one of the many times I explained the term.

Moreover, that term is not my invetion, but I heard it from

prof. A.Cohen @ TU-Eindhoven. I find it a very appropriate

charcterization of the misconception involved, namely that :

local (residues) --> global (integers) cannot work "in general"

implies the same "in a special case" (such as case_1 FLT mod p^2).

It is like saying: by Galois theory the solution in radicals

is not possible for "the general" polynomial of degree > 4,

and then to claim this is not possible either for special cases

(such as a degree 5 polynomial that is a product of lower degree

polynomials).

> > that one cannot use a local [residue] property to derive a global

> > [integer] conclusion. This holds *in general* all right, but NOT

> > in the *special* FLT case_1 with the syntax property 'EDS'

> > (Exponent p Distributes over a Sum).

>

> I don't recall you ever presenting any evidence for this;

> come to think of it, I don't ever recall your presenting a

> coherent statement of what you are claiming to have proved here.

Your memory, as shown above, is too short, unfortunately (for you).

> > This point is, by both gentlemen, ignored - a despicable state of

> > affairs for a NBMRTH list - thus in fact blocking any discussion.

>

> Oh dear, how sad, never mind. Mr Benschop, if you post to a moderated

> group or list, you must accept the fact that your article may be

> rejected. Live with it. Incidentally yours was not the only article

> in that thread not to be accepted.

No problem, really, I gave up trying to get any discussion there,

of course. The 'conceit' level is too high for me, see (again):

http://home.iae.nl/users/benschop/nr-th.htm

> >> [...]

> >> Incidentally, you failed to respond to my first question.

> >> Where exactly did you submit this paper?

> >

> > None of your business.

>

> Touch! I'll ask again: where have you submitted this paper,

> Mr Benschop? -- Robin Chapman

Not only is your memory bad, but you're deaf as well.

Let me spell it out to you: the NOYB journal.

Oct 14, 2002, 2:15:17 AM10/14/02

to

Robin Chapman wrote:

>

> Nico Benschop wrote:

>

> > [...]

> > Reading the process described in */nr-th.htm , especially the

> > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > they obviously (& on purpose?-) refrain from any answer to the

> > content of my suggestion re FLT case_1 : for *any* residue solution

> > mod p^2, the exponent p distributes over a sum, with as special

> > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > p-1, and given example for p=7) NOT extendable to integer p-th

> > powers -> breaking the Hensel lift,

> > contrary to common opinion, re quoted Bob S. in:

http://home.iae.nl/users/benschop/selmer.htm

PS:

FLT case_1: x^p + y^p = z^p [1] (odd prime p, and x,y,z coprime to p)

has no integer solution, is equivalent to:

no solution of [1] mod p^2 [ where exponent p distributes over a sum,

since each unit n^p mod p^2 is in an order p-1 cyclic subgroup of

the units group G(2) of order (p-1)p : each solution of [1] mod p^2

is in 'core' A(2) of order p-1 where n^p == n mod p^2, so:

(x+y)^p == x+y == x^p + y^p mod p^2 ]

can be extended to integers X, Y, Z with X==x, Y==y, Z==z mod p^2

such that X^p + Y^p = Z^p (sufficient is: for residues mod p^{2p} )

>

> Nico Benschop wrote:

>

> > [...]

> > Reading the process described in */nr-th.htm , especially the

> > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > they obviously (& on purpose?-) refrain from any answer to the

> > content of my suggestion re FLT case_1 : for *any* residue solution

> > mod p^2, the exponent p distributes over a sum, with as special

> > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > p-1, and given example for p=7) NOT extendable to integer p-th

> > powers -> breaking the Hensel lift,

http://home.iae.nl/users/benschop/selmer.htm

PS:

FLT case_1: x^p + y^p = z^p [1] (odd prime p, and x,y,z coprime to p)

has no integer solution, is equivalent to:

no solution of [1] mod p^2 [ where exponent p distributes over a sum,

since each unit n^p mod p^2 is in an order p-1 cyclic subgroup of

the units group G(2) of order (p-1)p : each solution of [1] mod p^2

is in 'core' A(2) of order p-1 where n^p == n mod p^2, so:

(x+y)^p == x+y == x^p + y^p mod p^2 ]

can be extended to integers X, Y, Z with X==x, Y==y, Z==z mod p^2

such that X^p + Y^p = Z^p (sufficient is: for residues mod p^{2p} )

>

> Aha! The Benschop mantra "Breaking The Hensel Lift".

> Never explained what that means, have you, Mr Benschop?

^^^^^^^^^^^^^^^^??

Short memory, mr. Chapman? ... (or just deaf, see further)

See http://mathforum.org/discuss/sci.math/a/m/363273/363276

(to which you replied, so don't say you didn't see it).

And that is only one of the many times I explained the term.

Moreover, that term is not my invetion, but I heard it from

prof. A.Cohen @ TU-Eindhoven. I find it a very appropriate

charcterization of the misconception involved, namely that :

local (residues) --> global (integers) cannot work "in general"

implies the same "in a special case" (such as case_1 FLT mod p^2).

It is like saying: by Galois theory the solution in radicals

is not possible for "the general" polynomial of degree > 4,

and then to claim this is not possible either for special cases

(such as a degree 5 polynomial that is a product of lower degree

polynomials).

> > that one cannot use a local [residue] property to derive a global

> > [integer] conclusion. This holds *in general* all right, but NOT

> > in the *special* FLT case_1 with the syntax property 'EDS'

> > (Exponent p Distributes over a Sum).

>

> I don't recall you ever presenting any evidence for this;

> come to think of it, I don't ever recall your presenting a

> coherent statement of what you are claiming to have proved here.

As shown above, your memory is too short, unfortunately (for you).

> > This point is, by both gentlemen, ignored - a despicable state of

> > affairs for a NBMRTH list - thus in fact blocking any discussion.

>

> Oh dear, how sad, never mind. Mr Benschop, if you post to a moderated

> group or list, you must accept the fact that your article may be

> rejected. Live with it. Incidentally yours was not the only article

> in that thread not to be accepted.

No problem, really, I gave up trying to get any discussion there,

of course. The 'conceit' level is too high for me, see (again):

http://home.iae.nl/users/benschop/nr-th.htm

> >> [...]

> >> Incidentally, you failed to respond to my first question.

> >> Where exactly did you submit this paper?

> >

> > None of your business.

>

> Touchy!

> I'll ask again: where have you submitted this paper, Mr Benschop?

> --

> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

> "His mind has been corrupted by colours, sounds and shapes."

> The League of Gentlemen ^^^^^^^^^^^^^^^^^^^^^^^^^^

Not only is your memory bad, but you're deaf (and/or blind) as well.

Let me spell it out for you: the NOYB journal.

Oct 15, 2002, 3:24:07 AM10/15/02

to

Nico Benschop <n.ben...@chello.nl> wrote in message news:<3DAA60A3...@chello.nl>...

> Robin Chapman wrote:

> >

> > Nico Benschop wrote:

> >

> > > [...]

> > > Reading the process described in */nr-th.htm , especially the

> > > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > > they obviously (& on purpose?-) refrain from any answer to the

> > > content of my suggestion re FLT case_1 : for *any* residue solution

> > > mod p^2, the exponent p distributes over a sum, with as special

> > > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > > p-1, and given example for p=7) NOT extendable to integer p-th

> > > powers -> breaking the Hensel lift,

> > > contrary to common opinion, re quoted Bob S. in:

> http://home.iae.nl/users/benschop/selmer.htm

> PS:

> FLT case_1: x^p + y^p = z^p [1] (odd prime p, and x,y,z coprime to p)

> has no integer solution, is equivalent to:

> no solution of [1] mod p^2 [ where exponent p distributes over a sum,

> since each unit n^p mod p^2 is in an order p-1 cyclic subgroup of

> the units group G(2) of order (p-1)p : each solution of [1] mod p^2

> is in 'core' A(2) of order p-1 where n^p == n mod p^2, so:

> (x+y)^p == x+y == x^p + y^p mod p^2 ]

> can be extended to integers X, Y, Z with X==x, Y==y, Z==z mod p^2

> such that X^p + Y^p = Z^p (sufficient is: for residues mod p^{2p}

Corr: in case of 0-extension, viz. X,Y < p^2 )

> Robin Chapman wrote:

> >

> > Nico Benschop wrote:

> >

> > > [...]

> > > Reading the process described in */nr-th.htm , especially the

> > > responses of Alf v.d. Poorten and Victor Miller (the moderator),

> > > they obviously (& on purpose?-) refrain from any answer to the

> > > content of my suggestion re FLT case_1 : for *any* residue solution

> > > mod p^2, the exponent p distributes over a sum, with as special

> > > case the cubic roots of 1 mod p^2 (for any prime p with 3 dividing

> > > p-1, and given example for p=7) NOT extendable to integer p-th

> > > powers -> breaking the Hensel lift,

> > > contrary to common opinion, re quoted Bob S. in:

> http://home.iae.nl/users/benschop/selmer.htm

> PS:

> FLT case_1: x^p + y^p = z^p [1] (odd prime p, and x,y,z coprime to p)

> has no integer solution, is equivalent to:

> no solution of [1] mod p^2 [ where exponent p distributes over a sum,

> since each unit n^p mod p^2 is in an order p-1 cyclic subgroup of

> the units group G(2) of order (p-1)p : each solution of [1] mod p^2

> is in 'core' A(2) of order p-1 where n^p == n mod p^2, so:

> (x+y)^p == x+y == x^p + y^p mod p^2 ]

> can be extended to integers X, Y, Z with X==x, Y==y, Z==z mod p^2

> such that X^p + Y^p = Z^p (sufficient is: for residues mod p^{2p}

More generally ('bootstrap' effect): for a solution of [1] mod p^k,

---> the 0-extended X,Y,Z < p^k (residues viewed as naturals)

---> show inequivalence X^p + Y^p \nequiv z^p mod p^{3k+1}.

Thus: 'quadratic' analysis suffices to show inequiv.

(you may draw your own conclusion, re FLT case_1 for integers;-)

See http://home.iae.nl/users/benschop/nf-abstr.htm

Oct 15, 2002, 2:10:24 PM10/15/02

to

n.ben...@chello.nl (Nico Benschop) wrote in message news:<caeab7cb.0210...@posting.google.com>...

> Corr: in case of 0-extension, viz. X,Y < p^2 )

>

> More generally ('bootstrap' effect): for a solution of [1] mod p^k,

> ---> the 0-extended X,Y,Z < p^k (residues viewed as naturals)

> ---> show inequivalence X^p + Y^p \nequiv z^p mod p^{3k+1}.

Proof?

> Thus: 'quadratic' analysis suffices to show inequiv.

> (you may draw your own conclusion, re FLT case_1 for integers;-)

I recall reading a MS of Benschop in which he claimed to prove

something of this nature. However his alleged "proof" was bogus.

But anyway, he has not answered my earlier questions.

Mr Benschop, are you going to withdraw your MS

"On Z mod (product of the first k primes), a Boolean lattice

of 2^k groups and Goldbach's conjecture"

now that you know that the main "proof" is unsound?

Also which journal did you submit it to?

Robin Chapman

Oct 15, 2002, 9:31:02 PM10/15/02

to

r...@maths.ex.ac.uk (Robin Chapman) wrote in message news:<a31ba6bf.02101...@posting.google.com>...

> While mathematicians may make errors, cranks don't make errors, they simply don't make sense. When reading a crank paper one comes across something which doesn't make sense. With a great deal of effort one may manage to extract some meaning from it, and when one has done that, it may even be correct.

Haven't you just contradicted yourself? First you say that it doesn't

make sense, then you admit that with effort you realize that it does.

So why not erase the "doesn't make sense" comments? And while you're

at it, wouldn't it be more helpful ("make sense") to propose an

alternate, clearer wording, once you understand what it is saying?

> Eventually one gives up, usually without identifying anything which the author would acknowledge as a mistake.

If you don't consider a temporary failure to communicate to be a

"mistake", then no, there is no "mistake".

> " Theorem 4.1 (Goldbach conjecture 'GC') Each 2n > 4 is the sum of

two odd primes. Proof. Apply Euclidean prime sieve (5) by induction

over k, with base values k <= 3 (lem3.2), using sums of *initial*

prime units I(k) subset G(k) for k >3. If GC fails for some 2n in I(k)

+ I(k) subset So(k), then all extensions 2n + cm_k (c > 0) would be

missing as well (lem3.1), not restoring residue 2n mod m_k for any

extension pair sum a + b = c > 0, which contradicts GR(k) [thm2.1].

Hence no 2n is missing from successive prime sum intervals [2p_{k+1}

... 4p_{k+1}] subset I(k) + I(k), which overlap by Bertrand's

postulate (7), proving Goldbach's conjecture."

> This is a fine example of my above thesis. I contend that it's just nonsense.

Have you taken the 2nd step of attempting to understand it, yet?

> What is certain is that their non sequiturs are rarely acknowledged as mistakes by cranks.

How have you ascertained that to be so? (I assume that, therefore,

you must have a formal definition of "crank" and have performed some

sort of statistical analysis of cranks, non sequiturs and the

acknowledgement thereof.)

> It is regrettable that the arXiv has been abused by various contributors, including Mr Benschop, who have uploaded worthless trash to the server. Mr Benschop, this MS does not contain, as claimed, a proof of Goldbach's conjecture, and the remainder is of no other interest. Please withdraw this MS from submission, and save a referee from the wasted effort of reading it.

In a professional environment, when one makes a proposal, anyone who

believes that they see flaws in that proposal gives an explanation as

to why they feel that way. The author then has the opportunity to

correct or refute the claim that these flaws exist. In no case is the

author subject to ridicule nor is the submission considered a "wasted

effort".

If a flaw is discovered in Martin Dunwoody's proof of the Poincare

Conjecture, should he be ridiculed and called a "crank"?

Webster defines a "crank" as an eccentric or bad-tempered person. I

don't see anything eccentric about working on unsolved problems of

mathematics, nor any indication of a temper in the above effort at a

proof. However, I cannot say the same about the response to that

effort.

> Robin Chapman

Charlie Volkstorf

Cambridge, MA

Oct 16, 2002, 4:45:26 AM10/16/02

to

Robin Chapman wrote:

>

> [...]

>

> [...]

> But anyway, he has not answered my earlier questions.

Your point on the precise form of the induction hypothesis

I did acknowledge as being useful, and I'm working on it.

The whole concept of "induction over k" - as suggested by the

described Euclidean prime sieve (= multi prime number code,

over m_k : the product of the first k primes) - may not be

necessary at all: just the range of k values relevant to

any given 2n may suffice.

<